If ${\text{m}}_{\text{1}}=10\text{kg,}{\text{m}}_{\text{2}}=4\text{kg}\text{and}{\text{m}}_{\text{3}}=2\text{kg}$ the tension in the string connecting ${\text{m}}_1\text{and}{\text{m}}_{\text{2}}$ is,

F.B.D of 10 kg

$100-{\text{T}}_{\text{1}}=10\text{a}$

F.B. D of 2 kg

${\text{T}}_{\text{2}}-20=2\text{a}$

F.B.D of 4 kg

${\text{T}}_{\text{1}}+20-{\text{T}}_{\text{2}}=4\text{a}$

From (2) & (3)$\begin{array}{c}100-{\text{T}}_{\text{1}}=10\text{a}\to \left(1\right)\\ {\text{T}}_{\text{1}}+20-{\text{T}}_{\text{2}}=4\text{a}\to \left(2\right)\\ \underset{\_}{{\text{T}}_2-20=2\text{a}\to \left(3\right)}\\ {\text{T}}_{\text{1}}+20-{\text{T}}_{\text{2}}=4\text{a}\\ \underset{\_}{20+{\text{T}}_{\text{2}}=2\text{a}}\\ {\text{T}}_{\text{1}}=6\text{a}\to \left(4\right)\end{array}$

Substitute eq (4) in eq(1)

$100-{\text{T}}_1=10\text{a}$

$\text{100}-\text{6a}=\text{10}\text{a}$

$\text{100}=\text{10}\text{a}+\text{6a}$

$\text{100}=\text{16}\text{a}$

$\text{10}\times \text{g}=\text{16}\times \text{a}$

$\frac{\text{10g}}{\text{16}}=\text{a}$

$\frac{\text{5g}}{\text{8}}=\text{a}$

$100-{\text{T}}_{\text{1}}=10\times \frac{5}{8}\text{g}$

$100-{\text{T}}_{\text{1}}=\frac{50}{8}\text{g}$

$100-{\text{T}}_{\text{1}}=\frac{25}{4}\text{g}$

$10\text{g}-{\text{T}}_{\text{1}}=\frac{25}{4}\text{g}$

$10\text{g}-\frac{25}{4}\text{g}={\text{T}}_{\text{1}}$

$\frac{40\text{g}-25\text{g}}{4}={\text{T}}_{\text{1}}$

$\frac{\text{15g}}{\text{4}}={\text{T}}_{\text{1}}$