Question Number 212160 by boblosh last updated on 04/Oct/24
Answered by A5T last updated on 04/Oct/24
![(x/2)+4y=4⇒x+8y=8...(i) (x/4)−((2y)/3)=(2/3)⇒3x−8y=8...(ii) [(1,8),(3,(−8)) ] [(x),(y) ]= [(8),(8) ] By Cramer′s rule: x=( determinant ((8,8),(8,(−8)))/ determinant ((1,8),(3,(−8))))=((−128)/(−32))=4 and y=( determinant ((1,8),(3,8))/ determinant ((1,8),(3,(−8))))=((−16)/(−32))=(1/2)](https://www.tinkutara.com/question/Q212172.png)
$$\frac{{x}}{\mathrm{2}}+\mathrm{4}{y}=\mathrm{4}\Rightarrow{x}+\mathrm{8}{y}=\mathrm{8}…\left({i}\right) \\ $$$$\frac{{x}}{\mathrm{4}}−\frac{\mathrm{2}{y}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\mathrm{3}{x}−\mathrm{8}{y}=\mathrm{8}…\left({ii}\right) \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}\\{\mathrm{8}}\end{bmatrix} \\ $$$${By}\:{Cramer}'{s}\:{rule}:\:{x}=\frac{\begin{vmatrix}{\mathrm{8}}&{\mathrm{8}}\\{\mathrm{8}}&{−\mathrm{8}}\end{vmatrix}}{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{vmatrix}}=\frac{−\mathrm{128}}{−\mathrm{32}}=\mathrm{4} \\ $$$${and}\:{y}=\frac{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{\mathrm{8}}\end{vmatrix}}{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{vmatrix}}=\frac{−\mathrm{16}}{−\mathrm{32}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$