# [SOLVED]Solve x³+3xy²=-49 and x²-8xy+y²=8y-17x

#### anemone

##### MHB POTW Director
Staff member
Solve for real solutions for the following system of equations:

$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$

#### Theia

##### Well-known member
I can find only $$x = -1, y = \pm 4$$ if I did algebra right.

Staff member

#### Theia

##### Well-known member
Given

$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$.

The 2nd equation gives:

$y^2 + (-8x-8)y + (17x+x^2) = 0.$

Discriminant of this equation doesn't give any restrictions for x. Thus one can write ($e = \pm 1$):

$y = e\sqrt{15x^2+15x+16}+4x+4.$

The first equation gives:

$y^2 = -\frac{x^3+49}{3x},$

and restricts $-\sqrt[3]{49} \le x < 0.$

After taking the 2nd power of the first expression for y and equating the $y^2$ expressions, one obtains

$e(8x + 8) \sqrt{15x^2 + 15x + 16} + 31x^2 + 47x + 32 = -\frac{x^3 + 49}{3x},$

where used the fact that $e^2=1$. Taking another 2nd power and putting all to left hand side, one obtains

$196x^6 + 588x^5 + 2793x^4 + 9212x^3 + 13818x^2 + 9408x + 2401 =0.$

Factoring the 6th degree equation is this time easy: $x=-1$ is a solution 4 times:

$49(x+1)^4 (4x^2-4x+49) = 0.$

Quadratic part gives only complex solution, so $x=-1$ is the only valid solution. Substituting it into e.g. 1st equation gives $y = \pm 4$.

Hence the only real solution is $x=-1, y = \pm4$.

#### anemone

##### MHB POTW Director
Staff member
Thanks for participating, Theia !

Factoring the 6th degree equation is this time easy: x=−1x=-1 is a solution 4 times:
I think the bigger picture that we need to address is why do we see and realize that $x=-1$ is a root of multiplicity 4.

#### Theia

##### Well-known member
I think the bigger picture that we need to address is why do we see and realize that $x=-1$ is a root of multiplicity 4.
I'd rather say, it's the result of trial and error:
plot -> see $x=-1$ is a root candidate -> substitute $x=-1$ into the equation -> see that it is a solution -> divide it out -> plot -> etc. This is how I did it and ended up to multiplicity 4.

#### anemone

##### MHB POTW Director
Staff member

My solution (I just solved it):
From $x^2-8xy+y^2=8y-17x$, we do some algebraic manipulation to get
$(y-4x)^2-16x^2+x^2=8y-17x\\(y-4x)^2=8y-17x+15x^2-(1)$
$x^2-8xy+y^2=8y-17x\\8xy=x^2+y^2-8y+17x-(2)$

From $x^3+3xy^2=-49$, in trying to relate the $y$ term to $y-4x$, we get
$x^3+3xy^2=-49\\x^3+3x(y-4x+4x)^2=-49\\x^3+3x[(y-4x)^2+8x(y-4x)+16x^2]=-49\\x^3+3x(y-4x)^2+24x^2(y-4x)+48x^2=-49\\49(x^3+1)+3x[(y-4x)^2+8x(y-4x)]=0\\49(x^3+1)+3x(8y-17x+15x^2+8xy-32x^2)=0\text{ from (1)}\\49(x+1)(x^2-x+1)+3x[8y(x+1)-17x(x+1)]=0\\(x+1)[49(x^2-x+1)+3x(8y-17x)]=0\\(x+1)[49x^2-49x+49+24xy-51x^2]=0\\(x+1)[-2x^2-49x+49+3(8xy)]=0\\(x+1)[-2x^2-49x+49+3x^2+3y^2-24y+51x]=0\text{ from (2)}\\(x+1)(x^2+2x+1+3y^2-24y+48)=0\\(x+1)[(x+1)^2+3(y-4)^2]=0$

This is possible if and only if $x=-1$.

When $x=-1$, we get
$-1-3y^2=-49\\y^2=16\\y=\pm 4$

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##### Well-known member

My solution (I just solved it):
From $x^2-8xy+y^2=8y-17x$, we do some algebraic manipulation to get
$(y-4x)^2-16x^2+x^2=8y-17x\\(y-4x)^2=8y-17x+15x^2-(1)$
$x^2-8xy+y^2=8y-17x\\8xy=x^2+y^2-8y+17x-(2)$

From $x^3+3xy^2=-49$, in trying to relate the $y$ term to $y-4x$, we get
$x^3+3xy^2=-49\\x^3+3x(y-4x+4x)^2=-49\\x^3+3x[(y-4x)^2+8x(y-4x)+16x^2]=-49\\x^3+3x(y-4x)^2+24x^2(y-4x)+48x^2=-49\\49(x^3+1)+3x[(y-4x)^2+8x(y-4x)]=0\\49(x^3+1)+3x(8y-17x+15x^2+8xy-32x^2)=0\text{ from (1)}\\49(x+1)(x^2-x+1)+3x[8y(x+1)-17x(x+1)]=0\\(x+1)[49(x^2-x+1)+3x(8y-17x)]=0\\(x+1)[49x^2-49x+49+24xy-51x^2]=0\\(x+1)[-2x^2-49x+49+3(8xy)]=0\\(x+1)[-2x^2-49x+49+3x^2+3y^2-24y+51x]=0\text{ from (2)}\\(x+1)(x^2+2x+1+3y^2-24y+48)=0\\(x+1)[(x+1)^2+3(y-4)^2]=0$

This is possible if and only if $x=-1$.

When $x=-1$, we get
$-1-3y^2=-49\\y^2=16\\y=\pm 1$
last line should be $y=\pm 4$

##### Well-known member
Both anemone and theia forgot to check that y = -4 is correct and y = 4 is erroneous

My mistake above statement is not true, must have made a calculation mistake.

I apologize.

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##### Well-known member
My solution trial and error

1st equation
$x^3+3x^2y = \frac{1}{2}((x+y)^3+(x-y)^3) = - 49$
or $(x+y)^3 + (x-y)^3 = - 98 = 27 - 125$
assuming x and y to be integers (x+y) = 3, (x-y) = -5 or x = -1, y = 4
or (x+y) =-5 , (x-y) 3= or x = -1, y = - 4

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#### Theia

##### Well-known member
Both anemone and theia forgot to check that y = -4 is correct and y = 4 is erroneous
Where is the error here? Could you please point it out? A screenshot from my CAS below:

#### anemone

##### MHB POTW Director
Staff member
Thanks kaliprasad for pointing out my honest mistake. I have edited the values of $y$.

Can you please check with your claim that the solution $(x, y)=(-1, 4)$ is incorrect, because it certainly is a valid answer.

Also, can you please elaborate the reasoning behind your assumption that the solutions to the system are integers, and how did you conclude those are the ONLY solutions?

##### Well-known member
Where is the error here? Could you please point it out? A screenshot from my CAS below:
View attachment 9794
you are right my mistake I apologize. I must have done some calculation mistake.

Thanks kaliprasad for pointing out my honest mistake. I have edited the values of $y$.
Can you please check with your claim that the solution $(x, y)=(-1, 4)$ is incorrect, because it certainly is a valid answer.