Question

Solve the following equations

(a) 4x - 3 = 2x + 1
(b) 2(x - 5) = 10
(c) (x + 2)/2 = 4
(d) 0.5x = 25

Answer 1

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Answer 2

SOLUTION :

(a) 4x - 3 = 2x + 1

⇒ 4x - 2x = 1 + 3     [ transporting 2x to LHS and -3 to RHS ]

⇒ 2x = 4

⇒ $\frac{2x}{ \bf{2} }$ =  $\frac{4}{ \bf{2} }$

⇒ x = 2

∴ x = 2 is the solution of the given equation [ 4x - 3 = 2x + 1  ]

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(b) 2(x - 5) = 10

⇒  $\frac{(2x-5)}{ \bf{2} }$ =  $\frac{10}{ \bf{2} }$     [ Dividing both sides by 2 ]

⇒ x - 5 = 5

⇒ x - 5 + 5 = 5 + 5     [ Adding 5 to both sides ]

⇒ x = 10  

∴ x = 10 is the solution of the given equation [  2(x - 5) = 10    ]

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(c) (x + 2)/2 = 4

⇒   $\frac{(x+2)}{ 2 } \times 2$ =  4 × 2     [ Multiplying both sides by 2 ]

⇒ x + 2 = 2

⇒ x + 2 - 2 = 2 - 2    [ Adding -2 to both sides ]

⇒ x = 0

∴ x = 0 is the solution of the given equation [  (x + 2)/2 = 4     ]

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(d) 0.5x = 25

⇒ $\frac{5}{10} \times x$ = 25

⇒ $\frac{5x}{10} \times \bf{10}$ = 25 × 10    [ Multiplying both sides by 10 ]

⇒ 5x = 250

⇒ $\frac{5x}{ \bf{5} }$ =  $\frac{250}{ \bf{5} }$     [ Dividing both sides by 5 ]

⇒ x = 50

∴ x = 50 is the solution of the given equation [   0.5x = 25     ]

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