Author Topic: solving help (Read 553 times) Tweet Share

johnhouse · « **on:** November 08, 2012, 07:27:24 pm »

solving using sim equations

3x+2y=4

x^2+xy-y^2=1

b^3 · « **Reply #1 on:** November 08, 2012, 07:36:12 pm »

Rearrange the first equation for $y$
$\begin{alignedat}{1}3x+2y & =4 \\ 2y & =4-3x \\ y & =\frac{4-3x}{2} \end{alignedat}$
Substitute into the second equation.
$\begin{alignedat}{1}x^{2}+xy-y^{2} & =1 \\ x^{2}+x\left(\frac{4-3x}{2}\right)-\left(\frac{4-3x}{2}\right)^{2} & =1 \\ x^{2}+\frac{4x}{2}-\frac{3x^{2}}{2}-\frac{16+2(4)(-3x)+9x^{2}}{4} & =1 \\ 4x^{2}+8x-6x^{2}-16+24x-9x^{2} & =4 \\ 11x^{2}-32x+20 & =0 \\ \left(11x-10\right)\left(x-2\right)&=0 \\11x-10=0&\mathrm{\:or\: }x-2=0 \\x=\frac{10}{11}&\mathrm{\:or\: }x=2 \end{alignedat}$
Now find the corresponding values of $y$
If $x=\frac{10}{11}$
$\begin{alignedat}{1}y & =\frac{4-\frac{10}{11}}{2} \\ & =\frac{\frac{44-10}{11}}{2} \\ & =\frac{34}{22} \\ & =\frac{17}{11} \end{alignedat}$
If $x=2$
$\begin{alignedat}{1}y & =\frac{4-2}{2} \\ & =1 \end{alignedat}$

So the solutions are $x=\frac{10}{11}$ and $\frac{17}{11}$ , or $x=2$ and $y=1$

johnhouse · « **Reply #2 on:** November 08, 2012, 07:55:14 pm »

thanks so much !

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Author Topic: solving help (Read 553 times) Tweet Share

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solving help

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