Alternative methods for solving a system of one linear one non linear simultaneous equationsMethods for solving linear systemsMethods of characteristic for system of first order linear hyperbolic partial differential equations: reference and examplesSolving a specific system of n non linear equationsCRT - non-linear system of equationsMethods for solving a $4$ system of equationSolving a system of 4 non-linear equationsSolving systems of linear equations containing integersSolving linear equations involving many variablesPossibility of solving a certain 4D system with 4 equations with certain methodsSolve a system of non-linear equations
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Alternative methods for solving a system of one linear one non linear simultaneous equations
Methods for solving linear systemsMethods of characteristic for system of first order linear hyperbolic partial differential equations: reference and examplesSolving a specific system of n non linear equationsCRT - non-linear system of equationsMethods for solving a $4$ system of equationSolving a system of 4 non-linear equationsSolving systems of linear equations containing integersSolving linear equations involving many variablesPossibility of solving a certain 4D system with 4 equations with certain methodsSolve a system of non-linear equations
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Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
algebra-precalculus systems-of-equations roots quadratics
edited Jul 24 at 0:23
Michael Rozenberg
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asked Jul 24 at 0:09
KanturaKantura
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add a comment |
$begingroup$
Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
algebra-precalculus systems-of-equations roots quadratics
edited Jul 24 at 0:23
Michael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
asked Jul 24 at 0:09
KanturaKantura
1,1807 silver badges20 bronze badges
$endgroup$
add a comment |
$begingroup$
Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
algebra-precalculus systems-of-equations roots quadratics
edited Jul 24 at 0:23
Michael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
asked Jul 24 at 0:09
KanturaKantura
1,1807 silver badges20 bronze badges
$endgroup$
Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
algebra-precalculus systems-of-equations roots quadratics
algebra-precalculus systems-of-equations roots quadratics
edited Jul 24 at 0:23
Michael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
asked Jul 24 at 0:09
KanturaKantura
1,1807 silver badges20 bronze badges
edited Jul 24 at 0:23
Michael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
asked Jul 24 at 0:09
KanturaKantura
1,1807 silver badges20 bronze badges
edited Jul 24 at 0:23
Michael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
edited Jul 24 at 0:23
Michael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
edited Jul 24 at 0:23
Michael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
122k20 gold badges105 silver badges210 bronze badges
asked Jul 24 at 0:09
KanturaKantura
1,1807 silver badges20 bronze badges
asked Jul 24 at 0:09
KanturaKantura
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asked Jul 24 at 0:09
KanturaKantura
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1,1807 silver badges20 bronze badges
add a comment |
add a comment |
10 Answers
10
active
oldest
votes
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In general, the set of equations:
$$sum_k=1^Nx_k^p = S_p$$
for $1leq pleq N$, can be solved by considering the function:
$$f(x) = -sum_p=1^Nlogleft(1-fracx_pxright) tag1$$
The expansion of $f(x)$ around infinity is given by:
$$f(x) = sum_r=1^inftyfracS_rr x^r$$
We can thus write down $f(x)$ to order $x^-2$ as:
$$f(x) = frac5x + frac132 x^2 + mathcalOleft(x^-3right)tag2$$
From (1) it follows that $x^2 expleft[-f(x)right]$ is a second degree polynomial that has the solutions as its roots. Using (2) it follows that:
$$expleft[-f(x)right] = 1 - frac5x + frac6x^2 + mathcalOleft(x^-3right)$$
It thus follows that:
$$(x-x_1)(x-x_2) = x^2 - 5 x + 6$$
So, the solutions are $x_1=2$ and $x_2 = 3$ and vice versa.
answered Jul 24 at 0:38
Count IblisCount Iblis
8,9602 gold badges16 silver badges37 bronze badges
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5
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Is there any reference to this method, I need to understand it further if possible. Thank you.
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– NoChance
Jul 24 at 11:08
2
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That is a sweet trick. But unfortunately, it only covers the special case where all variables are treated symmetrically.
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– Paul Sinclair
Jul 24 at 16:26
add a comment |
$begingroup$
We have $$(x+y)^2=13+2xy,$$ which gives
$$xy=6$$ and by the Viete's theorem $x$ and $y$ are roots of the equation:
$$t^2-5t+6=0$$ or
$$(t-2)(t-3)=0,$$ which gives the answer:
$$(2,3),(3,2)$$
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
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i was wondering where did $t^2-5t+6=0$ come from? Thank you.
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– NoChance
Jul 24 at 4:24
3
$begingroup$
@NoChance See here: en.wikipedia.org/wiki/Vieta%27s_formulas If $x+y=frac51$ and $xy=frac61$ so $x$ and $y$ they are roots of the equation: $1t^2-5t+6=0$
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– Michael Rozenberg
Jul 24 at 4:47
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Got it. Many thanks for your clear explanation.
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– NoChance
Jul 24 at 11:05
1
$begingroup$
@NoChance You are welcome!
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– Michael Rozenberg
Jul 24 at 11:24
add a comment |
$begingroup$
You can use some symmetries (but I'm not sure if that makes any difference)
$$
2 x y =
(x + y) ^2 - (x^2 + y^2) =
25 - 13 = 12,
$$
express the difference
$$
(x - y)^2 =
(x^2 + y^2) - 2 x y =
1,
$$
and get a system of linear equations
$$
beginaligned
x + y &= 5,\
x - y &= pm 1,
endaligned
$$
that yields $x = 3$ and $y = 2$ or $x=2$ and $y=3$
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
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The quadratic equation can be used.
Given:
x + y = 5, then y = 5 -4
Given
x^2 + y^2 = 13
then x^2 + (4-x)^2 = 13
and x^2 + x^2 - 10x + 25 -13 = 0
2x^2 + (-10x) + 12 = 0
Then the co-factors are a = 2, b = -10, c = 12
y = [-b (+-) sqrt(b^2 - 4ac)]/[2a] <-- Quadratic Formula
y = [-(-10) (+-) sqrt((-10)^2 - 4(2*12))]/(2*2)
y = [10 (+-) sqrt(100-96)]/4
y = [10 + 2]/4 and y = [10-2]/4
y = 12/4 and y = 8/4
y = 3 and y = 2
given x + y = 5
When y = 3, x + 3 = 5, x = 5-3, x = 2
when y = 2, x+2 = 5, x = 5-2, x = 3
Answers: x = 3, y = 2 and x = 2, y = 3
Try your answers in all of the original equations and against any given or implied restrictions to make sure they work. They do! (Always check for 'extraneous' answers.)
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
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Using LaTex, you could simply enclose each equation between $...$ and the answer would look much better.
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– NoChance
Jul 24 at 14:47
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I'm new to this site and don't know what LaTex is. I apologize for how the post looks but hopefully the method is clear even if the format is hard to read.
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– Science_1
Jul 24 at 15:00
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No problem. To make the equations look like text book type setting a somewhat simple formatting can be used. For example instead of writing a=1, you write it between $..$ like $a=1$. A good description is here:math.meta.stackexchange.com/questions/5020/…
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– NoChance
Jul 24 at 15:12
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Thank you for this, NoChance! I will study your reference so my future answers can look as nice as everyone else's. Cheers!
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– Science_1
Jul 24 at 15:22
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It seems to me that this solution is precisely what the OP called "the most basic method". The questions is about other methods.
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– Arnaud D.
