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Solve the following equations:

i. 3x + 12 = 2x – 4

ii. 5 (x – 3) = 3 (x + 2)

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i. 3x + 12 = 2x – 4 

∴ 3x + 12 – 12 = 2x – 4 – 12 

…[Subtracting 12 from both the sides] 

∴ 3x = 2x – 16 

∴ 3x – 2x = 2x – 16 – 2x 

…[Subtracting 2x from both the sides] 

∴ x = – 16

ii. 5 (x – 3) = 3 (x + 2) 

∴ 5x – 15 = 3x + 6 

∴ 5x – 15 + 15 = 3x + 6 + 15 

…[Adding 15 on both the sides] 

∴ 5x = 3x + 21 

∴ 5x – 3x = 3x + 21 – 3x …[Subtracting 3x from both the sides] 

∴ 2x = 21

\(\therefore \cfrac{2x}{2} =\cfrac{21}{2}\) …[Dividing both the sides by 2]

∴ x = \(\cfrac{21}{2}\)

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