# Solve the following equations and also check your result in each case: 5x/3 - (x - 1)/4 = (x - 3)/5

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

$\frac{5x}3-\frac{(x-1)}4$ = $\frac{x-3}5$

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$\frac{5x}3-\frac{(x-1)}4$ = $\frac{x-3}5$

On transposing $\frac{x-3}5$ to LHS, we get

On taking LCM of 3 and 4; substituting x = $-\frac{51}{73},$ we get

We got LHS=RHS