Question

Solve and check the solution in the following equations
(a) x + 7 = 9
(b) x/4 = 25
(c) 9y = -135
(d) 15 - x = 4
(e) 3(x - 3) = 15
(f) 7y + 3 = 9

Class IV
LINEAR EQUATIONS

Answer 1

SOLUTION :

(a) x + 7 = 9

⇒ x + 7 - 7 = 9 - 7      [ Adding -7 to both sides ]

⇒ x = 2

∴ x = 2 is the solution of the given equation [ x + 7 = 9 ]

VERIFICATION

LHS = x + 7

       = 2 + 7            [ put the value of x = 2 ]

       = 9 = RHS

∴ LHS = RHS  (verified)

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(b) x/4 = 25

⇒ $\frac{x}{4} \times \bf{4}$ = 25 × 4   [ Multiplying both sides by 4 ]

⇒ x = 100

∴ x = 100 is the solution of the given equation [ x/4 = 25 ]

VERIFICATION

LHS = x/4

       = $\frac{100}{4}$         [ put the value of x = 100 ]

       = 25 = RHS

∴ LHS = RHS  (verified)

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(c) 9y = -135

⇒  $\frac{9y}{ \bf{9} }$ =  $\frac{-135}{ \bf{9} }$      [ Dividing both sides by 9 ]

⇒ y = -15

∴ y = -15 is the solution of the given equation [ 9y = -135  ]

VERIFICATION

LHS =  9y

       = 9 × -15        [ put the value of y = -15 ]

       = -135 = RHS

∴ LHS = RHS  (verified)

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(d) 15 - x = 4  

⇒ - x + 15 - 15 = 4 - 15   [ Adding -15 to both sides ]

⇒ -x = -11

⇒ x = 11   [ as minus(-) divided by minus(-) will be plus(+) ]

∴ x = 11 is the solution of the given equation [ 15 - x = 4    ]

VERIFICATION

LHS =  15 - x

       = 15 - 11                [ put the value of x = 11 ]

       = 4 = RHS

∴ LHS = RHS  (verified)

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(e) 3(x - 3) = 15

⇒  $\frac{3(3-x)}{ \bf{3} }$ =  $\frac{15}{ \bf{3} }$      [ Dividing both sides by 3 ]

⇒ x - 3 = 5

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

⇒ x = 8

∴ x = 8 is the solution of the given equation [ 3(x - 3) = 15   ]

VERIFICATION

LHS =  3(x - 3)

       = 3(8 - 3)          [ put the value of x = 8 ]

       = 3 × 8 - 3 × 3

       = 24 - 9

       = 15 = RHS

∴ LHS = RHS  (verified)

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(f) 7y + 3 = 9

⇒ 7y + 3 - 3 = 9 - 3    [ Adding -3 to both sides ]

⇒ 7y = 6

⇒  $\frac{7y}{ \bf{7} }$ =  $\frac{6}{ \bf{7} }$

⇒ y =  $\frac{6}{7}$

∴ y = $\frac{6}{7}$ is the solution of the given equation [ 7y + 3 = 9  ]

VERIFICATION

LHS = 7y + 3

       =  $7 \times \frac{6}{7} + 3$             [ put the value of y =  $\frac{6}{7}$ ]

       = 6 + 3

       = 9 = RHS

∴ LHS = RHS  (verified)

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Answer 2

SOLUTION :


(a) x + 7 = 9


⇒ x + 7 - 7 = 9 - 7      [ Adding -7 to both sides ]
⇒ x = 2
∴ x = 2 is the solution of the given equation [ x + 7 = 9 ]

VERIFICATION
LHS = x + 7
       = 2 + 7            [ put the value of x = 2 ]
       = 9 = RHS

∴ LHS = RHS  (verified)

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(b) x/4 = 25


⇒ \frac{x}{4} \times \bf{4}4x​×4 = 25 × 4   [ Multiplying both sides by 4 ]
⇒ x = 100
∴ x = 100 is the solution of the given equation [ x/4 = 25 ]

VERIFICATION
LHS = x/4
       = \frac{100}{4}4100​         [ put the value of x = 100 ]
       = 25 = RHS

∴ LHS = RHS  (verified)


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(c) 9y = -135


⇒  \frac{9y}{ \bf{9} }99y​ =  \frac{-135}{ \bf{9} }9−135​      [ Dividing both sides by 9 ]
⇒ y = -15
∴ y = -15 is the solution of the given equation [ 9y = -135  ]

VERIFICATION
LHS =  9y
       = 9 × -15        [ put the value of y = -15 ]
       = -135 = RHS

∴ LHS = RHS  (verified)

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(d) 15 - x = 4  

⇒ - x + 15 - 15 = 4 - 15   [ Adding -15to both sides ]
⇒ -x = -11
⇒ x = 11   [ as minus(-) divided by minus(-) will be plus(+) ]
∴ x = 11 is the solution of the given equation [ 15 - x = 4    ]

VERIFICATION
LHS =  15 - x
       = 15 - 11                [ put the value of x = 11 ]
       = 4 = RHS

∴ LHS = RHS  (verified)

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(e) 3(x - 3) = 15


⇒  \frac{3(3-x)}{ \bf{3} }33(3−x)​ =  \frac{15}{ \bf{3} }315​      [ Dividing both sides by 3 ]
⇒ x - 3 = 5
⇒ x - 3 + 3 = 5 + 3   [ Adding 3 to both sides ]
⇒ x = 8
∴ x = 8 is the solution of the given equation [ 3(x - 3) = 15   ]

VERIFICATION
LHS =  3(x - 3)
       = 3(8 - 3)          [ put the value of x = 8 ]
       = 3 × 8 - 3 × 3
       = 24 - 9
       = 15 = RHS

∴ LHS = RHS  (verified)

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(f) 7y + 3 = 9

⇒ 7y + 3 - 3 = 9 - 3    [ Adding -3 to both sides ]
⇒ 7y = 6
⇒  \frac{7y}{ \bf{7} }77y​ =  \frac{6}{ \bf{7} }76​
⇒ y =  \frac{6}{7}76​
∴ y = \frac{6}{7}76​ is the solution of the given equation [ 7y + 3 = 9  ]

VERIFICATION
LHS = 7y + 3
       =  7 \times \frac{6}{7} + 37×76​+3             [ put the value of y =  \frac{6}{7}76​ ]
       = 6 + 3
       = 9 = RHS

∴ LHS = RHS  (verified)

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