Question

Solve the equation. Check your solution.

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

Answer 1

Step 1 : Solve the equation

We have, $-\frac{1}{3}\left(4+z\right)=-\frac{5}{6}$

$$\begin{gathered} \Rightarrow 4+\mathrm{z}=\left(\frac{\cancel{-}5}{{}_2\cancel{6}}\right)\times \cancel{-}\cancel{3}[\mathrm{Multiply}\mathrm{both}\mathrm{sides}\mathrm{by}-3] \\ \Rightarrow 4+\mathrm{z}=\frac{5}{2} \\ \Rightarrow \mathrm{z}=-4+\frac{5}{2}[\mathrm{Subtract}4\mathrm{from}\mathrm{both}\mathrm{sides}] \\ \therefore \mathrm{z}=-\frac{3}{2}\left[\mathrm{Simplify}\right] \end{gathered}$$

Step 2 : Check the solution

Put the value of $z$ in the given equation

$$\begin{gathered} \Rightarrow -\frac{1}{3}\left(4-\frac{3}{2}\right)=-\frac{5}{6} \\ \Rightarrow -\frac{1}{3}\left(\frac{8-3}{2}\right)=-\frac{5}{6} \\ \Rightarrow -\frac{1}{3}\times \frac{5}{2}=-\frac{5}{6} \\ \Rightarrow \frac{-5}{6}=\frac{-5}{6} \\ \therefore \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

Hence, the value of $z$ is $-\frac{3}{2}$.