 RS Aggarwal Class 6 Solutions Chapter 9 - Linear Equations In One Variable Ex 9A (9.1)

RS Aggarwal Class 6 Chapter 9 - Linear Equations In One Variable Ex 9A (9.1) Solutions Free PDF

Solve each of the following equations and verify the answer in each case:

Question 1:

x + 5 = 12

Solution:

x + 5 = 12

Subtracting 5 from both the sides:

$\Rightarrow$ x + 5 – 5 = 12 – 5

$\Rightarrow$ x = 7

Verification:

Substituting x = 7 in the L.H.S.

$\Rightarrow$ 7 + 5 = 12 = R.H.S.

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

Hence, verified.

Question 2:

x + 3 = -2

Solution:

x + 3 = -2

Subtracting 3 from both the sides:

$\Rightarrow$ x + 3 – 3 = -2 – 3

$\Rightarrow$ x = -5

Verification:

Substituting x = -5 in the L.H.S.

$\Rightarrow$ -5 + 3 = -2 = R.H.S.

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

Hence, verified.

Question 3:

x – 7 = 6

Solution:

x – 7 = 6

Adding 7 on both the sides:

$\Rightarrow$ x – 7+7= 6+7

$\Rightarrow$ x = 13

Verification:

Substituting x = 13 in the L.H.S.

$\Rightarrow$ 13 –7 = 6 = R.H.S.

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

Hence, verified.

Question 4:

x – 2 = -5

Solution:

x – 2 = -5

Adding 2 on both the sides:

$\Rightarrow$ x – 2 + 2 = -5 + 2

$\Rightarrow$ x = -3

Verification:

Substituting x = -3 in the L.H.S.

$\Rightarrow$ -3 – 2 = -5 = R.H.S.

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

Hence, verified.

Question 5:

3x – 5 = 13

Solution:

3x – 5 = 13

$\Rightarrow$ 3x – 5 + 5 = 13 + 5 [Adding 5 on both the sides]

$\Rightarrow$3x = 18

$\Rightarrow$$\frac{3x}{3}=\frac{18}{3}$ [Dividing both the sides by 3]

$\Rightarrow$ x = 6

Verification:

Substituting x = 6 in the L.H.S.

$\Rightarrow$3 x 6 – 5 = 18 – 5 = 13 = R.H.S.

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

Hence, verified.

Question 6:

4x + 7 = 15

Solution:

4x + 7 = 15

$\Rightarrow$ 4x + 7 – 7 = 15 – 7 [Subtracting 7 from both the sides]

$\Rightarrow$ 4x = 8

$\Rightarrow$$\frac{4x}{4}=\frac{8}{4}$ [Dividing both the sides by 4]

$\Rightarrow$ x = 2

Verification:

Substituting x = 2 in the L.H.S.

$\Rightarrow$4 x 2+ 7= 8 + 7= 15 = R.H.S.

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

Hence, verified.

Question 7:

$\frac{x}{5}=12$

Solution:

$\frac{x}{5}=12$

$\Rightarrow$$\frac{x}{5}\times 5=12\times 5$ [Multiplying both the sides by 5]

$\Rightarrow$ x = 60

Verification:

Substituting x = 60 in the L.H.S. :

$\Rightarrow$$\frac{60}{5}$ = 12 = R.H.S.

$\Rightarrow$ L.H.S. = R.H.S.

Hence, verified.

Question 8:

$\frac{3x}{5}=15$

Solution:

$\frac{3x}{5}=15$

$\Rightarrow$$\frac{3x}{5}\times 5=15\times 5$ [Multiplying both the sides by 5]

$\Rightarrow$ 3x = 75

$\Rightarrow$$\frac{3x}{3}=\frac{75}{3}$ [Dividing both the sides by 3]

$\Rightarrow$ x = 25

Verification:

Substituting x = 25 in the L.H.S. :

$\Rightarrow$$\frac{3\times 25}{3}=15$= R.H.S.

$\Rightarrow$ L.H.S. = R.H.S.

Hence, verified.

Question 9:

5x – 3 = x + 17

Solution:

5x – 3 = x + 17

$\Rightarrow$ 5x – x = 17 + 3 [Transposing x to the L.H.S. and 3 to the R.H.S.]

$\Rightarrow$ 4 x = 20

$\Rightarrow$ $\frac{4x}{4}=\frac{20}{4}$ [Dividing both the sides by 4]

$\Rightarrow$ x = 5

Verification:

Substituting x = 5 on both the sides:

L.H.S. : 5(5) – 3

= 25 – 3

= 22

R.H.S. : 5 + 17 = 22

$\Rightarrow$ L.H.S. = R.H.S.

