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

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In summary, the conversation involves solving a system of equations and requesting that Theia post their solution instead of just the final results. Theia provides their solution using trial and error and later corrects their mistake with the help of another participant.
  • #1
anemone
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Solve for real solutions for the following system of equations:

$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$
 
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  • #2
I can find only \( x = -1, y = \pm 4 \) if I did algebra right.
 
  • #3
Hey Theia, please kindly post your solution here instead of just the final results so we can learn from your solution! :cool:
 
  • #4
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 $x=-1, y = \pm4$.
 
  • #5
Thanks for participating, Theia!

Theia said:
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. :)
 
  • #6
anemone said:
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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  • #7
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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  • #8
anemone said:
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 $y=\pm 4$
 
  • #9
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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  • #10
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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  • #11
kaliprasad said:
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
 
  • #12
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?
 
  • #13
Theia said:
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.
 
  • #14
anemone said:
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
 

FAQ: Solve x³+3xy²=-49 and x²-8xy+y²=8y-17x

What are the solutions to the given equations?

The solutions to the given equations are x = -7, y = 1 or x = 1, y = -3.

How do you solve a system of equations with two variables?

To solve a system of equations with two variables, you can use substitution, elimination, or graphing methods. In this case, we can use substitution by solving one equation for one variable and substituting it into the other equation.

Can these equations be solved using the quadratic formula?

No, these equations cannot be solved using the quadratic formula because they are not in the form of ax²+bx+c=0. They are in the form of ax³+bx²+cx+d=0, which requires a different method of solving.

How many solutions can a system of two equations have?

A system of two equations can have one solution, no solution, or infinitely many solutions. In this case, there are two distinct solutions.

What is the importance of solving systems of equations in real-life situations?

Solving systems of equations is important in real-life situations because it allows us to find the values of multiple variables that satisfy a set of equations. This can be useful in various fields such as engineering, economics, and physics.

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