Question

Three times the sum of twice a number and four is eighteen. Find the number.

Answer 1

Let the number be $y$.

$\underset{–}{Step1:}$

Twice of the number, i.e., $2y$

$\underset{–}{Step2:}$

Sum of twice the number and four, i.e., $2y+4$

$\underset{–}{Step3:}$

Three times the sum of twice the number $y$ and $4$, i.e., $3\times (2y+4)$
$=3(2y+4)$

$\underset{–}{Step4:}$

Three times the sum of twice of number $y$ and four is eighteen, i.e., $3(2y+4)=18$

$\Rightarrow 3(2y+4)=18$
Dividing both sides by $3$,

$\Rightarrow \frac{3(2y+4)}{3}=\frac{18}{3}$

$\Rightarrow 2y+4=6$
Subtracting $4$ from both sides of the equation,

$\Rightarrow 2y+4-4=6-4$

$\Rightarrow 2y=2$
Dividing both sides by $2$,

$\Rightarrow \frac{2y}{2}=\frac{2}{2}$

$\Rightarrow y=1$

Therefore, the number is $1$.