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# 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.