Question

Evaluate:
$\frac{(z+5)}{6}-\frac{(z+1)}{9}=\frac{(z+3)}{4}$

Answer 1

We have,

$\frac{(z+5)}{6}-\frac{(z+1)}{9}=\frac{(z+3)}{4}$

$\frac{3(z+5)-2(z+1)}{18}=\frac{(z+3)}{4}$

$\frac{3z+15-2z-2}{18}=\frac{(z+3)}{4}$

$\frac{z+13}{18}=\frac{z+3}{4}$

$\frac{z+13}{9}=\frac{z+3}{2}$

$2z+26=9z+27$

$2z-9z=27-26$

$-7z=1$

$z=-\frac{1}{7}$

Hence, this is the answer.