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

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anemone

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Feb 14, 2012
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Solve for real solutions for the following system of equations:

$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$
 

Theia

Well-known member
Mar 30, 2016
92
I can find only \( x = -1, y = \pm 4 \) if I did algebra right.
 
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anemone

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Feb 14, 2012
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Hey Theia , please kindly post your solution here instead of just the final results so we can learn from your solution! :cool:
 

Theia

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Mar 30, 2016
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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 $x=-1, y = \pm4$.
 
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anemone

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Feb 14, 2012
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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
Mar 30, 2016
92
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. :giggle:
 
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anemone

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

Well-known member
Mar 31, 2013
1,283
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 $y=\pm 4$
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
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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kaliprasad

Well-known member
Mar 31, 2013
1,283
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
 
Last edited:

Theia

Well-known member
Mar 30, 2016
92
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:
Screenshot_20200508-204941_MaximaOnAndroid.jpg
 

anemone

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Feb 14, 2012
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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?
 

kaliprasad

Well-known member
Mar 31, 2013
1,283

kaliprasad

Well-known member
Mar 31, 2013
1,283
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