Jul 24 at 15:31
|
show 1 more comment
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Compute the Gröbner basis of your system. Let us start by writing this with zeroes on the right of the equals signs. beginalign*
0 &= x+y-5 \
0 &= x^2 + y^2 - 13 text.
endalign*
We pick a variable ordering. Let us choose $x < y$. (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering. We compute the first $s$-polynomial. We need the GCD of the leading terms
$$ gcd(x, x^2) = x^2 $$
and using this we get
beginalign*
0 &= fracx^2x(x+y-5) - fracx^2x^2(x^2 + y^2 - 13) \
&= x^2 + xy - 5x -(x^2 + y^2 - 13) \
&= xy - y^2 -5x + 13 text.
endalign*
Now $gcd(xy, x) = xy$ and
beginalign*
0 &= fracxyx(x+y-5) - fracxyxy(xy - y^2 -5x + 13) \
&= xy + y^2 - 5y -(xy - y^2 -5x + 13) \
&= 2y^2 +5x -5y -13
endalign*
and since we have a relation for $x$ and $y$ both of degree $1$, beginalign*
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5) \
&= 2y^2 +5x - 5y - 13 -5x -5y + 25 \
&= 2y^2 -10y + 12 \
&= 2(y^2 - 5y + 6) text,
endalign*
and since twice a thing is zero means the thing is zero, we have
$$ y^2 - 5y + 6 = 0 text. $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
beginalign*
x^2 + y^2 - 13 &= 0 \
xy - y^2 -5x + 13 &= 0 \
y^2 - 5y + 6 &= 0 \
x+y-5 &= 0 text.
endalign*
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone. Solving that polynomial, $y = 2$ or $y = 3$. Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
beginalign*
x^2 - 9 &= 0 \
-3x + 9 &= 0 \
0 &= 0 \
x-3 &= 0 \
y -2 &= 0 text,
endalign*
giving the solution $(x,y) = (3,2)$, or
beginalign*
x^2 - 4 &= 0 \
-2x + 4 &= 0 \
0 &= 0 \
x-2 &= 0 \
y - 3 &= 0 text,
endalign*
giving the solution $(x,y) = (2,3)$.
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
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Let $u^2-5u+c$ be the polynomial whose roots are $x$ and $y$, i.e.
$$u^2-5u+c=(u-x)(u-y)=u^2-u(x+y)+xy.$$
Then
beginalign*
x^2-5x+c&=0\
y^2-5y+c&=0.
endalign*
Adding the two equations and using the facts given we get
$$13-25+2c=0 implies c=6.$$
Thus we have $u^2-5u+6$ as our polynomial, so $x=2,y=3$ or vice versa.
answered Jul 24 at 3:52
Anurag AAnurag A
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Let's use some geometry.
I tried as simple approach as I was able to muster
We can do it because it's easy to see that the only possible solutions will always contain x > 0 and y > 0: if it's not so then at least one of them will be greater than 5 as follows from the first equation and then it's square is greater than 25 which contradicts with the second equation. Let it be x <= y for simplicity.
Your equations tell this picture:
The areas of rectangles R are equal and also their area is equal to the area of outer square without squares X and Y all divided by 2, so R = (25 - 13)/2 = 6
Then by square symmetry we also have:
So, area of S is the area of outer square minus area R four times, thus S = 25 - 4*R = 25 - 4*6 = 1, but the side of S (which is 1 since S is a square) is also the difference between the sides of squares Y and X (which are y and x) and therefore x + 1 = y
Remembering now our first figure and x + y = 5 we get x = 2 and y = 3.
By symmetry of course, if (x, y) is a solution, then (y, x) is too, so x = 3 and y = 2 also solves the original equations. This permutation is also easily illustrated on the figures above (as they don't change if x and y are just swapped).
answered Jul 24 at 12:56
Alma DoAlma Do
1214 bronze badges
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Another method, good for double-checking your answers on a test, is to graph the two equations on a TI-84 or similar calculator, then examine the graph to see where the lines overlap.
On the calculator, under the [y=] button, set
Y1 = 5-x
Y2 = sqrt(13-x^2)
Y3 = -sqrt(13-x^2)
Then press [graph].
When the graph is displayed, press [2nd][trace] to get into the calculation menu.
Choose #5, intersect.
Select the lines that intersect and choose a 'guess' point that is close to the intersection. The calculator will come back with the answer, x = 3, y = 2
Do the same for the other intersection and the calculator will come back with x =2 , y = 3.
This doesn't always work where you don't have nice, text-book solutions, but when it does work, boy is it nice!
answered Jul 24 at 15:21
Science_1Science_1
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You can use polynomial division to eliminate a variable.
$$(x^2+y^2-13) - x(x+y-5) = y^2 -xy+5x-13$$
$$(y^2 -xy+5x-13) - (-y+5)(x+y-5) = 2y^2-10y+12$$
Solve the equation 2$y^2-10y+12=0$, then plug the values of $y$ into the linear equation.
Generalization: Given a system of polynomial equations in $2$ variables, if one of the equations has one of the variables occurring only as a linear term, then you can eliminate that variable by polynomial division to get a polynomial equation in the remaining unknown. The utility of this is somewhat suspect due to the unsolvability of many univariate polynomials.
Bigger generalizations:
Groebner bases https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis
Elimination theory https://en.wikipedia.org/wiki/Elimination_theory
If you don't need exact answers, but only decimal approximations up to a specific precision, skip all this and look up Newton's method.
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
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The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 and (-6 + -4) = -10.
2x^2 + (-10x) + 12 = 0
but, as we found, -10x = (-6x + -4x) so by substitution..
2x^2 + (-6x + -4x) + 12 = 0
Regrouping we get..
[2x^2 - 6x] + [-4x + 12] = 0
Factoring like terms out we get..
2x(x-3) + -4(x-3) = 0
Factoring (x-3) out we get..
(x-3)(2x-4) = 0
The zeros occur where x-3 = 0 and where 2x-4 = 0
x-3 = 0, x = 3
2x - 4 = 0, 2x = 4, x = 4/2, x = 2
Given x+y = 5
when x = 3, 3+y=5, y = 5-3, y = 2
when x = 2, 2+y=5, y = 5-2, y = 3
Answers:
x = 3, y = 2 and x = 2, y = 3
answered Jul 24 at 14:56
Science_1Science_1
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10 Answers
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10 Answers
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active
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$begingroup$
In general, the set of equations:
$$sum_k=1^Nx_k^p = S_p$$
for $1leq pleq N$, can be solved by considering the function:
$$f(x) = -sum_p=1^Nlogleft(1-fracx_pxright) tag1$$
The expansion of $f(x)$ around infinity is given by:
$$f(x) = sum_r=1^inftyfracS_rr x^r$$
We can thus write down $f(x)$ to order $x^-2$ as:
$$f(x) = frac5x + frac132 x^2 + mathcalOleft(x^-3right)tag2$$
From (1) it follows that $x^2 expleft[-f(x)right]$ is a second degree polynomial that has the solutions as its roots. Using (2) it follows that:
$$expleft[-f(x)right] = 1 - frac5x + frac6x^2 + mathcalOleft(x^-3right)$$
It thus follows that:
$$(x-x_1)(x-x_2) = x^2 - 5 x + 6$$
So, the solutions are $x_1=2$ and $x_2 = 3$ and vice versa.
answered Jul 24 at 0:38
Count IblisCount Iblis
8,9602 gold badges16 silver badges37 bronze badges
$endgroup$
5
$begingroup$
Is there any reference to this method, I need to understand it further if possible. Thank you.
$endgroup$
– NoChance
Jul 24 at 11:08
2
$begingroup$
That is a sweet trick. But unfortunately, it only covers the special case where all variables are treated symmetrically.
$endgroup$
– Paul Sinclair
Jul 24 at 16:26
add a comment |
$begingroup$
In general, the set of equations:
$$sum_k=1^Nx_k^p = S_p$$
for $1leq pleq N$, can be solved by considering the function:
$$f(x) = -sum_p=1^Nlogleft(1-fracx_pxright) tag1$$
The expansion of $f(x)$ around infinity is given by:
$$f(x) = sum_r=1^inftyfracS_rr x^r$$
We can thus write down $f(x)$ to order $x^-2$ as:
$$f(x) = frac5x + frac132 x^2 + mathcalOleft(x^-3right)tag2$$
From (1) it follows that $x^2 expleft[-f(x)right]$ is a second degree polynomial that has the solutions as its roots. Using (2) it follows that:
$$expleft[-f(x)right] = 1 - frac5x + frac6x^2 + mathcalOleft(x^-3right)$$
It thus follows that:
$$(x-x_1)(x-x_2) = x^2 - 5 x + 6$$
So, the solutions are $x_1=2$ and $x_2 = 3$ and vice versa.
answered Jul 24 at 0:38
Count IblisCount Iblis
8,9602 gold badges16 silver badges37 bronze badges
$endgroup$
5
$begingroup$
Is there any reference to this method, I need to understand it further if possible. Thank you.
$endgroup$
– NoChance
Jul 24 at 11:08
2
$begingroup$
That is a sweet trick. But unfortunately, it only covers the special case where all variables are treated symmetrically.
