Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

1. Solve each of the following equation.

Question (i).

6x + 10 = – 2

Answer:

Given equation is 6x + 10 = – 2

Transposing + 10 from L.H.S to R.H.S

we get

6x = -2 – 10

or 6x = -12

Dividing both sides by 6, we get

\(\frac{6 x}{6}=\frac{-12}{6}\)

or x = – 2, which is the required solution.

To check Put x = – 2 in the LHS of the equation 6x + 10 = – 2

L.H.S. = 6x + 10

= 6 × -2 + 10

= -12 + 10

= – 2 = R.H.S.

∴ L.H.S. = R.H.S.

Question (ii).

2y – 3 = 2

Answer:

Given equation is 2y – 3 = 2

Transposing – 3 from L.H.S. to R.H.S,

we get

2y = 2 + 3

or 2y = 5

Dividing both sides by 2, we get:

\(\frac{2 y}{2}=\frac{5}{2}\)

or y = \(\frac {5}{2}\), which is the required solution

To check. Put y = \(\frac {5}{2}\) in the L.H.S of the equation 2y – 3 = 2

L.H.S = 2y – 3 = 2 × \(\frac {5}{2}\) – 3

= 5 – 3 = 2 = R.H.S.

∴ L.H.S. = R.H.S.

Question (iii).

\(\frac{a}{5}\) + 3 = 2

Answer:

Given equation is \(\frac{a}{5}\) + 3 = 2

Transposing + 3 from L.H.S to R.H.S., we get

\(\frac{a}{5}\) = 2 – 3

or \(\frac{a}{5}\) = -1

Multiplying both sides, by 5, we get

5 × \(\frac{a}{5}\) = 5 × – 1

or a = – 5, which is the required solution.

To Check: Put a = – 5 in the L.H.S of the equation

\(\frac{a}{5}\) + 3 = 2,

L.H.S. = \(\frac{a}{5}\) + 3

= \(\frac {-5}{5}\) + 3

= – 1 + 3

= 2 = R.H.S.

∴ L.H.S. = R.H.S.

Question (iv).

\(\frac{3 x}{2}=\frac{2}{3}\)

Answer:

Given equation is \(\frac{3 x}{2}=\frac{2}{3}\)

Multiplying both sides by 2, we get

2 × \(\frac{3 x}{2}\) = 2 × \(\frac {2}{3}\)

or 3x = \(\frac {4}{3}\)

Dividing both sides by 3 we get

\(\frac{3 x}{3}=\frac{4}{3} \times \frac{1}{3}\)

or x = \(\frac {4}{9}\), which is the required solution.

To Check. Put x = \(\frac {4}{9}\) in the L.H.S. of equation \(\frac{3 x}{2}=\frac{2}{3}\)

L.H.S. = \(\frac{3 x}{2}=\frac{3}{2} \times \frac{4}{9}\) = \(\frac {2}{3}\) = R.H.S.

∴L.H.S. = R.H.S.

Question (v).

\(\frac {5}{2}\)x = -5

Answer:

Given equation is \(\frac {5}{2}\) x = – 5

Multiplying both sides by 2, we get

2 × \(\frac {5}{2}\) x = 2 × – 5

or 5x = – 10

Dividing both sides by 5, we get

\(\frac{5 x}{5}=\frac{-10}{5}\)

or x = – 2, which is the required solution.

To Check. Put x = – 2 in L.H.S. of the equation \(\frac {5}{2}\)x = – 5

L.H.S. = \(\frac {5}{2}\)x = \(\frac {5}{2}\) × -2

= – 5 = R.H.S.

∴ L.H.S. = R.H.S.

Question (vi).

2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)

Answer:

Given equation is 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)

Subtract \(\frac {5}{2}\) from both sides, we get

2x + \(\frac {5}{2}\) – \(\frac {5}{2}\)

= \(\frac {37}{2}\) – \(\frac {5}{2}\)

or 2x = \(\frac{37-5}{2}\)

or 2x = \(\frac {32}{2}\)

or 2x = 16

Dividing both sides by 2, we get

\(\frac{2 x}{2}=\frac{16}{2}\)

or x = 8, which is the required solution.

To Check. Put x = 8 in the L.H.S. of the equation 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)

L.H.S. = 2x + \(\frac {5}{2}\)

= 2 × 8 + \(\frac {5}{2}\)

= 16 + \(\frac {5}{2}\)

= \(\frac{32+5}{2}\)

= \(\frac {37}{2}\) = R.H.S.