Hence, verified.

Question 10:

$2x-\frac{1}{2}=3$

Solution:

$2x-\frac{1}{2}=3$

$\Rightarrow$$2x-\frac{1}{2}+\frac{1}{2}=3+\frac{1}{2}$ [Adding $\frac{1}{2}$ on both the sides]

$\Rightarrow$$2x=\frac{6+1}{2}$

$\Rightarrow$ $2x=\frac{7}{2}$

$\Rightarrow$$\frac{2x}{2}=\frac{7}{2\times 2}$ [Dividing both the sides by 2]

$\Rightarrow$ $x=\frac{7}{4}$

Verification:

Substituting x = $\frac{7}{4}$ in the L.H.S. :

$2\left ( \frac{7}{4} \right )-\frac{1}{2}$

$=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3=R.H.S.$

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

Hence, verified.

Question 11:

3(x + 6) = 24

Solution:

3(x + 6) = 24

$\Rightarrow$$3\times x+3\times 6=24$ [On expanding the brackets]

$\Rightarrow$ 3x + 18 = 24

$\Rightarrow$ 3x + 18 – 18 = 24 – 18 [Subtracting 18 from both the sides]

$\Rightarrow$ 3x = 6

$\Rightarrow$$\frac{3x}{3}=\frac{6}{3}$ [Dividing both the sides by 3]

$\Rightarrow$ x = 2

Verification:

Substituting x = 2 in the L.H.S. :

3(2 + 6) = 3 x 8 = 24 = R.H.S.

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

Hence, verified.

Question 12:

6x + 5 = 2x + 17

Solution:

6x + 5 = 2x + 17

$\Rightarrow$ 6x – 2x = 17 – 5 [Transposing 2x to the L.H.S. and 5 to the R.H.S.]

$\Rightarrow$ 4x = 12

$\Rightarrow$ $\frac{4x}{4}=\frac{12}{4}$ [Dividing both the sides by 4]

$\Rightarrow$ x = 3

Verification:

Substituting x = 3 on both the sides:

L.H.S. : 6(3) + 5

= 18 + 5

= 23

R.H.S. : 2(3) + 17

= 6 + 17

= 23

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

Hence, verified.

Question 13:

$\frac{x}{4}-8=1$

Solution:

$\frac{x}{4}-8=1$

$\Rightarrow$$\frac{x}{4}-8+8=1+8$ [Adding 8 on both the sides]

$\Rightarrow$$\frac{x}{4}=9$

$\Rightarrow$$\frac{x}{4}\times 4=9\times 4$ [Multiplying both the sides by 4]

or, x = 36

Verification:

Substituting x = 36 in the L.H.S. :

or, $\frac{36}{4}-8$ = 9 – 8 = 1 = R.H.S.

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

Hence, verified.

Question 14:

$\frac{x}{2}=\frac{x}{3}+1$

Solution:

$\frac{x}{2}=\frac{x}{3}+1$

$\Rightarrow$$\frac{x}{2}-\frac{x}{3}=1$ [Transposing $\frac{x}{3}$ to the L.H.S.]

$\Rightarrow$$\frac{3x-2x}{6}=1$

$\Rightarrow$$\frac{x}{6}=1$

$\Rightarrow$$\frac{x}{6}\times 6=1\times 6$ [Multiplying both the sides by 6]

or, x = 6

Vrification:

Substituting x = 6 on both the sides:

L.H.S. : $\frac{6}{2}=3$

R.H.S. : $\frac{6}{3}+1=2+1=3$

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

Hence, verified.

Question 15:

3(x + 2) – 2(x – 1) = 7

Solution:

3(x + 2) – 2(x – 1) = 7

$\Rightarrow$$3\times x+3\times 2-2\times x-2\times (-1)=7$ [On expanding the brackets]

or, 3x + 6 – 2x + 2 = 7

or, x + 8 = 7

or, x + 8 – 8 = 7 – 8 [Subtracting 8 from both the sides]

or, x = -1

Verification:

Substituting x = -1 in the L.H.S. :

3(-1 + 2) – 2(-1 – 1)

or, 3(1) – 2(-2)

or, 3 + 4 = 7 = R.H.S.

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

Hence, verified.

Question 16:

5(x – 1) + 2(x + 3) + 6 = 0

Solution:

5(x – 1) + 2(x + 3) + 6 = 0

$\Rightarrow$ 5x – 5 + 2x + 6 = 0 [Expanding within the brackets]

$\Rightarrow$ 7x + 7 = 0

$\Rightarrow$ x + 1 = 0 [Dividing by 7]

$\Rightarrow$ x = -1

Verification:

Putting x = -1 in the L.H.S. :

5(-1 – 1) + 2(-1 + 3) + 6

= 5(-2) + 2(2) + 6

= -10 + 5 + 6 = 0 = R.H.S.

