Question

Solve the following equations.
$\text{(i)}17p-2=49$
$\text{(ii)}2m+7=9$
$\text{(iii)}3x+12=2x-4$

Answer 1

$\text{(i)}17p-2=49$
$17p-2=49\text{Adding 2 to both sides}$
$\Rightarrow 17p-2+2=49+2$
$\Rightarrow 17p=51$
Dividing both sides by 17
$\Rightarrow \frac{17p}{17}=\frac{51}{17}\Rightarrow p=3$
Hence, $p=3$ is the solution of the equation.
$\text{(ii)}2m+7=9$
$2m+7=9$ Subtracting 7 from both sides
$\Rightarrow 2m+7-7=9-7$
$\Rightarrow 2m=2$
Dividing both sides by 2
$\Rightarrow \frac{2m}{2}=\frac{2}{2}$
$\Rightarrow m=1$
$\text{Hence,}m=1\text{is the solution of the equation}$
$\text{(iii)}3x+12=2x-4$
$3x+12=2x-4$
Subtracting 12 from both sides
$\Rightarrow 3x+12-12=2x-4-12$
$\Rightarrow 3x=2x-16$
Subtracting 2x from both sides
$\Rightarrow 3x-2x=2x-16-2x$
$\Rightarrow x=-16$
Hence, x = −16 is the solution of the equation.