Question

anyone know how to do in substitution method ....​

Answer 1

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$\tt \: \left. \begin{cases} { a ^ { 2 } x-b ^ { 2 } y = a ^ { 2 } -2b ^ { 2 } } \\ { b ^ { 2 } x+a ^ { 2 } y=b ^ { 2 } +20 } \end{cases} \right.$

$\huge \bold{ \underline{ \underline{ \red{Solution}}}}$

$\tt \: xa^{2}-yb^{2}=-2b^{2}+a^{2}$

• To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$\tt \: a^{2}x+\left(-b^{2}\right)y=a^{2}-2b^{2}, \\ \tt \: b^{2}x+a^{2}y=b^{2}+20$

• Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.$\tt \: a^{2}x+\left(-b^{2}\right)y=a^{2}-2b^{2}$

$\tt \: a^{2}x=b^{2}y+a^{2}-2b^{2} \\ \\ \tt \: x=\frac{1}{a^{2}}\left(b^{2}y+a^{2}-2b^{2}\right) \\ \\ \tt \: x=\frac{b^{2}}{a^{2}}y-\frac{2b^{2}}{a^{2}}+1$

• Substitute $\tt \: \frac{a^{2}-2b^{2}+b^{2}y}{a^{2}}$for x in the other equation, $\tt \: b^{2}x+a^{2}y=b^{2}+20.$

$\tt \: b^{2}\left(\frac{b^{2}}{a^{2}}y-\frac{2b^{2}}{a^{2}}+1\right)+a^{2}y=b^{2}+20 \\ \\ \tt \: \frac{b^{4}}{a^{2}}y-\frac{2b^{4}}{a^{2}}+b^{2}+a^{2}y=b^{2}+20 \\ \\ \tt \: \left(\frac{b^{4}}{a^{2}}+a^{2}\right)y-\frac{2b^{4}}{a^{2}}+b^{2}=b^{2}+20 \\ \\ \tt \: \left(\frac{b^{4}}{a^{2}}+a^{2}\right)y=\frac{2b^{4}}{a^{2}}+20 \\ \\ \tt \:y=\frac{2\left(10a^{2}+b^{4}\right)}{a^{4}+b^{4}}$

• Substitute $\tt \: \frac{2\left(10a^{2}+b^{4}\right)}{b^{4}+a^{4}} \: for \: y \: in \: x=\frac{b^{2}}{a^{2}}y-\frac{2b^{2}}{a^{2}}+1.$Because the resulting equation contains only one variable, you can solve for x directly.

$\tt \: x=\frac{b^{2}}{a^{2}}\times \left(\frac{2\left(10a^{2}+b^{4}\right)}{a^{4}+b^{4}}\right)-\frac{2b^{2}}{a^{2}}+1 \\ \\ \tt \:x=\frac{2b^{2}\left(10a^{2}+b^{4}\right)}{a^{2}\left(a^{4}+b^{4}\right)}-\frac{2b^{2}}{a^{2}}+1 \\ \\ \tt \: x=\frac{a^{4}-2a^{2}b^{2}+b^{4}+20b^{2}}{a^{4}+b^{4}} \\ \\ \tt \bold{ x=\frac{a^{4}-2a^{2}b^{2}+b^{4}+20b^{2}}{a^{4}+b^{4}},}\\ \\ \tt \: \bold{ y=\frac{2\left(10a^{2}+b^{4}\right)}{a^{4}+b^{4}}}$

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Answer 2

Answer:

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