$endgroup$
– Paul Sinclair
Jul 24 at 16:26
add a comment |
$begingroup$
In general, the set of equations:
$$sum_k=1^Nx_k^p = S_p$$
for $1leq pleq N$, can be solved by considering the function:
$$f(x) = -sum_p=1^Nlogleft(1-fracx_pxright) tag1$$
The expansion of $f(x)$ around infinity is given by:
$$f(x) = sum_r=1^inftyfracS_rr x^r$$
We can thus write down $f(x)$ to order $x^-2$ as:
$$f(x) = frac5x + frac132 x^2 + mathcalOleft(x^-3right)tag2$$
From (1) it follows that $x^2 expleft[-f(x)right]$ is a second degree polynomial that has the solutions as its roots. Using (2) it follows that:
$$expleft[-f(x)right] = 1 - frac5x + frac6x^2 + mathcalOleft(x^-3right)$$
It thus follows that:
$$(x-x_1)(x-x_2) = x^2 - 5 x + 6$$
So, the solutions are $x_1=2$ and $x_2 = 3$ and vice versa.
answered Jul 24 at 0:38
Count IblisCount Iblis
8,9602 gold badges16 silver badges37 bronze badges
$endgroup$
In general, the set of equations:
$$sum_k=1^Nx_k^p = S_p$$
for $1leq pleq N$, can be solved by considering the function:
$$f(x) = -sum_p=1^Nlogleft(1-fracx_pxright) tag1$$
The expansion of $f(x)$ around infinity is given by:
$$f(x) = sum_r=1^inftyfracS_rr x^r$$
We can thus write down $f(x)$ to order $x^-2$ as:
$$f(x) = frac5x + frac132 x^2 + mathcalOleft(x^-3right)tag2$$
From (1) it follows that $x^2 expleft[-f(x)right]$ is a second degree polynomial that has the solutions as its roots. Using (2) it follows that:
$$expleft[-f(x)right] = 1 - frac5x + frac6x^2 + mathcalOleft(x^-3right)$$
It thus follows that:
$$(x-x_1)(x-x_2) = x^2 - 5 x + 6$$
So, the solutions are $x_1=2$ and $x_2 = 3$ and vice versa.
answered Jul 24 at 0:38
Count IblisCount Iblis
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answered Jul 24 at 0:38
Count IblisCount Iblis
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answered Jul 24 at 0:38
Count IblisCount Iblis
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answered Jul 24 at 0:38
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5
$begingroup$
Is there any reference to this method, I need to understand it further if possible. Thank you.
$endgroup$
– NoChance
Jul 24 at 11:08
2
$begingroup$
That is a sweet trick. But unfortunately, it only covers the special case where all variables are treated symmetrically.
$endgroup$
– Paul Sinclair
Jul 24 at 16:26
add a comment |
5
$begingroup$
Is there any reference to this method, I need to understand it further if possible. Thank you.
$endgroup$
– NoChance
Jul 24 at 11:08
2
$begingroup$
That is a sweet trick. But unfortunately, it only covers the special case where all variables are treated symmetrically.
$endgroup$
– Paul Sinclair
Jul 24 at 16:26
5
5
$begingroup$
Is there any reference to this method, I need to understand it further if possible. Thank you.
$endgroup$
– NoChance
Jul 24 at 11:08
$begingroup$
Is there any reference to this method, I need to understand it further if possible. Thank you.
$endgroup$
– NoChance
Jul 24 at 11:08
2
2
$begingroup$
That is a sweet trick. But unfortunately, it only covers the special case where all variables are treated symmetrically.
$endgroup$
– Paul Sinclair
Jul 24 at 16:26
$begingroup$
That is a sweet trick. But unfortunately, it only covers the special case where all variables are treated symmetrically.
$endgroup$
– Paul Sinclair
Jul 24 at 16:26
add a comment |
$begingroup$
We have $$(x+y)^2=13+2xy,$$ which gives
$$xy=6$$ and by the Viete's theorem $x$ and $y$ are roots of the equation:
$$t^2-5t+6=0$$ or
$$(t-2)(t-3)=0,$$ which gives the answer:
$$(2,3),(3,2)$$
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
$endgroup$
$begingroup$
i was wondering where did $t^2-5t+6=0$ come from? Thank you.
$endgroup$
– NoChance
Jul 24 at 4:24
3
$begingroup$
@NoChance See here: en.wikipedia.org/wiki/Vieta%27s_formulas If $x+y=frac51$ and $xy=frac61$ so $x$ and $y$ they are roots of the equation: $1t^2-5t+6=0$
$endgroup$
– Michael Rozenberg
Jul 24 at 4:47
$begingroup$
Got it. Many thanks for your clear explanation.
$endgroup$
– NoChance
Jul 24 at 11:05
1
$begingroup$
@NoChance You are welcome!
$endgroup$
– Michael Rozenberg
Jul 24 at 11:24
add a comment |
$begingroup$
We have $$(x+y)^2=13+2xy,$$ which gives
$$xy=6$$ and by the Viete's theorem $x$ and $y$ are roots of the equation:
$$t^2-5t+6=0$$ or
$$(t-2)(t-3)=0,$$ which gives the answer:
$$(2,3),(3,2)$$
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
$endgroup$
$begingroup$
i was wondering where did $t^2-5t+6=0$ come from? Thank you.
$endgroup$
– NoChance
Jul 24 at 4:24
3
$begingroup$
@NoChance See here: en.wikipedia.org/wiki/Vieta%27s_formulas If $x+y=frac51$ and $xy=frac61$ so $x$ and $y$ they are roots of the equation: $1t^2-5t+6=0$
$endgroup$
– Michael Rozenberg
Jul 24 at 4:47
$begingroup$
Got it. Many thanks for your clear explanation.
$endgroup$
– NoChance
Jul 24 at 11:05
1
$begingroup$
@NoChance You are welcome!
$endgroup$
– Michael Rozenberg
Jul 24 at 11:24
add a comment |
$begingroup$
We have $$(x+y)^2=13+2xy,$$ which gives
$$xy=6$$ and by the Viete's theorem $x$ and $y$ are roots of the equation:
$$t^2-5t+6=0$$ or
$$(t-2)(t-3)=0,$$ which gives the answer:
$$(2,3),(3,2)$$
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
$endgroup$
We have $$(x+y)^2=13+2xy,$$ which gives
$$xy=6$$ and by the Viete's theorem $x$ and $y$ are roots of the equation:
$$t^2-5t+6=0$$ or
$$(t-2)(t-3)=0,$$ which gives the answer:
$$(2,3),(3,2)$$
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
answered Jul 24 at 0:18
Michael RozenbergMichael Rozenberg
122k20 gold badges105 silver badges210 bronze badges
122k20 gold badges105 silver badges210 bronze badges
$begingroup$
i was wondering where did $t^2-5t+6=0$ come from? Thank you.
$endgroup$
– NoChance
Jul 24 at 4:24
3
$begingroup$
@NoChance See here: en.wikipedia.org/wiki/Vieta%27s_formulas If $x+y=frac51$ and $xy=frac61$ so $x$ and $y$ they are roots of the equation: $1t^2-5t+6=0$
$endgroup$
– Michael Rozenberg
Jul 24 at 4:47
$begingroup$
Got it. Many thanks for your clear explanation.
$endgroup$
– NoChance
Jul 24 at 11:05
1
$begingroup$
@NoChance You are welcome!
$endgroup$
– Michael Rozenberg
Jul 24 at 11:24
add a comment |
$begingroup$
i was wondering where did $t^2-5t+6=0$ come from? Thank you.
$endgroup$
– NoChance
Jul 24 at 4:24
3
$begingroup$
@NoChance See here: en.wikipedia.org/wiki/Vieta%27s_formulas If $x+y=frac51$ and $xy=frac61$ so $x$ and $y$ they are roots of the equation: $1t^2-5t+6=0$
$endgroup$
– Michael Rozenberg
Jul 24 at 4:47
$begingroup$
Got it. Many thanks for your clear explanation.
$endgroup$
– NoChance
Jul 24 at 11:05
1
$begingroup$
@NoChance You are welcome!
$endgroup$
– Michael Rozenberg
Jul 24 at 11:24
$begingroup$
i was wondering where did $t^2-5t+6=0$ come from? Thank you.
