# Solve the following equations and also check your result in each case: (45 - 2x)/15 -(4x + 10)/5 = (15 - 14x)/9

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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