Question

Solve the following equations
(a) 2x + 1 = 5
(b) 6y - 5 = 19
(c) 3 + 4y = -5
(d) (2/3)p = 6
(e) 5y + 10 = 4y - 10
(f) (1/2)x + 3 = 5

Answer 1

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Answer 2

SOLUTION :

(a) 2x + 1 = 5

⇒ 2x + 1 - 1 = 5 - 1   [ adding -1 to both sides ]

⇒ 2x = 4

⇒ $\frac{2x}{ \bf{2} }$ = $\frac{4}{ \bf{2} }$  [Divide both sides by 2 ]

⇒ 1x = 2

∴ x = 2 is the solution of the given equation (2x + 1 = 5)

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(b) 6y - 5 = 19

⇒ 6y - 5 + 5 = 19 + 5  [ adding 5 to both sides ]

⇒ 6y = 24

⇒ $\frac{6y}{ \bf{6} }$ = $\frac{24}{ \bf{6} }$  [ Divide both sides by 6 ]

⇒ y = 4

∴ y = 4 is the solution if the given equation (6y - 5 = 19)

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(c) 3 + 4y = -5

⇒ 4y + 3 - 3 = -5 - 3   [ Adding -3 to both sides ]

⇒ 4y = -8

⇒ $\frac{4y}{ \bf{4} }$ = $\frac{-8}{ \bf{4} }$  [Divide both sides by 4 ]

⇒ y = -2

∴ y = -2 is the solution of the given equation (3 + 4y = -5)

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(d) (2/3)p = 6

⇒ $\frac{2}{3} \times p$ = 6

⇒ $\frac{2p}{3}$ = 6

⇒ $\frac{2p}{3} \times \bf{3}$ = 6 × 3   [ Multiply both sides by 3 ]

⇒ 2p = 18

⇒ $\frac{2p}{ \bf{2} }$ =  $\frac{18}{ \bf{2} }$

⇒ p = 9

∴ p = 9  is the solution of the given equation ( (2/3)p = 6 )

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(e) 5y + 10 = 4y - 10

⇒ 5y + 10 - 10 = 4y - 10 - 10  [Adding -10 to both sides ]

⇒ 5y = 4y - 20

⇒ 5y - 4y = -20 + 4y - 4y   [ Adding -4y to both sides ]

⇒ y = -20

∴ y = -20 is the solution of the given equation (5y + 10 = 4y - 10)

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(f) (1/2)x + 3 = 5

⇒  $\frac{1}{2} \times x$ + 3 = 5

⇒ $\frac{x}{2}$ + 3 = 5

⇒ $\frac{x}{2}$ + 3 - 3 = 5 - 3  [ Adding -3 to both sides ]

⇒ $\frac{x}{2}$ = 2

⇒ $\frac{x}{2} \times \bf{2}$ = 2 × 2

⇒ x = 4

∴ x = 4 is the solution of the given equation ( (1/2)x + 3 = 5 )

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