$endgroup$
– NoChance
Jul 24 at 4:24
$begingroup$
i was wondering where did $t^2-5t+6=0$ come from? Thank you.
$endgroup$
– NoChance
Jul 24 at 4:24
3
3
$begingroup$
@NoChance See here: en.wikipedia.org/wiki/Vieta%27s_formulas If $x+y=frac51$ and $xy=frac61$ so $x$ and $y$ they are roots of the equation: $1t^2-5t+6=0$
$endgroup$
– Michael Rozenberg
Jul 24 at 4:47
$begingroup$
@NoChance See here: en.wikipedia.org/wiki/Vieta%27s_formulas If $x+y=frac51$ and $xy=frac61$ so $x$ and $y$ they are roots of the equation: $1t^2-5t+6=0$
$endgroup$
– Michael Rozenberg
Jul 24 at 4:47
$begingroup$
Got it. Many thanks for your clear explanation.
$endgroup$
– NoChance
Jul 24 at 11:05
$begingroup$
Got it. Many thanks for your clear explanation.
$endgroup$
– NoChance
Jul 24 at 11:05
1
1
$begingroup$
@NoChance You are welcome!
$endgroup$
– Michael Rozenberg
Jul 24 at 11:24
$begingroup$
@NoChance You are welcome!
$endgroup$
– Michael Rozenberg
Jul 24 at 11:24
add a comment |
$begingroup$
You can use some symmetries (but I'm not sure if that makes any difference)
$$
2 x y =
(x + y) ^2 - (x^2 + y^2) =
25 - 13 = 12,
$$
express the difference
$$
(x - y)^2 =
(x^2 + y^2) - 2 x y =
1,
$$
and get a system of linear equations
$$
beginaligned
x + y &= 5,\
x - y &= pm 1,
endaligned
$$
that yields $x = 3$ and $y = 2$ or $x=2$ and $y=3$
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
$endgroup$
add a comment |
$begingroup$
You can use some symmetries (but I'm not sure if that makes any difference)
$$
2 x y =
(x + y) ^2 - (x^2 + y^2) =
25 - 13 = 12,
$$
express the difference
$$
(x - y)^2 =
(x^2 + y^2) - 2 x y =
1,
$$
and get a system of linear equations
$$
beginaligned
x + y &= 5,\
x - y &= pm 1,
endaligned
$$
that yields $x = 3$ and $y = 2$ or $x=2$ and $y=3$
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
$endgroup$
add a comment |
$begingroup$
You can use some symmetries (but I'm not sure if that makes any difference)
$$
2 x y =
(x + y) ^2 - (x^2 + y^2) =
25 - 13 = 12,
$$
express the difference
$$
(x - y)^2 =
(x^2 + y^2) - 2 x y =
1,
$$
and get a system of linear equations
$$
beginaligned
x + y &= 5,\
x - y &= pm 1,
endaligned
$$
that yields $x = 3$ and $y = 2$ or $x=2$ and $y=3$
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
$endgroup$
You can use some symmetries (but I'm not sure if that makes any difference)
$$
2 x y =
(x + y) ^2 - (x^2 + y^2) =
25 - 13 = 12,
$$
express the difference
$$
(x - y)^2 =
(x^2 + y^2) - 2 x y =
1,
$$
and get a system of linear equations
$$
beginaligned
x + y &= 5,\
x - y &= pm 1,
endaligned
$$
that yields $x = 3$ and $y = 2$ or $x=2$ and $y=3$
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
answered Jul 24 at 0:23
guestguest
1961 silver badge3 bronze badges
1961 silver badge3 bronze badges
add a comment |
add a comment |
$begingroup$
The quadratic equation can be used.
Given:
x + y = 5, then y = 5 -4
Given
x^2 + y^2 = 13
then x^2 + (4-x)^2 = 13
and x^2 + x^2 - 10x + 25 -13 = 0
2x^2 + (-10x) + 12 = 0
Then the co-factors are a = 2, b = -10, c = 12
y = [-b (+-) sqrt(b^2 - 4ac)]/[2a] <-- Quadratic Formula
y = [-(-10) (+-) sqrt((-10)^2 - 4(2*12))]/(2*2)
y = [10 (+-) sqrt(100-96)]/4
y = [10 + 2]/4 and y = [10-2]/4
y = 12/4 and y = 8/4
y = 3 and y = 2
given x + y = 5
When y = 3, x + 3 = 5, x = 5-3, x = 2
when y = 2, x+2 = 5, x = 5-2, x = 3
Answers: x = 3, y = 2 and x = 2, y = 3
Try your answers in all of the original equations and against any given or implied restrictions to make sure they work. They do! (Always check for 'extraneous' answers.)
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
$endgroup$
$begingroup$
Using LaTex, you could simply enclose each equation between $...$ and the answer would look much better.
$endgroup$
– NoChance
Jul 24 at 14:47
$begingroup$
I'm new to this site and don't know what LaTex is. I apologize for how the post looks but hopefully the method is clear even if the format is hard to read.
$endgroup$
– Science_1
Jul 24 at 15:00
$begingroup$
No problem. To make the equations look like text book type setting a somewhat simple formatting can be used. For example instead of writing a=1, you write it between $..$ like $a=1$. A good description is here:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– NoChance
Jul 24 at 15:12
$begingroup$
Thank you for this, NoChance! I will study your reference so my future answers can look as nice as everyone else's. Cheers!
$endgroup$
– Science_1
Jul 24 at 15:22
$begingroup$
It seems to me that this solution is precisely what the OP called "the most basic method". The questions is about other methods.
$endgroup$
– Arnaud D.
Jul 24 at 15:31
|
show 1 more comment
$begingroup$
The quadratic equation can be used.
Given:
x + y = 5, then y = 5 -4
Given
x^2 + y^2 = 13
then x^2 + (4-x)^2 = 13
and x^2 + x^2 - 10x + 25 -13 = 0
2x^2 + (-10x) + 12 = 0
Then the co-factors are a = 2, b = -10, c = 12
y = [-b (+-) sqrt(b^2 - 4ac)]/[2a] <-- Quadratic Formula
y = [-(-10) (+-) sqrt((-10)^2 - 4(2*12))]/(2*2)
y = [10 (+-) sqrt(100-96)]/4
y = [10 + 2]/4 and y = [10-2]/4
y = 12/4 and y = 8/4
y = 3 and y = 2
given x + y = 5
When y = 3, x + 3 = 5, x = 5-3, x = 2
when y = 2, x+2 = 5, x = 5-2, x = 3
Answers: x = 3, y = 2 and x = 2, y = 3
Try your answers in all of the original equations and against any given or implied restrictions to make sure they work. They do! (Always check for 'extraneous' answers.)
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
$endgroup$
$begingroup$
Using LaTex, you could simply enclose each equation between $...$ and the answer would look much better.
$endgroup$
– NoChance
Jul 24 at 14:47
$begingroup$
I'm new to this site and don't know what LaTex is. I apologize for how the post looks but hopefully the method is clear even if the format is hard to read.
$endgroup$
– Science_1
Jul 24 at 15:00
$begingroup$
No problem. To make the equations look like text book type setting a somewhat simple formatting can be used. For example instead of writing a=1, you write it between $..$ like $a=1$. A good description is here:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– NoChance
Jul 24 at 15:12
$begingroup$
Thank you for this, NoChance! I will study your reference so my future answers can look as nice as everyone else's. Cheers!
$endgroup$
– Science_1
Jul 24 at 15:22
$begingroup$
It seems to me that this solution is precisely what the OP called "the most basic method". The questions is about other methods.
$endgroup$
– Arnaud D.
Jul 24 at 15:31
|
show 1 more comment
$begingroup$
The quadratic equation can be used.
