Question

Solve for $x$:

$\frac{x-4}{7}-\frac{x+4}{5}=\frac{x+3}{7}$

Answer 1

Given, $\frac{x-4}{7}-\frac{x+4}{5}=\frac{x+3}{7}$

$\Rightarrow$ $\frac{5(x-4)-7(x+4)}{35}=\frac{x+3}{7}$ ....[Cross-multiplying the denominators on the LHS]

$\Rightarrow$ $\frac{5x-20-7x-28}{35}=\frac{x+3}{7}$ ...[By Distribution Law]

$\Rightarrow$ $\frac{-2x-48}{35}=\frac{x+3}{7}$ ...[On simplifying]

$\Rightarrow$ $-2x-48=\frac{x+3}{7}\times 35$

$\Rightarrow$ $-2x-48=(x+3)\times 5$

$\Rightarrow$ $-2x-48=5x+15$ ...[By Distribution Law]

$\Rightarrow$ $-2x-5x=15+48$ ...[Transposing $x$ terms to one side]

$\Rightarrow$ $-7x=63$ ...[On simplifying]

$\Rightarrow$ $x=\frac{63}{-7}=-9$.