Question

Solve the following equation:

$28=4+3(t+5)$

Answer 1

Step:Solving the equation.

Given :- $28=4+3(t+5)$

⇒ $28=4+3t+15$ (Expanding the bracket on R.H.S. )

$28=19+3t$

⇒ $-3t=19-28$ (Transporting $28$ to R.H.S. and $3t$ on L.H.S.)

⇒ $-3t=-9$

⇒ $\frac{-3t}{-3}=\frac{-9}{-3}$ (Dividing both side of equation by $-5$)

⇒ $t=3$

Therefore, $t=3$ is the required solution