Given:
x + y = 5, then y = 5 -4
Given
x^2 + y^2 = 13
then x^2 + (4-x)^2 = 13
and x^2 + x^2 - 10x + 25 -13 = 0
2x^2 + (-10x) + 12 = 0
Then the co-factors are a = 2, b = -10, c = 12
y = [-b (+-) sqrt(b^2 - 4ac)]/[2a] <-- Quadratic Formula
y = [-(-10) (+-) sqrt((-10)^2 - 4(2*12))]/(2*2)
y = [10 (+-) sqrt(100-96)]/4
y = [10 + 2]/4 and y = [10-2]/4
y = 12/4 and y = 8/4
y = 3 and y = 2
given x + y = 5
When y = 3, x + 3 = 5, x = 5-3, x = 2
when y = 2, x+2 = 5, x = 5-2, x = 3
Answers: x = 3, y = 2 and x = 2, y = 3
Try your answers in all of the original equations and against any given or implied restrictions to make sure they work. They do! (Always check for 'extraneous' answers.)
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
$endgroup$
The quadratic equation can be used.
Given:
x + y = 5, then y = 5 -4
Given
x^2 + y^2 = 13
then x^2 + (4-x)^2 = 13
and x^2 + x^2 - 10x + 25 -13 = 0
2x^2 + (-10x) + 12 = 0
Then the co-factors are a = 2, b = -10, c = 12
y = [-b (+-) sqrt(b^2 - 4ac)]/[2a] <-- Quadratic Formula
y = [-(-10) (+-) sqrt((-10)^2 - 4(2*12))]/(2*2)
y = [10 (+-) sqrt(100-96)]/4
y = [10 + 2]/4 and y = [10-2]/4
y = 12/4 and y = 8/4
y = 3 and y = 2
given x + y = 5
When y = 3, x + 3 = 5, x = 5-3, x = 2
when y = 2, x+2 = 5, x = 5-2, x = 3
Answers: x = 3, y = 2 and x = 2, y = 3
Try your answers in all of the original equations and against any given or implied restrictions to make sure they work. They do! (Always check for 'extraneous' answers.)
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
edited Jul 24 at 15:29
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
answered Jul 24 at 14:34
Science_1Science_1
172 bronze badges
172 bronze badges
$begingroup$
Using LaTex, you could simply enclose each equation between $...$ and the answer would look much better.
$endgroup$
– NoChance
Jul 24 at 14:47
$begingroup$
I'm new to this site and don't know what LaTex is. I apologize for how the post looks but hopefully the method is clear even if the format is hard to read.
$endgroup$
– Science_1
Jul 24 at 15:00
$begingroup$
No problem. To make the equations look like text book type setting a somewhat simple formatting can be used. For example instead of writing a=1, you write it between $..$ like $a=1$. A good description is here:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– NoChance
Jul 24 at 15:12
$begingroup$
Thank you for this, NoChance! I will study your reference so my future answers can look as nice as everyone else's. Cheers!
$endgroup$
– Science_1
Jul 24 at 15:22
$begingroup$
It seems to me that this solution is precisely what the OP called "the most basic method". The questions is about other methods.
$endgroup$
– Arnaud D.
Jul 24 at 15:31
|
show 1 more comment
$begingroup$
Using LaTex, you could simply enclose each equation between $...$ and the answer would look much better.
$endgroup$
– NoChance
Jul 24 at 14:47
$begingroup$
I'm new to this site and don't know what LaTex is. I apologize for how the post looks but hopefully the method is clear even if the format is hard to read.
$endgroup$
– Science_1
Jul 24 at 15:00
$begingroup$
No problem. To make the equations look like text book type setting a somewhat simple formatting can be used. For example instead of writing a=1, you write it between $..$ like $a=1$. A good description is here:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– NoChance
Jul 24 at 15:12
$begingroup$
Thank you for this, NoChance! I will study your reference so my future answers can look as nice as everyone else's. Cheers!
$endgroup$
– Science_1
Jul 24 at 15:22
$begingroup$
It seems to me that this solution is precisely what the OP called "the most basic method". The questions is about other methods.
$endgroup$
– Arnaud D.
Jul 24 at 15:31
$begingroup$
Using LaTex, you could simply enclose each equation between $...$ and the answer would look much better.
$endgroup$
– NoChance
Jul 24 at 14:47
$begingroup$
Using LaTex, you could simply enclose each equation between $...$ and the answer would look much better.
$endgroup$
– NoChance
Jul 24 at 14:47
$begingroup$
I'm new to this site and don't know what LaTex is. I apologize for how the post looks but hopefully the method is clear even if the format is hard to read.
$endgroup$
– Science_1
Jul 24 at 15:00
$begingroup$
I'm new to this site and don't know what LaTex is. I apologize for how the post looks but hopefully the method is clear even if the format is hard to read.
$endgroup$
– Science_1
Jul 24 at 15:00
$begingroup$
No problem. To make the equations look like text book type setting a somewhat simple formatting can be used. For example instead of writing a=1, you write it between $..$ like $a=1$. A good description is here:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– NoChance
Jul 24 at 15:12
$begingroup$
No problem. To make the equations look like text book type setting a somewhat simple formatting can be used. For example instead of writing a=1, you write it between $..$ like $a=1$. A good description is here:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– NoChance
Jul 24 at 15:12
$begingroup$
Thank you for this, NoChance! I will study your reference so my future answers can look as nice as everyone else's. Cheers!
$endgroup$
– Science_1
Jul 24 at 15:22
$begingroup$
Thank you for this, NoChance! I will study your reference so my future answers can look as nice as everyone else's. Cheers!
$endgroup$
– Science_1
Jul 24 at 15:22
$begingroup$
It seems to me that this solution is precisely what the OP called "the most basic method". The questions is about other methods.
$endgroup$
– Arnaud D.
Jul 24 at 15:31
$begingroup$
It seems to me that this solution is precisely what the OP called "the most basic method". The questions is about other methods.
$endgroup$
– Arnaud D.
Jul 24 at 15:31
|
show 1 more comment
$begingroup$
Compute the Gröbner basis of your system. Let us start by writing this with zeroes on the right of the equals signs. beginalign*
0 &= x+y-5 \
0 &= x^2 + y^2 - 13 text.
endalign*
We pick a variable ordering. Let us choose $x < y$. (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering. We compute the first $s$-polynomial. We need the GCD of the leading terms
$$ gcd(x, x^2) = x^2 $$
and using this we get
beginalign*
0 &= fracx^2x(x+y-5) - fracx^2x^2(x^2 + y^2 - 13) \
&= x^2 + xy - 5x -(x^2 + y^2 - 13) \
&= xy - y^2 -5x + 13 text.
endalign*
Now $gcd(xy, x) = xy$ and
beginalign*
0 &= fracxyx(x+y-5) - fracxyxy(xy - y^2 -5x + 13) \
&= xy + y^2 - 5y -(xy - y^2 -5x + 13) \
&= 2y^2 +5x -5y -13
endalign*
and since we have a relation for $x$ and $y$ both of degree $1$, beginalign*
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5) \
&= 2y^2 +5x - 5y - 13 -5x -5y + 25 \
&= 2y^2 -10y + 12 \
&= 2(y^2 - 5y + 6) text,
endalign*
and since twice a thing is zero means the thing is zero, we have
$$ y^2 - 5y + 6 = 0 text. $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
beginalign*
x^2 + y^2 - 13 &= 0 \
xy - y^2 -5x + 13 &= 0 \
y^2 - 5y + 6 &= 0 \
x+y-5 &= 0 text.
endalign*
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone. Solving that polynomial, $y = 2$ or $y = 3$. Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
beginalign*
x^2 - 9 &= 0 \
-3x + 9 &= 0 \
0 &= 0 \
x-3 &= 0 \
y -2 &= 0 text,
endalign*
giving the solution $(x,y) = (3,2)$, or
beginalign*
x^2 - 4 &= 0 \
-2x + 4 &= 0 \
0 &= 0 \
x-2 &= 0 \
y - 3 &= 0 text,
endalign*
giving the solution $(x,y) = (2,3)$.