Hence, verified.

Question 17:

6(1 – 4x) + 7(2 + 5x) = 53

Solution:

6(1 – 4x) + 7(2 + 5x) = 53

or, $6\times 1-6\times 4x+7\times 2+7\times 5x=53$ [On expanding the brackets]

or, 6 – 24x + 14 + 35x = 53

or, 11x + 20 = 53

or, 11x + 20 – 20 = 53 – 20 [Subtracting 20 from both the sides]

or, 11x = 33

or, $\frac{11x}{11}=\frac{33}{11}$ [Dividing both the sides by 11]

or, x = 3

Verification:

Substituting x = 3 in the L.H.S. :

6(1 – 4 x 3) + 7(2 + 5 x 3)

= 6(1 – 12) + 7(2 + 15)

= 6(-11) + 7(17)

= -66 + 119 = 53 = R.H.S.

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

Hence, verified.

Question 18:

16(3x – 5) – 10(4x – 8) = 40

Solution:

16(3x – 5) – 10(4x – 8) = 40

or, $16\times 3x-16\times 5-10\times 4x-10\times (-8)=40$ [On expanding the brackets]

or, 48x – 80 – 40x + 80 = 40

or, 8x = 40

or, $\frac{8x}{8}=\frac{40}{8}$ [Dividing both the sides by 8]

or, x = 5

Verification:

Substituting x = 5 in the L.H.S. :

16(3 x 5 – 5) – 10(4 x 5 – 8)

= 16(15 – 5) – 10(20 – 8)

= 16(10) – 10(12)

= 160 – 120 = 40 = R.H.S.

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

Hence, verified.

Question 19:

3(x + 6) + 2(x + 3) = 64

Solution:

3(x + 6) + 2(x + 3) = 64

$\Rightarrow$$3\times x+3\times 6+2\times x+2\times 3=64$ [On expanding the brackets]

$\Rightarrow$ 3x + 18 + 2x + 6 = 64

$\Rightarrow$ 5x + 24 = 64

$\Rightarrow$ 5 x + 24 – 24 = 64 – 24 [Subtracting 24 from both the sides]

$\Rightarrow$ 5x = 40

$\Rightarrow$$\frac{5x}{5}=\frac{40}{5}$

$\Rightarrow$ x = 8

Verification:

Substituting x = 8 in the L.H.S. :

3(8 + 6) + 2(8 + 3)

= 3(14) + 2(11)

= 42 + 22 = 64 = R.H.S.

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

Hence, verified.

Question 20:

3(2 – 5x) – 2(1 – 6x) = 1

Solution:

3(2 – 5x) – 2(1 – 6x) = 1

or, $3\times 2+3\times (-5x)-2\times 1-2\times (-6x)=1$ [On expanding the brackets]

or, 6 – 15x – 2 + 12x = 1

or, 4 – 3x = 1

or, 3 = 3x

or, x = 1

Verification:

Substituting x = 1 in the L.H.S. :

3(2 – 5 x 1) – 2(1 – 6 x 1)

= 3(2 – 5) – 2(1 – 6)

= 3(-3) – 2(-5)

= -9 + 10 = 1 = R.H.S.

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

Hence, verified.

Question 21:

$\frac{n}{4}-5=\frac{n}{6}+\frac{1}{2}$

Solution:

$\frac{n}{4}-5=\frac{n}{6}+\frac{1}{2}$

or, $\frac{n}{4}-\frac{n}{6}=\frac{1}{2}+5$ [Transposing n/6 to the L.H.S. and 5 to the R.H.S.]

or,$\frac{3n-2n}{12}=\frac{1+10}{2}$

or, $\frac{n}{12}=\frac{11}{2}$

or, $\frac{n}{12}\times 12=\frac{11}{2}\times 12$ [Dividing both the sides by 12]

or, n = 66

Verification:

Substituting n = 66 on both the sides:

L.H.S. :

$\frac{66}{4}-5=\frac{33}{2}=5=\frac{33-10}{2}=\frac{23}{2}$

R.H.S. :

$\frac{66}{6}+\frac{1}{2}=11+\frac{1}{2}=\frac{22+1}{2}=\frac{23}{2}$

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

Hence, verified.

