Question

Solve this linear equation in two variables


$$\begin{gathered} 6x+8y=\frac{1}{10} \\ 26x+48y=\frac{1}{2} \end{gathered}$$

Answer 1

Dear Student,
$$\begin{gathered} 6x+8y=\frac{1}{10} \\ 3x+4y=\frac{1}{20}.............\left(i\right) \\ 26x+48y=\frac{1}{2} \\ 13x+24y=\frac{1}{4}.................\left(ii\right) \\ multilyeq\left(i\right)by13andeq\left(ii\right)by3,andsubtract\left(i\right)from\left(ii\right) \\ weget, \\ 39x+72y-\left(39x+52y\right)=\left(\frac{3}{4}\right)-\left(\frac{13}{20}\right) \\ 39x+72y-39x-52y=\left(\frac{3}{4}\right)-\left(\frac{13}{20}\right) \\ 20y=\frac{15-13}{20} \\ 20y=\frac{2}{20} \\ y=\frac{1}{200} \\ Nowputtingthevalueofyinequation\left(i\right)weget \\ 3x+4\times \frac{1}{200}=\frac{1}{20} \\ 3x+\frac{1}{50}=\frac{1}{20} \\ 3x=\frac{1}{20}-\frac{1}{50} \\ 3x=\frac{5-2}{100} \\ x=\frac{1}{100} \\ hence\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{100}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{200}} \end{gathered}$$
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