S.chand publication New Learning Composite mathematics solution of class 7 Chapter 6 Linear Equations in One Variable Exercise 6A

 Exercise 6A

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Q1 | Ex-6A |Class 8 |S.Chand |Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 1

(a) 2x + 7 = 11

Sol :

2x=11-7

2x=4

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

∴x=2

(b) 8 + 5y = 23

Sol :

5y=23-8

5y=15

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

∴y=3

(c) 4x – 9 = 7

Sol :

4x=7+9

4x=16

$x=\frac{16}{4}=4$

∴x=4

(d) $\frac{y}{7}+1= 3$

Sol :

$\frac{y}{7}+1=3$

$\frac{y}{7}=3-1$

$\frac{y}{7}=2$

∴y=14

(e) $6 + \frac{t}{4} = 9$

Sol :

$6+\frac{t}{4}=9$

$\frac{t}{4}=9-6$

$\frac{t}{4}=3$

∴t=12

(f) $\frac{x}{3} + 9 = 1$

Sol :

$\frac{x}{3}+9=1$

$\frac{x}{3}=1-9$

$\frac{x}{3}=-8$

∴x=-24

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Q2 | Ex-6A |Class 8 |S.Chand |Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 2

(a) 2x + 1 = x + 7

Sol :

2x-x=7-1

∴x=6

(b) 5x + 12 = 3x + 20

Sol :

5x-3x=20-12

2x=8

$x=\frac{8}{2}=4$

∴x=4

(c) 3y = 8y – 15

Sol :

3y-8y=-15

-5y=-15

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

∴y=3

(d) 6p – 18 = 5p – 7

Sol :

6p-18=5p-7

6p-5p=-7+18

∴p=11

(e) t + 10 = 20 – t

Sol :

t+t=20-10

2t=10

∴$t=\frac{10}{2}=5$

(f) 7z – 18 = 28 + 2z

Sol :

7z-2z=-28+18

5z=-10

$z=-\frac{10}{5}=-2$

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Q3 | Ex-6A |Class 8 |S.Chand |Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 3

(a) 5x + 8 – 2x = 14

Sol :

5x-2x=14-8

3x=6

$x=\frac{6}{3}=2$

(b) 5(x – 7) = -10

Sol :

5x-35=-10

5x=-10+35

$x=\frac{25}{5}=5$

(c) -7x – 11 + 3x = 5

Sol :

-4x=5+11

-4x=16

$x=\frac{16}{-4}=-4$

(d) 6 = -2(7 – c)

Sol :

6=-14+2c

2c=6+14

2c=20

c=10

(e) 5(1 – 2w) + 8w = 15

Sol :

5-10w+8w=15

-2w=15-5

-2w=10

$-2w=\frac{10}{2}$

∴w=-5

(f) 17 = 3(p – 5) + 8

Sol :

17=39-15+8

3p=17+15-8

3p=24

$p=\frac{24}{3}=8$

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Q4 | Ex-6A |Class 8 |S.Chand |Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 4

(a) 3x – 12 + 4x = 6x + 6

Sol :

3x+4x-6x=6+12

x=18

(b) 3(2x – 1) – 2x = -3

Sol :

6x-3-2x=-3

6x-2x=-3+3

4x=0

x=0

(c) 12 – 6x = 6(4 + 2x) + 6

Sol :

12-6x=24+12x+6

-6x-12x=24+6-12

-18x=18

x=-1

(d) 6 + 2(3x – 6) = 3(x + 2)

Sol :

6+6x-12=3x+6

6x-3x=6-6+12

3x=12

$x=\frac{12}{3}=4$

(e) 4(2x – 6) + 3 (x – 3) = 0

Sol :

8x-24+3x-9=0

8x+3x=24+9

11x=33

$x=\frac{33}{11}=3$

(f) 3(2y – 4) + 3y = 5(y – 4)

Sol :

6y-12+3y=5y-20

6y+3y-5y=-20+12

4y=-8

$y=\frac{-8}{4}=-2$

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Q5 | Ex-6A |Class 8 |S.Chand |Composite maths |Linear Equations in One Variable |Ch-6 |myhelper

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Question 5

8 + 2(5m – 8) + 3(m + 2) = 12 + 4(2m – 1)

Sol :

8+10m-16+3m+6=12+8m-4

10m+3m-8m=12-4-8+16-6

5m=+10

$m=-\frac{10}{5}=2$