∴ L.H.S. = R.H.S.

2. Solve the following equation

Question (i).

5 (x + 1) = 25

Answer:

Given equation is 5 (x + 1) = 25

Dividing both sides by 5 we get

\(\frac{5(x+1)}{5}=\frac{25}{5}\)

or x + 1 = 5

Transposing 1 from L.H.S. to R.H.S. we get

x = 5 – 1

or x = 4, which is the required solution.

To Check. Put x = 4 in the L.H.S. of the equation 5 (x + 1) = 25

L.H.S. = 5 (x + 1)

= 5 (4 + 1)

= 5 (5)

= 25 = R.H.S.

∴ L.H.S. = R.H.S.

Question (ii).

2 (3x – 1) = 10

Answer:

Given equation is 2 (3x – 1) = 10

Dividing both sides by 2, we get

\(\frac{2(3 x-1)}{2}=\frac{10}{2}\)

or 3x – 1 = 5

Transposing – 1 from L.H.S. to R.H.S we get

3x = 5 + 1

3x = 6

Dividing both sides by 3, we get \(\frac{3 x}{3}=\frac{6}{3}\)

or x = 2, which is the required solution.

To Check. Put x = 2, in the L.H.S. of the equation 2 (3x – 1) = 10

L.H.S. = 2 (3x – 1) = 10

L.H.S = 2 (3x – 1) = 2 (3 × 2 – 1)

= 2 (6 – 1)

= 2 × 5

= 10 = R.H.S.

∴L.H.S. = R.H.S.

Question (iii).

4 (2 – x) = 8

Answer:

Given equation is 4 (2 – x) = 8

Dividing both sides by 4, we get

\(\frac{4(2-x)}{4}=\frac{8}{4}\)

or 2 – x= 2

Transposing 2 from L.H.S. to R.H.S. we get

-x = 2 – 2

or – x = 0

Multiplying both sides by – 1, we get

-x × – 1 = x – 1

or x = 0, which is the required solution.

To Check. Put x = 0 in the L.H.S. of the equation 4 (2 – x) = 8

L.H.S. = 4 (2 – x) = 4 (2 – 0)

= 4 × 2

= 8 = R.H.S.

∴ L.H.S. = R.H.S.

Question (iv).

– 4 (2 + x) = 8.

Answer:

Given equation is – 4 (2 + x) = 8

Dividing both sides by – 4, we get

\(\frac{-4(2+x)}{-4}=\frac{8}{-4}\)

Transposing 2 from L.H.S. to R.H.S. we get :

x = – 2 – 2

or x = – 4, which is the required solution

To Check. Put x = – 4 in the L.H.S. of equation – 4 (2 + x) = 8

L.H.S. = – 4 (2 + x) = – 4 [2 + (- 4)]

= – 4 (2 – 4)

= – 4 (- 2)

= 8 = R.H.S.

∴ L.H.S. = R.H.S.

3. Solve the following equations :

Question (i).

4 = 5 (x – 2)

Answer:

Given equation is 4 = 5 (x – 2)

or 4 = 5x – 10

Transposing 5x to L.H.S. and 4 to R.H.S.,

we get

– 5x = – 4 – 10

or – 5x = – 14

Dividing both sides by – 5, we get

\(\frac{-5 x}{-5}=\frac{-14}{-5}\)

or, x = \(\frac {14}{5}\), which is the required solution.

To Check. Put x = \(\frac {14}{5}\) in the R.H.S. of the equation 4 = 5 (x – 2)

R.H.S. = 5 (x – 2) = 5\(\left(\frac{14}{5}-2\right)\)

= 5\(\left(\frac{14-10}{5}\right)\)

= 5 \(\left(\frac{4}{5}\right)\)

= 4 = L.H.S.

∴ L.H.S. = R.H.S.

Question (ii).

– 4 = 5 (x – 2)

Answer:

Given equation is – 4 = 5 (x – 2)

or – 4 = 5x – 10

Transposing -4 to R.H.S and 5x to L.H.S

we get

-5x = 4 – 10 or -5x = -6

Dividing both sides by – 5 we get

\(\frac{-5 x}{-5}=\frac{-6}{-5}\)

or x = \(\frac {6}{5}\), which is the required solution.