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
$endgroup$
add a comment |
$begingroup$
Compute the Gröbner basis of your system. Let us start by writing this with zeroes on the right of the equals signs. beginalign*
0 &= x+y-5 \
0 &= x^2 + y^2 - 13 text.
endalign*
We pick a variable ordering. Let us choose $x < y$. (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering. We compute the first $s$-polynomial. We need the GCD of the leading terms
$$ gcd(x, x^2) = x^2 $$
and using this we get
beginalign*
0 &= fracx^2x(x+y-5) - fracx^2x^2(x^2 + y^2 - 13) \
&= x^2 + xy - 5x -(x^2 + y^2 - 13) \
&= xy - y^2 -5x + 13 text.
endalign*
Now $gcd(xy, x) = xy$ and
beginalign*
0 &= fracxyx(x+y-5) - fracxyxy(xy - y^2 -5x + 13) \
&= xy + y^2 - 5y -(xy - y^2 -5x + 13) \
&= 2y^2 +5x -5y -13
endalign*
and since we have a relation for $x$ and $y$ both of degree $1$, beginalign*
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5) \
&= 2y^2 +5x - 5y - 13 -5x -5y + 25 \
&= 2y^2 -10y + 12 \
&= 2(y^2 - 5y + 6) text,
endalign*
and since twice a thing is zero means the thing is zero, we have
$$ y^2 - 5y + 6 = 0 text. $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
beginalign*
x^2 + y^2 - 13 &= 0 \
xy - y^2 -5x + 13 &= 0 \
y^2 - 5y + 6 &= 0 \
x+y-5 &= 0 text.
endalign*
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone. Solving that polynomial, $y = 2$ or $y = 3$. Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
beginalign*
x^2 - 9 &= 0 \
-3x + 9 &= 0 \
0 &= 0 \
x-3 &= 0 \
y -2 &= 0 text,
endalign*
giving the solution $(x,y) = (3,2)$, or
beginalign*
x^2 - 4 &= 0 \
-2x + 4 &= 0 \
0 &= 0 \
x-2 &= 0 \
y - 3 &= 0 text,
endalign*
giving the solution $(x,y) = (2,3)$.
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
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$begingroup$
Compute the Gröbner basis of your system. Let us start by writing this with zeroes on the right of the equals signs. beginalign*
0 &= x+y-5 \
0 &= x^2 + y^2 - 13 text.
endalign*
We pick a variable ordering. Let us choose $x < y$. (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering. We compute the first $s$-polynomial. We need the GCD of the leading terms
$$ gcd(x, x^2) = x^2 $$
and using this we get
beginalign*
0 &= fracx^2x(x+y-5) - fracx^2x^2(x^2 + y^2 - 13) \
&= x^2 + xy - 5x -(x^2 + y^2 - 13) \
&= xy - y^2 -5x + 13 text.
endalign*
Now $gcd(xy, x) = xy$ and
beginalign*
0 &= fracxyx(x+y-5) - fracxyxy(xy - y^2 -5x + 13) \
&= xy + y^2 - 5y -(xy - y^2 -5x + 13) \
&= 2y^2 +5x -5y -13
endalign*
and since we have a relation for $x$ and $y$ both of degree $1$, beginalign*
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5) \
&= 2y^2 +5x - 5y - 13 -5x -5y + 25 \
&= 2y^2 -10y + 12 \
&= 2(y^2 - 5y + 6) text,
endalign*
and since twice a thing is zero means the thing is zero, we have
$$ y^2 - 5y + 6 = 0 text. $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
beginalign*
x^2 + y^2 - 13 &= 0 \
xy - y^2 -5x + 13 &= 0 \
y^2 - 5y + 6 &= 0 \
x+y-5 &= 0 text.
endalign*
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone. Solving that polynomial, $y = 2$ or $y = 3$. Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
beginalign*
x^2 - 9 &= 0 \
-3x + 9 &= 0 \
0 &= 0 \
x-3 &= 0 \
y -2 &= 0 text,
endalign*
giving the solution $(x,y) = (3,2)$, or
beginalign*
x^2 - 4 &= 0 \
-2x + 4 &= 0 \
0 &= 0 \
x-2 &= 0 \
y - 3 &= 0 text,
endalign*
giving the solution $(x,y) = (2,3)$.
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
$endgroup$
Compute the Gröbner basis of your system. Let us start by writing this with zeroes on the right of the equals signs. beginalign*
0 &= x+y-5 \
0 &= x^2 + y^2 - 13 text.
endalign*
We pick a variable ordering. Let us choose $x < y$. (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering. We compute the first $s$-polynomial. We need the GCD of the leading terms
$$ gcd(x, x^2) = x^2 $$
and using this we get
beginalign*
0 &= fracx^2x(x+y-5) - fracx^2x^2(x^2 + y^2 - 13) \
&= x^2 + xy - 5x -(x^2 + y^2 - 13) \
&= xy - y^2 -5x + 13 text.
endalign*
Now $gcd(xy, x) = xy$ and
beginalign*
0 &= fracxyx(x+y-5) - fracxyxy(xy - y^2 -5x + 13) \
&= xy + y^2 - 5y -(xy - y^2 -5x + 13) \
&= 2y^2 +5x -5y -13
endalign*
and since we have a relation for $x$ and $y$ both of degree $1$, beginalign*
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5) \
&= 2y^2 +5x - 5y - 13 -5x -5y + 25 \
&= 2y^2 -10y + 12 \
&= 2(y^2 - 5y + 6) text,
endalign*
and since twice a thing is zero means the thing is zero, we have
$$ y^2 - 5y + 6 = 0 text. $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
beginalign*
x^2 + y^2 - 13 &= 0 \
xy - y^2 -5x + 13 &= 0 \
y^2 - 5y + 6 &= 0 \
x+y-5 &= 0 text.
endalign*
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone. Solving that polynomial, $y = 2$ or $y = 3$. Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
beginalign*
x^2 - 9 &= 0 \
-3x + 9 &= 0 \
0 &= 0 \
x-3 &= 0 \
y -2 &= 0 text,
endalign*
giving the solution $(x,y) = (3,2)$, or
beginalign*
x^2 - 4 &= 0 \
-2x + 4 &= 0 \
0 &= 0 \
x-2 &= 0 \
y - 3 &= 0 text,
endalign*
giving the solution $(x,y) = (2,3)$.
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
answered Jul 24 at 23:16
Eric TowersEric Towers
35.6k2 gold badges23 silver badges71 bronze badges
35.6k2 gold badges23 silver badges71 bronze badges
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$begingroup$
Let $u^2-5u+c$ be the polynomial whose roots are $x$ and $y$, i.e.
$$u^2-5u+c=(u-x)(u-y)=u^2-u(x+y)+xy.$$
Then
beginalign*
x^2-5x+c&=0\
y^2-5y+c&=0.
endalign*
Adding the two equations and using the facts given we get
$$13-25+2c=0 implies c=6.$$
Thus we have $u^2-5u+6$ as our polynomial, so $x=2,y=3$ or vice versa.
answered Jul 24 at 3:52
Anurag AAnurag A
29.4k1 gold badge23 silver badges52 bronze badges
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$begingroup$
Let $u^2-5u+c$ be the polynomial whose roots are $x$ and $y$, i.e.
$$u^2-5u+c=(u-x)(u-y)=u^2-u(x+y)+xy.$$
Then
beginalign*
x^2-5x+c&=0\
y^2-5y+c&=0.
endalign*
Adding the two equations and using the facts given we get
$$13-25+2c=0 implies c=6.$$
Thus we have $u^2-5u+6$ as our polynomial, so $x=2,y=3$ or vice versa.
answered Jul 24 at 3:52
Anurag AAnurag A
29.4k1 gold badge23 silver badges52 bronze badges
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$begingroup$
Let $u^2-5u+c$ be the polynomial whose roots are $x$ and $y$, i.e.
$$u^2-5u+c=(u-x)(u-y)=u^2-u(x+y)+xy.$$
Then
beginalign*
x^2-5x+c&=0\
y^2-5y+c&=0.
endalign*
Adding the two equations and using the facts given we get
$$13-25+2c=0 implies c=6.$$
Thus we have $u^2-5u+6$ as our polynomial, so $x=2,y=3$ or vice versa.
answered Jul 24 at 3:52
Anurag AAnurag A
29.4k1 gold badge23 silver badges52 bronze badges
$endgroup$
Let $u^2-5u+c$ be the polynomial whose roots are $x$ and $y$, i.e.
$$u^2-5u+c=(u-x)(u-y)=u^2-u(x+y)+xy.$$
Then
beginalign*
x^2-5x+c&=0\
y^2-5y+c&=0.
endalign*
Adding the two equations and using the facts given we get
$$13-25+2c=0 implies c=6.$$
Thus we have $u^2-5u+6$ as our polynomial, so $x=2,y=3$ or vice versa.
answered Jul 24 at 3:52
Anurag AAnurag A
29.4k1 gold badge23 silver badges52 bronze badges
edited Jul 24 at 5:21
answered Jul 24 at 3:52
Anurag AAnurag A
29.4k1 gold badge23 silver badges52 bronze badges
answered Jul 24 at 3:52
Anurag AAnurag A
29.4k1 gold badge23 silver badges52 bronze badges
answered Jul 24 at 3:52
Anurag AAnurag A
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29.4k1 gold badge23 silver badges52 bronze badges
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$begingroup$
Let's use some geometry.
