Question

Solve the following equations and check your result:

$\frac{2b}{3}-\frac{3b}{8}=\frac{7}{12}$

Answer 1

Step 1: Finding the solution of a given equation

$$\begin{gathered} \Rightarrow \frac{2b}{3}-\frac{3b}{8}=\frac{7}{12}\left[\mathrm{Given}\right] \\ \Rightarrow \frac{2b\times 8-3b\times 3}{3\times 8}=\frac{7}{12} \\ \Rightarrow \frac{16b-9b}{24}=\frac{7}{12} \\ \Rightarrow \frac{16b-9b}{24}\times 24=\frac{7}{12}\times 24[\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}24] \\ \Rightarrow 16b-9b=7\times 2 \\ \Rightarrow 7b=14 \\ \Rightarrow \frac{7b}{7}=\frac{14}{7}[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}7] \\ \therefore b=2 \end{gathered}$$

Step 2: Checking the solution of the equation

we will check our results by putting $\mathrm{b}=2$ in the given equation:

So, L.H.S $=\frac{2b}{3}-\frac{3b}{8}$

$$\begin{gathered} =\frac{2\times 2}{3}-\frac{3\times 2}{8} \\ =\frac{4}{3}-\frac{6}{8} \\ =\frac{4\times 8-6\times 3}{3\times 8} \\ =\frac{32-18}{24} \\ =\frac{14}{24} \\ =\frac{7}{12} \end{gathered}$$

R.H.S$=\frac{7}{12}$

$\therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}$

Hence, the solution of the given equation is $2$.