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# 8.$$\frac { 2x }{ 3 } + 1 = \frac { 7x }{ 15 } + 3$$

Given that:$$\frac { 2x }{ 3 } + 1 = \frac { 7x }{ 15 } + 3$$
15$$(\frac { 2x }{ 3 } + 1)$$ = 15$$(\frac { 7x }{ 15 } + 3)\quad$$[multiplying both the sides by 15]
$$\Rightarrow \frac { 2x }{ 3 } × 15 + 1 ×15 = \frac { 7x }{ 15 } ×15 + 3 ×15$$

$$\Rightarrow$$ 2x × 5 + 15 = 7x + 45

$$\Rightarrow$$ 10x + 15 = 7x + 45

$$\Rightarrow$$ 10x – 7x = 45 – 15 [Transposing 7x to LHS and 15 to RHS]

$$\Rightarrow$$3x = 30

$$\Rightarrow$$ x = 30 ÷ 3 = 10 [Transposing 3 to RHS]

Thus, the required solution is x = 10.

Checking

Substitute x=10 in the given equation, we have

$$\frac { 2×10 }{ 3 } + 1 = \frac { 7×10 }{ 15 } + 3$$

$$\Rightarrow \frac { 20 }{ 3 } + 1 = \frac { 70 }{ 15 } + 3$$

$$\Rightarrow \frac { 20 + 3}{ 3 } = \frac { 70 + 45}{ 15 }$$

$$\Rightarrow \frac { 23}{ 3 } = \frac { 115}{ 15 }$$

$$\Rightarrow$$ LHS= RHS

Hence,the required solution is x = 10.