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I tried solving this system of equations multiple ways: $3x+2y=8, \, x^2-2y^2+xy=4$

I tried substituting $x$, substituting $y$, equating the two by making the second equation equal to 8, but the closest I got was by breaking down the $-2y^2$ to $-y^2-y^2$:

From first equation, $y=4-\frac{3x}{2}$. $$x^2-y^2-y^2+xy=4$$ $$(x+y)(x-y)-y(x-y)=4$$ $$(x-y)(x+y-y)=4$$ Then substitute y for the value found in first equation. This gave me $x=4$ or $x=16/5$.

The actual answer is $x=2, y=1$ or $x=18/5, y=-7/5$

How do I solve it step by step?

1__
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Camembert123
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    Hello!Your question seems interesting but I think that me,and everyone else,would apriciate it,much more,if it would have been written in latex.If you don't know how to do it,you can find tutorials on the internet.If you still don't succed you can tell me and I will be happy to help you! – alien2003 Feb 28 '21 at 19:43
  • This is a nonlinear system. Don't expect it to have a closed form solution as it possibly (likely) doesn't have one. – K.defaoite Mar 04 '21 at 02:17

2 Answers2

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Since $y=4-\frac{3x}{2}$, you can express the second equation in terms of $x$ only by substituting $y$ and then solving the resulting quadratic. So we have

$$x^2-2y^2+xy=x^2-2(4-\frac{3x}{2})^2+x(4-\frac{3x}{2})$$ $$=x^2-2(16-12x+\frac{9x^2}{4})+4x-\frac{3x^2}{2}=x^2-\frac{9x^2}{2}-\frac{3x^2}{2}+24x+4x-32$$ $$=-5x^2+28x-32=4$$

and rearranging gives

$$5x^2-28x+36=(5x-18)(x-2)=0$$

etc.

Alessio K
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hint

The second equation can be put in the form

$$(x+\frac y2)^2-\frac{y^2}{4}-2y^2=4$$

or

$$(2x+y)^2-9y^2=16$$

the first will be

$$6x+4y=16=3(2x+y)+y$$ If we put $ X=2x+y $, the system becomes $$X^2-9y^2=16$$ $$3X+y=16$$ then

$$X^2-9(16-3X)^2=16$$

hamam_Abdallah
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