Question

# $\frac{\left(2-7x\right)}{\left(1-5x\right)}=\frac{\left(3+7x\right)}{\left(4+5x\right)}$. Solve for$x$.

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Solution

## Find the value of$x$.$\frac{\left(2-7x\right)}{\left(1-5x\right)}=\frac{\left(3+7x\right)}{\left(4+5x\right)}$By applying alternendo,$\frac{\left(2-7x\right)}{\left(3+7x\right)}=\frac{\left(1-5x\right)}{\left(4+5x\right)}$Now applying Componendo and Dividendo,$\frac{\left(2-7x\right)+\left(3+7x\right)}{\left(2-7x\right)-\left(3+7x\right)}=\frac{\left(1-5x\right)+\left(4+5x\right)}{\left(1-5x\right)-\left(4+5x\right)}\phantom{\rule{0ex}{0ex}}\frac{2-7x+3+7x}{2-7x-3-7x}=\frac{1-5x+4+5x}{1-5x-4-5x}\phantom{\rule{0ex}{0ex}}\frac{5}{-1-14x}=\frac{5}{-3-10x}\phantom{\rule{0ex}{0ex}}5\left(-3-10x\right)=5\left(-1-14x\right)\phantom{\rule{0ex}{0ex}}-15-50x=-5-70x\phantom{\rule{0ex}{0ex}}70x-50x=-5+15\phantom{\rule{0ex}{0ex}}20x=10\phantom{\rule{0ex}{0ex}}x=\frac{10}{20}\phantom{\rule{0ex}{0ex}}x=\frac{1}{2}$Hence, the value of $x$ is$\frac{1}{2}$.

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