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- Feb 14, 2012

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Solve for real solutions for the following system of equations:

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

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

- Thread starter anemone
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- Feb 14, 2012

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Solve for real solutions for the following system of equations:

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

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

- Mar 30, 2016

- 92

I can find only \( x = -1, y = \pm 4 \) if I did algebra right.

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- Mar 30, 2016

- 92

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

The 2nd equation gives:

\[ y^2 + (-8x-8)y + (17x+x^2) = 0. \]

Discriminant of this equation doesn't give any restrictions for

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

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 $x=-1, y = \pm4$.

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- Feb 14, 2012

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

- Mar 30, 2016

- 92

I'd rather say, it's the result of trial and error: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.

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.

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- Feb 14, 2012

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Thanks
Theia
for your clarification! And thanks for your solution!

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

\[ -1-3y^2=-49\\y^2=16\\y=\pm 4 \]

My solution (I just solved it):

\[ (y-4x)^2-16x^2+x^2=8y-17x\\(y-4x)^2=8y-17x+15x^2-(1) \]

\[ x^2-8xy+y^2=8y-17x\\8xy=x^2+y^2-8y+17x-(2) \]

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

\[ -1-3y^2=-49\\y^2=16\\y=\pm 4 \]

Last edited:

- Mar 31, 2013

- 1,279

Thanks Theia for your clarification! And thanks for your solution!

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 $y=\pm 4$

- Mar 31, 2013

- 1,279

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.

Last edited:

- Mar 31, 2013

- 1,279

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

$x^3+3x^2y = \frac{1}{2}((x+y)^3+(x-y)^3) = - 49$

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

Last edited:

- Mar 30, 2016

- 92

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- Feb 14, 2012

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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?

- Mar 31, 2013

- 1,279

you are right my mistake I apologize. I must have done some calculation mistake.Where is the error here? Could you please point it out? A screenshot from my CAS below:

View attachment 9794

- Mar 31, 2013

- 1,279

Hello Anemone

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?

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