I tried as simple approach as I was able to muster
We can do it because it's easy to see that the only possible solutions will always contain x > 0 and y > 0: if it's not so then at least one of them will be greater than 5 as follows from the first equation and then it's square is greater than 25 which contradicts with the second equation. Let it be x <= y for simplicity.
Your equations tell this picture:
The areas of rectangles R are equal and also their area is equal to the area of outer square without squares X and Y all divided by 2, so R = (25 - 13)/2 = 6
Then by square symmetry we also have:
So, area of S is the area of outer square minus area R four times, thus S = 25 - 4*R = 25 - 4*6 = 1, but the side of S (which is 1 since S is a square) is also the difference between the sides of squares Y and X (which are y and x) and therefore x + 1 = y
Remembering now our first figure and x + y = 5 we get x = 2 and y = 3.
By symmetry of course, if (x, y) is a solution, then (y, x) is too, so x = 3 and y = 2 also solves the original equations. This permutation is also easily illustrated on the figures above (as they don't change if x and y are just swapped).
answered Jul 24 at 12:56
Alma DoAlma Do
1214 bronze badges
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$begingroup$
Let's use some geometry.
I tried as simple approach as I was able to muster
We can do it because it's easy to see that the only possible solutions will always contain x > 0 and y > 0: if it's not so then at least one of them will be greater than 5 as follows from the first equation and then it's square is greater than 25 which contradicts with the second equation. Let it be x <= y for simplicity.
Your equations tell this picture:
The areas of rectangles R are equal and also their area is equal to the area of outer square without squares X and Y all divided by 2, so R = (25 - 13)/2 = 6
Then by square symmetry we also have:
So, area of S is the area of outer square minus area R four times, thus S = 25 - 4*R = 25 - 4*6 = 1, but the side of S (which is 1 since S is a square) is also the difference between the sides of squares Y and X (which are y and x) and therefore x + 1 = y
Remembering now our first figure and x + y = 5 we get x = 2 and y = 3.
By symmetry of course, if (x, y) is a solution, then (y, x) is too, so x = 3 and y = 2 also solves the original equations. This permutation is also easily illustrated on the figures above (as they don't change if x and y are just swapped).
answered Jul 24 at 12:56
Alma DoAlma Do
1214 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's use some geometry.
I tried as simple approach as I was able to muster
We can do it because it's easy to see that the only possible solutions will always contain x > 0 and y > 0: if it's not so then at least one of them will be greater than 5 as follows from the first equation and then it's square is greater than 25 which contradicts with the second equation. Let it be x <= y for simplicity.
Your equations tell this picture:
The areas of rectangles R are equal and also their area is equal to the area of outer square without squares X and Y all divided by 2, so R = (25 - 13)/2 = 6
Then by square symmetry we also have:
So, area of S is the area of outer square minus area R four times, thus S = 25 - 4*R = 25 - 4*6 = 1, but the side of S (which is 1 since S is a square) is also the difference between the sides of squares Y and X (which are y and x) and therefore x + 1 = y
Remembering now our first figure and x + y = 5 we get x = 2 and y = 3.
By symmetry of course, if (x, y) is a solution, then (y, x) is too, so x = 3 and y = 2 also solves the original equations. This permutation is also easily illustrated on the figures above (as they don't change if x and y are just swapped).
answered Jul 24 at 12:56
Alma DoAlma Do
1214 bronze badges
$endgroup$
Let's use some geometry.
I tried as simple approach as I was able to muster
We can do it because it's easy to see that the only possible solutions will always contain x > 0 and y > 0: if it's not so then at least one of them will be greater than 5 as follows from the first equation and then it's square is greater than 25 which contradicts with the second equation. Let it be x <= y for simplicity.
Your equations tell this picture:
The areas of rectangles R are equal and also their area is equal to the area of outer square without squares X and Y all divided by 2, so R = (25 - 13)/2 = 6
Then by square symmetry we also have:
So, area of S is the area of outer square minus area R four times, thus S = 25 - 4*R = 25 - 4*6 = 1, but the side of S (which is 1 since S is a square) is also the difference between the sides of squares Y and X (which are y and x) and therefore x + 1 = y
Remembering now our first figure and x + y = 5 we get x = 2 and y = 3.
By symmetry of course, if (x, y) is a solution, then (y, x) is too, so x = 3 and y = 2 also solves the original equations. This permutation is also easily illustrated on the figures above (as they don't change if x and y are just swapped).
answered Jul 24 at 12:56
Alma DoAlma Do
1214 bronze badges
answered Jul 24 at 12:56
Alma DoAlma Do
1214 bronze badges
answered Jul 24 at 12:56
Alma DoAlma Do
1214 bronze badges
answered Jul 24 at 12:56
Alma DoAlma Do
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1214 bronze badges
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$begingroup$
Another method, good for double-checking your answers on a test, is to graph the two equations on a TI-84 or similar calculator, then examine the graph to see where the lines overlap.
On the calculator, under the [y=] button, set
Y1 = 5-x
Y2 = sqrt(13-x^2)
Y3 = -sqrt(13-x^2)
Then press [graph].
When the graph is displayed, press [2nd][trace] to get into the calculation menu.
Choose #5, intersect.
Select the lines that intersect and choose a 'guess' point that is close to the intersection. The calculator will come back with the answer, x = 3, y = 2
Do the same for the other intersection and the calculator will come back with x =2 , y = 3.
This doesn't always work where you don't have nice, text-book solutions, but when it does work, boy is it nice!
answered Jul 24 at 15:21
Science_1Science_1
172 bronze badges
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$begingroup$
Another method, good for double-checking your answers on a test, is to graph the two equations on a TI-84 or similar calculator, then examine the graph to see where the lines overlap.
On the calculator, under the [y=] button, set
Y1 = 5-x
Y2 = sqrt(13-x^2)
Y3 = -sqrt(13-x^2)
Then press [graph].
When the graph is displayed, press [2nd][trace] to get into the calculation menu.
Choose #5, intersect.
Select the lines that intersect and choose a 'guess' point that is close to the intersection. The calculator will come back with the answer, x = 3, y = 2
Do the same for the other intersection and the calculator will come back with x =2 , y = 3.
This doesn't always work where you don't have nice, text-book solutions, but when it does work, boy is it nice!
answered Jul 24 at 15:21
Science_1Science_1
172 bronze badges
$endgroup$
add a comment |
$begingroup$
Another method, good for double-checking your answers on a test, is to graph the two equations on a TI-84 or similar calculator, then examine the graph to see where the lines overlap.
On the calculator, under the [y=] button, set
Y1 = 5-x
Y2 = sqrt(13-x^2)
Y3 = -sqrt(13-x^2)
Then press [graph].
When the graph is displayed, press [2nd][trace] to get into the calculation menu.
Choose #5, intersect.
Select the lines that intersect and choose a 'guess' point that is close to the intersection. The calculator will come back with the answer, x = 3, y = 2
Do the same for the other intersection and the calculator will come back with x =2 , y = 3.
This doesn't always work where you don't have nice, text-book solutions, but when it does work, boy is it nice!
answered Jul 24 at 15:21
Science_1Science_1
172 bronze badges
$endgroup$
Another method, good for double-checking your answers on a test, is to graph the two equations on a TI-84 or similar calculator, then examine the graph to see where the lines overlap.
On the calculator, under the [y=] button, set
Y1 = 5-x
Y2 = sqrt(13-x^2)
Y3 = -sqrt(13-x^2)
Then press [graph].
When the graph is displayed, press [2nd][trace] to get into the calculation menu.
Choose #5, intersect.
Select the lines that intersect and choose a 'guess' point that is close to the intersection. The calculator will come back with the answer, x = 3, y = 2
Do the same for the other intersection and the calculator will come back with x =2 , y = 3.
