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Question Number 118896 by bobhans last updated on 20/Oct/20

Answered by benjo_mathlover last updated on 20/Oct/20

$$\to \left(\begin{array}{c}2b\\ b2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}{ac}^2+c\\ c-1\end{array}\right)$$$$\to \left(\begin{array}{c}x\\ y\end{array}\right)=\frac{1}{4-b^2}\left(\begin{array}{c}2-b\\ -b2\end{array}\right)\left(\begin{array}{c}{ac}^2+c\\ c-1\end{array}\right)$$$$\to \left(\begin{array}{c}x\\ y\end{array}\right)=\frac{1}{4-b^2}\left(\begin{array}{c}2{ac}^2+2c-bc+b\\ -{abc}^2-bc+2c-2\end{array}\right)$$$$$$

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