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3.\(x+7-\frac{8x}3=\frac{17}6-\frac{5x}2\)


Answer :

Given that,\(x+7-\frac{8x}3=\frac{17}6-\frac{5x}2\) LCM of denominators 3,6 and 2 =6

∴\(6×x+7-6×\frac{8x}3=6×\frac{17}6-6×\frac{5x}2\quad \)[Multiplying both the sides by 6]

\(\Rightarrow 6x+42-16x=17-15x\)

\(\Rightarrow -10x + 42 = 17 – 15x \)

\(\Rightarrow-10x + 15x = 17 – 42 \)[Transposing 15x to LHS and 42 to RHS]

\(\Rightarrow 5x = -25 \)

\(\Rightarrow x = -25 ÷ 5 \) [Transposing 5 to RHS]

\(\Rightarrow x = -5 \)

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