This doesn't always work where you don't have nice, text-book solutions, but when it does work, boy is it nice!
answered Jul 24 at 15:21
Science_1Science_1
172 bronze badges
answered Jul 24 at 15:21
Science_1Science_1
172 bronze badges
answered Jul 24 at 15:21
Science_1Science_1
172 bronze badges
answered Jul 24 at 15:21
Science_1Science_1
172 bronze badges
172 bronze badges
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$begingroup$
You can use polynomial division to eliminate a variable.
$$(x^2+y^2-13) - x(x+y-5) = y^2 -xy+5x-13$$
$$(y^2 -xy+5x-13) - (-y+5)(x+y-5) = 2y^2-10y+12$$
Solve the equation 2$y^2-10y+12=0$, then plug the values of $y$ into the linear equation.
Generalization: Given a system of polynomial equations in $2$ variables, if one of the equations has one of the variables occurring only as a linear term, then you can eliminate that variable by polynomial division to get a polynomial equation in the remaining unknown. The utility of this is somewhat suspect due to the unsolvability of many univariate polynomials.
Bigger generalizations:
Groebner bases https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis
Elimination theory https://en.wikipedia.org/wiki/Elimination_theory
If you don't need exact answers, but only decimal approximations up to a specific precision, skip all this and look up Newton's method.
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
1
$endgroup$
add a comment |
$begingroup$
You can use polynomial division to eliminate a variable.
$$(x^2+y^2-13) - x(x+y-5) = y^2 -xy+5x-13$$
$$(y^2 -xy+5x-13) - (-y+5)(x+y-5) = 2y^2-10y+12$$
Solve the equation 2$y^2-10y+12=0$, then plug the values of $y$ into the linear equation.
Generalization: Given a system of polynomial equations in $2$ variables, if one of the equations has one of the variables occurring only as a linear term, then you can eliminate that variable by polynomial division to get a polynomial equation in the remaining unknown. The utility of this is somewhat suspect due to the unsolvability of many univariate polynomials.
Bigger generalizations:
Groebner bases https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis
Elimination theory https://en.wikipedia.org/wiki/Elimination_theory
If you don't need exact answers, but only decimal approximations up to a specific precision, skip all this and look up Newton's method.
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
1
$endgroup$
add a comment |
$begingroup$
You can use polynomial division to eliminate a variable.
$$(x^2+y^2-13) - x(x+y-5) = y^2 -xy+5x-13$$
$$(y^2 -xy+5x-13) - (-y+5)(x+y-5) = 2y^2-10y+12$$
Solve the equation 2$y^2-10y+12=0$, then plug the values of $y$ into the linear equation.
Generalization: Given a system of polynomial equations in $2$ variables, if one of the equations has one of the variables occurring only as a linear term, then you can eliminate that variable by polynomial division to get a polynomial equation in the remaining unknown. The utility of this is somewhat suspect due to the unsolvability of many univariate polynomials.
Bigger generalizations:
Groebner bases https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis
Elimination theory https://en.wikipedia.org/wiki/Elimination_theory
If you don't need exact answers, but only decimal approximations up to a specific precision, skip all this and look up Newton's method.
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
1
$endgroup$
You can use polynomial division to eliminate a variable.
$$(x^2+y^2-13) - x(x+y-5) = y^2 -xy+5x-13$$
$$(y^2 -xy+5x-13) - (-y+5)(x+y-5) = 2y^2-10y+12$$
Solve the equation 2$y^2-10y+12=0$, then plug the values of $y$ into the linear equation.
Generalization: Given a system of polynomial equations in $2$ variables, if one of the equations has one of the variables occurring only as a linear term, then you can eliminate that variable by polynomial division to get a polynomial equation in the remaining unknown. The utility of this is somewhat suspect due to the unsolvability of many univariate polynomials.
Bigger generalizations:
Groebner bases https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis
Elimination theory https://en.wikipedia.org/wiki/Elimination_theory
If you don't need exact answers, but only decimal approximations up to a specific precision, skip all this and look up Newton's method.
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
1
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
1
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
1
answered Jul 25 at 1:49
QTXe3jBAn7tXnScAYAVat3qKQTXe3jBAn7tXnScAYAVat3qK
1
1
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$begingroup$
The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 and (-6 + -4) = -10.
2x^2 + (-10x) + 12 = 0
but, as we found, -10x = (-6x + -4x) so by substitution..
2x^2 + (-6x + -4x) + 12 = 0
Regrouping we get..
[2x^2 - 6x] + [-4x + 12] = 0
Factoring like terms out we get..
2x(x-3) + -4(x-3) = 0
Factoring (x-3) out we get..
(x-3)(2x-4) = 0
The zeros occur where x-3 = 0 and where 2x-4 = 0
x-3 = 0, x = 3
2x - 4 = 0, 2x = 4, x = 4/2, x = 2
Given x+y = 5
when x = 3, 3+y=5, y = 5-3, y = 2
when x = 2, 2+y=5, y = 5-2, y = 3
Answers:
x = 3, y = 2 and x = 2, y = 3
answered Jul 24 at 14:56
Science_1Science_1
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$begingroup$
The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 and (-6 + -4) = -10.
2x^2 + (-10x) + 12 = 0
but, as we found, -10x = (-6x + -4x) so by substitution..
2x^2 + (-6x + -4x) + 12 = 0
Regrouping we get..
[2x^2 - 6x] + [-4x + 12] = 0
Factoring like terms out we get..
2x(x-3) + -4(x-3) = 0
Factoring (x-3) out we get..
(x-3)(2x-4) = 0
The zeros occur where x-3 = 0 and where 2x-4 = 0
x-3 = 0, x = 3
2x - 4 = 0, 2x = 4, x = 4/2, x = 2
Given x+y = 5
when x = 3, 3+y=5, y = 5-3, y = 2
when x = 2, 2+y=5, y = 5-2, y = 3
Answers:
x = 3, y = 2 and x = 2, y = 3
answered Jul 24 at 14:56
Science_1Science_1
172 bronze badges
$endgroup$
add a comment |
$begingroup$
The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 and (-6 + -4) = -10.
2x^2 + (-10x) + 12 = 0
but, as we found, -10x = (-6x + -4x) so by substitution..
2x^2 + (-6x + -4x) + 12 = 0
Regrouping we get..
[2x^2 - 6x] + [-4x + 12] = 0
Factoring like terms out we get..
2x(x-3) + -4(x-3) = 0
Factoring (x-3) out we get..
(x-3)(2x-4) = 0
The zeros occur where x-3 = 0 and where 2x-4 = 0
x-3 = 0, x = 3
2x - 4 = 0, 2x = 4, x = 4/2, x = 2
Given x+y = 5
when x = 3, 3+y=5, y = 5-3, y = 2
when x = 2, 2+y=5, y = 5-2, y = 3
Answers:
x = 3, y = 2 and x = 2, y = 3
answered Jul 24 at 14:56
Science_1Science_1
172 bronze badges
$endgroup$
The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 and (-6 + -4) = -10.
2x^2 + (-10x) + 12 = 0
but, as we found, -10x = (-6x + -4x) so by substitution..
2x^2 + (-6x + -4x) + 12 = 0
Regrouping we get..
[2x^2 - 6x] + [-4x + 12] = 0
Factoring like terms out we get..
2x(x-3) + -4(x-3) = 0
Factoring (x-3) out we get..
(x-3)(2x-4) = 0
The zeros occur where x-3 = 0 and where 2x-4 = 0
x-3 = 0, x = 3
2x - 4 = 0, 2x = 4, x = 4/2, x = 2
Given x+y = 5
when x = 3, 3+y=5, y = 5-3, y = 2
when x = 2, 2+y=5, y = 5-2, y = 3
Answers:
x = 3, y = 2 and x = 2, y = 3
answered Jul 24 at 14:56
Science_1Science_1
172 bronze badges
edited Jul 24 at 15:08
answered Jul 24 at 14:56
Science_1Science_1
172 bronze badges
answered Jul 24 at 14:56
Science_1Science_1
172 bronze badges
answered Jul 24 at 14:56
Science_1Science_1
172 bronze badges
172 bronze badges
add a comment |
add a comment |
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