Sarthaks Test
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Solve each of the following equations and also check your result in each case:

\(\frac{(45-2x)}{15}-\frac{(4x+10)}{5}=\frac{(15-14x)}{9}\)

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\(\frac{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}\)

On transposing \(\frac{15-14x}9\) to LHS, we get

\(\frac{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}\) = 0

On substituting x  = \(\frac{15}{14},\) we get

We got LHS=RHS

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