To Check. Put x = \(\frac {6}{5}\) in the R.H.S. of the equation – 4 = 5 (x – 2)

L.H.S. = 5 (x – 2)

= 5\(\left(\frac{6}{5}-2\right)\)

= 5\(\left(\frac{6-10}{5}\right)\)

= 5\(\left(\frac{-4}{5}\right)\)

= -4 = L.H.S.

L.H.S. = R.H.S.

Question (iii).

4 + 5 (p – 1) = 34

Answer:

Given equation is 4 + 5(p – 1) = 34

Transposing 4 to R.H.S. we get

5(p – 1) = 34 – 4

5(p – 1) = 30

Dividing both sides, by 5, we get

\(\frac{5(p-1)}{5}=\frac{30}{5}\)

p – 1=6

Transposing -1 to R.H.S. we get

p = 6 + 1

p = 7 which is the required solution.

To Check : Put p = 7 in L.H.S. of the equation 4 + 5 (p – 1) = 34

L.H.S. = 4 + 5 (p – 1)

= 4 + 5 (7 – 1)

= 4 + 5 (6)

= 4 + 30

= 34 = R.H.S.

∴ L.H.S. = R.H.S.

Question (iv).

6y – 1 = 2y + 1.

Answer:

Given equation is 6y – 1 = 2y + 1

Transposing – 1 to R.H.S. and 2y to L.H.S,

we get

6y – 2y = 1 + 1

or 4y = 2 or y = \(\frac {2}{4}\)

or y = \(\frac {1}{2}\), which is the required solution.

To Check Put y = \(\frac {1}{2}\) in both L.H.S. and R.H.S. of the equation

6y – 1 = 2y + 1

L.H.S. = 6y – 1 = 6 × \(\frac {1}{2}\) – 1 = 3 – 1 = 2

R.H.S. = 2y + 1 = 2 × \(\frac {1}{2}\) + 1 = 1 + 1 = 2.

∴ L.H.S. = R.H.S.

4.

Question (i).

Construct 3 equations starting with x = 2

Answer:

First Equation.

(i) Start with x = 2

Multiplying both sides by 10

10x = 20

Adding 2 to both sides

10x + 2 = 20 + 2

or 10x + 2 = 22

This has resulted in an equation.

Second Equation. Start with x = 2

Divide both sides by 5

∴ \(\frac{x}{5}=\frac{2}{5}\)

This has resulted in an equation.

Third Equation. Start with x = 2

Multiply both sides by 5, we get

5x = 5 × 2

or 5x = 10

Subtracting 3 from both sides, we get

5x – 4 = 10 – 3

or 5x – 3 = 7

This has resulted in an equation.

Question (ii).

Construct 3 equation starting with x = – 2

Answer:

First Equation. Start with x = – 2

Multiplying both sides with 3, we get

3x = – 6

This has resulted in an equation

Second Equation. Start with x = – 2

Multiplying both sides with 3, we get 3x = -6

Adding 7 to both sides, we get 3x + 7

= -6 + 7 or 3x + 7 = 1

This has resulted in an equation.

Third Equation. Start with x = – 2

Multiplying both side with 2 we get 3x = – 6

Adding 10 to both sides we get

3x+ 10 = -6 + 10

or 3x + 10 = 4

This has resulted in an equation.

Multiple Choice Questions :

5. If 7x + 4 = 39, then x is equal to :

(a) 6

(b) -4

(c) 5

(d) 8

Answer:

(c) 5

6. If 8m – 8 = 56 then m is equal to :

(a) -4

(b) -2

(c) -14

(d) 8

Answer:

(d) 8

7. Which of the following number satisfies the equation – 6 + x = -18 ?

(a) 10

(b) – 13

(c) – 12

(d) – 16.

Answer:

(a) 10

8. If \(\frac{x}{2}\) = 14, then the value of 2x + 6 is equal to :

(a) 62

(b) -64

(c) 16

(d) -62.

Answer:

(a) 62

9. If 3 subtracted from twice a number is 5, then the number is :

(a) -4

(b) -2

(c) 2

(d) 4

Answer:

(d) 4

10. If 5 added to thrice an integer is – 7, then the integer is :

(a) – 6

(b) – 5

(c) -4

(d) 4

Answer:

(c) -4