# [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:

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

• • topsquark, Theia, MarkFL and 1 other person

#### Theia

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

• MarkFL, anemone and Ackbach

#### anemone

##### MHB POTW Director
Staff member
Hey Theia , please kindly post your solution here instead of just the final results so we can learn from your solution! • MarkFL and Ackbach

#### 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 $-\sqrt{49} \le x < 0.$

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, Ackbach and MarkFL

#### 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 $y=\pm 4$

• • anemone and Jameson

##### 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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• anemone

##### 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 and mathmari

#### Theia

##### Well-known member
• anemone and Ackbach

#### 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?

• MarkFL and Ackbach

##### 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.

##### Well-known 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?
Hello Anemone
I have not mentioned that solution is unique.
I tried a solution and because an integer solution existed it was found
if integer solution did not exist my method would not have found one.,
this was just an illustration of trial and error method

• Ackbach