Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 212160 by boblosh last updated on 04/Oct/24

Answered by A5T last updated on 04/Oct/24

$$\frac{x}{2}+4y=4\Rightarrow x+8y=8...\left(i\right)$$$$\frac{x}{4}-\frac{2y}{3}=\frac{2}{3}\Rightarrow 3x-8y=8...\left(ii\right)$$$$\left[\begin{array}{cc}1& 8\\ 3& -8\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ 8\end{array}\right]$$$$By{Cramer}^{\prime }srule:x=\frac{\left|\begin{array}{cc}8& 8\\ 8& -8\end{array}\right|}{\left|\begin{array}{cc}1& 8\\ 3& -8\end{array}\right|}=\frac{-128}{-32}=4$$$$andy=\frac{\left|\begin{array}{cc}1& 8\\ 3& 8\end{array}\right|}{\left|\begin{array}{cc}1& 8\\ 3& -8\end{array}\right|}=\frac{-16}{-32}=\frac{1}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com