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anemone
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Solve for real solutions for the following system of equations:
$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$
$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$
Theia said:Factoring the 6th degree equation is this time easy: x=−1x=-1 is a solution 4 times:
I'd rather say, it's the result of trial and error: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. :)
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
\[ -1-3y^2=-49\\y^2=16\\y=\pm 1 \]
Where is the error here? Could you please point it out? A screenshot from my CAS below:kaliprasad said:Both anemone and theia forgot to check that y = -4 is correct and y = 4 is erroneous
Theia said:Where is the error here? Could you please point it out? A screenshot from my CAS below:
View attachment 9794
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?
The solutions to the given equations are x = -7, y = 1 or x = 1, y = -3.
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.
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.
A system of two equations can have one solution, no solution, or infinitely many solutions. In this case, there are two distinct solutions.
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.