Question 22:

$\frac{2m}{3}+8=\frac{m}{2}-1$

Solution:

$\frac{2m}{3}+8=\frac{m}{2}-1$

or, $\frac{2m}{3}-\frac{m}{2}=-1-8$ [Transposing m/2 to the L.H.S. and 8 to the R.H.S.]

or, $\frac{4m-3m}{6}=-9$

or, $\frac{m}{6}=-9$

or, $\frac{m}{6}\times 6=-9\times 6$ [Multiplying both the sides by 6]

or, m = -54

Verification:

Substituting x = -54 on both the sides:

L.H.S. :

$\frac{2(-54)}{3}+8=\frac{-54}{2}-1$

= $\frac{-108}{3}+8$

= -36 + 8

= -28

R.H.S. :

$\frac{-54}{2}-1$

= -27 – 1

= -28

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

Hence, verified

Question 23:

$\frac{2x}{5}-\frac{3}{2}=\frac{x}{2}+1$

Solution:

$\frac{2x}{5}-\frac{3}{2}=\frac{x}{2}+1$

or,$\frac{2x}{5}-\frac{x}{2}=1+\frac{3}{2}$ [Transposing x/2 to the L.H.S. and 3/2 to R.H.S.]

or,$\frac{4x-5x}{10}=\frac{2+3}{2}$

or, $\frac{-x}{10}=\frac{5}{2}$

or, $\frac{-x}{10}\left ( -10 \right )=\frac{5}{2}\times \left ( -10 \right )$ [Multiplying both the sides by -10]

or, x = -25

Verification:

Substituting x = -25 on both the sides:

L.H.S. :$\frac{2(-25)}{5}-\frac{3}{2}$

= $\frac{-50}{5}-\frac{3}{2}$

= $-10-\frac{3}{2}=\frac{-23}{2}$

R.H.S. : $\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$

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

Hence, verified.

Question 24:

$\frac{x-3}{5}-2=\frac{2x}{5}$

Solution:

$\frac{x-3}{5}-2=\frac{2x}{5}$

or, $\frac{x}{5}-\frac{3}{5}-2=\frac{2x}{5}$

or, $-\frac{3}{5}-2=\frac{2x}{5}-\frac{x}{5}$ [Transposing x/5 to the R.H.S.]

or, $\frac{-3-10}{5}=\frac{x}{5}$

or, $\frac{-13}{5}=\frac{x}{5}$

or, $\frac{-13}{5}(5)=\frac{x}{5}\times (5)$ [Multiplying both the sides by 5]

or, x = -13

Verification:

Substituting x = -13 on both the sides:

L.H.S. : $\frac{-13-3}{5}-2$

= $\frac{-16}{5}-2=\frac{-16-10}{5}=\frac{-26}{5}$

R.H.S. : $\frac{2\times (-13)}{5}=\frac{-26}{5}$

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

Hence, verified.

Question 25:

$\frac{3x}{10}-4=14$

Solution:

$\frac{3x}{10}-4=14$

or, $\frac{3x}{10}-4+4=14+4$ [Adding 4 on both the sides]

or, $\frac{3x}{10}18$

or, $\frac{3x}{10}\times 10=18\times 10$ [Multiplying both the sides by 10]

or, 3x = 180

or, $\frac{3x}{3}=\frac{180}{3}$ [Dividing both the sides by 3]

or, x = 60

Verification:

Substituting x = 60 on the L.H.S. :

$\frac{3\times 60}{10}-4$

= $\frac{180}{10}-4=18-4=14=R.H.S.$

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

Hence, verified.

Question 26:

$\frac{3}{4}(x-1)=x-3$

Solution:

$\frac{3}{4}(x-1)=x-3$

$\Rightarrow$$\frac{3}{4}\times x-\frac{3}{4}\times 1=x-3$ [On expanding the brackets]

$\Rightarrow$$\frac{3x}{4}-\frac{3}{4}=x-3$

$\Rightarrow$$\frac{3x}{4}-x=-3+\frac{3}{4}$ [Transposing x to the L.H.S. and $-\frac{3}{4}$ to the R.H.S.]

$\Rightarrow$$\frac{3x-4x}{4}=\frac{-12+3}{4}$

$\Rightarrow$$\frac{-x}{4}=\frac{-9}{4}$

$\Rightarrow$$\frac{-x}{4}\times (-4)=\frac{-9}{4}\times (-4)$ [Multiplying both the sides by -4]

or, x = 9

Verification:

Substituting x = 9 on both the sides:

L.H.S. : $\frac{3}{4}(9-1)$

=$\frac{3}{4}(8)$

= 6

R.H.S. : 9 – 3 = 6

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

Hence, verified.

Practise This Question

Choose the pair of equations which satisfy the point (1,-1)