4. 4z + 3 = 6 + 2z

Given, 4z + 3 = 6 + 2z
\(\Rightarrow\) 4z – 2z + 3 = 6 (Transposing 2z to LHS)

\(\Rightarrow\)2z + 3 = 6

\(\Rightarrow\)2z = 6 – 3 (Transposing 3 to RHS)

\(\Rightarrow\)2z = 3

\(\Rightarrow\) z = \(\frac { 3 }{ 2 }\)

Hence z = \(\frac { 3 }{ 2 }\) is the required solution.

Checking:

4z + 3 = 6 + 2z

Substituting z = \(\frac { 3 }{ 2 }\) in the given equation, we have

LHS = 4z + 3 = 4 × \(\frac { 3 }{ 2 }\) + 3 = 6 + 3 = 9

RHS = 6 + 2z = 6 + 2 × \(\frac { 3 }{ 2 }\) = 6 + 3 = 9

LHS = RHS

Hence z = \(\frac { 3 }{ 2 }\) is the required solution.