Question

If $\frac{x+4}{2}+\frac{x-3}{3}=\frac{x+2}{4}+\frac{x+3}{5}$, find the value of $x$.

Answer 1

$\frac{3(x+4)+2(x-3)}{6}=\frac{5(x+2)+(x+3)}{20}$

$[3x+12+2x-6]\times 20=[5x+10+4x+12]\times 6$

$(5x+6)20=(9x+22)\times 6$

$100x+120=54x+132$

$46x=12\Rightarrow x=\frac{6}{23}$