# How to solve the following system of equations.

$2x^2-3xy+2y^2=2\frac{3}{4}\\x^2-4xy+y^2+\frac{1}{2}=0$

I tried all the methods that I know, but I could't isolate $x$ or $y$ to form one equation.

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Have you tried rewriting the equations in terms of $u = x^2 + y^2$ and $v = xy$? – hardmath Feb 19 '14 at 11:03

By multiplying the second by $-2$ you get: $$2x^2-3xy+2y^2=\frac{6}{4}$$ $$-2x^2+8xy-2y^2=1$$ Then adding them you get: $$5xy=\frac{10}{4} \Rightarrow xy=\frac{1}{2} \Rightarrow x=\frac{1}{2y}$$

Substituting this in the second equation you get: $$\frac{1}{4y^2}-2+y^2+\frac{1}{2}=0 \Rightarrow \frac{1}{4y^2}+y^2=\frac{3}{2}$$

Multiplying this equation by $4y^2$ you get: $$1+4y^4=6y^2 \Rightarrow 4y^4-6y^2+1=0$$

Let $u=y^2$:$$4u^2-6u+1=0$$ Then continue by solving this quadratic polynomial...

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I believe the $2 \frac{3}{4}$ in the first equation for $x,y$ should be interpreted as a mixed fraction, not a product, so it equals $\frac{11}{4}$, not $\frac{6}{4}$ as you've assumed. – hardmath Feb 19 '14 at 11:18
Thank you. @Mary Please look at my answer and tell me if it's correct wrong. – David Sebastian Feb 19 '14 at 12:01

Hint

Multiply the second by $2$, substract the result from the first. So, $x^2$ is gone and you express $x$ as a function of $y$. Plug it into one of the equations to get one equation in $y$.

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If you can look at my answer, it'd be a big help. Thanks. – David Sebastian Feb 19 '14 at 12:02

As has been said in other answers, first solve for $xy$

$$2x^2-3xy+2y^2=\frac{11}4$$ $$-2x^2+8xy-2y^2=1$$ $$5xy=\frac{15}4$$ $$xy=\frac34$$

Now take the original second equation and manipulate it.

$$x^2-4xy+y^2+2xy=-\frac12+2(\frac34)$$ $$(x-y)^2=1$$ $$x-y=\pm1$$ $$x^2-4xy+y^2+6xy=-\frac12+6(\frac34)$$ $$(x+y)^2=4$$ $$x+y=\pm2$$

So there are 4 pairs of linear equations for 4 different solutions.

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Multiply second equation by $2$, $$2x^2-8xy+2y^2+1=0$$ Substract the first equation by the second, $$2x^2-3xy+2y^2-\frac{11}{4}-(2x^2-8xy+2y^2+1)=0\\2x^2-3xy-2y^2-\frac{11}{4}-2x^2+8xy-2y^2-1=0\\5xy=\frac{15}{4}\\x=\frac{3}{4y}$$ Substitute $x=\frac{3}{4y}$ in the first equation, $$2x^2-3xy+2y^2=\frac{11}{4}\\2(\frac{3}{4y})^2-3(\frac{3}{4y})y+2y^2-\frac{11}{4}=0\\\frac{9}{8y^2}+2y^2-5=0\\16y^4-40y^2+9=0$$Now use the quadratic Formula to solve $16y^4-40y^2+9=0$,$$y^2=\frac{-(-40)\pm\sqrt{(-40)^2-4.16.9}}{2.16}\\y^2=\frac{40\pm\sqrt{1600-4.16.9}}{2.16}\\y^2=\frac{40\pm\sqrt{16(100-4.9)}}{2.16}\\y^2=\frac{40\pm4\sqrt{64}}{2.16}=\frac{40\pm4.8}{2.16}=\frac{5\pm4}{4}\\y^2=\frac{5+4}{4}=\frac{9}{4}\Rightarrow y=\pm\frac{3}{2}\\or,~~y^2=\frac{5-4}{4}=\frac{1}{4}\Rightarrow y=\pm\frac{1}{2}$$

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$$2(\frac{3}{4y})^2-3(\frac{3}{4y})y+2y^2=\frac{11}{4}$$ $$2\frac{9}{16y^2}-\frac{9}{4}+2y^2=\frac{11}{4}$$ $$\frac{9}{8y^2}+2y^2=\frac{11}{4}+\frac{9}{4}$$ $$\frac{9}{8y^2}+2y^2=\frac{20}{4}$$ Then by multiplying it by $8y^2$ you get: $$9+16y^4=40y^2 \Rightarrow 16y^4-40y^2+9=0$$ Now you can continue as you did before.. – Mary Star Feb 19 '14 at 12:23
Thanks a lot Mary. I will correct the solution. :) – David Sebastian Feb 19 '14 at 12:29
At the quadratic formula it is like that: $$y^2=\frac{40 \pm \sqrt{1600-4 \cdot 16 \cdot 9}}{2 \cdot 16}$$ $$y^2=\frac{40 \pm \sqrt{16(100-4 \cdot 9)}}{2 \cdot 4 \cdot 4}$$ $$y^2=\frac{40 \pm 4 \sqrt{100-36}}{2 \cdot 4 \cdot 4}$$ $$y^2=\frac{10 \pm \sqrt{64}}{8}$$ $$y^2=\frac{10 \pm 8}{8}$$ $$y^2=\frac{18}{8}=\frac{9}{4} \Rightarrow y= \pm \frac{3}{2}$$ $$\text{Or } y^2=\frac{2}{8}=\frac{1}{4} \Rightarrow y= \pm \frac{1}{2}$$ – Mary Star Feb 19 '14 at 14:05
Thank you Mary. Seems like i'm making silly mistakes. – David Sebastian Feb 19 '14 at 19:03