Selina Textbook Solutions: Linear Equations in one Variable | Mathematics Class 8 ICSE PDF Download

 Page 1


14. Linear Equations in one Variable 
(With Problems Based on Linear equations) 
EXERCISE 14(A) 
Solve the following equations: 
Question 1. 
20 = 6 + 2x 
Solution: 
20 = 6 + 2x 
20 – 6 = 2x 
14 = 2x 
7 = x 
x = 7 
Question 2. 
15 + x = 5x + 3 
Solution: 
15 – 3 = 5x – x 
12 = 4x 
3 = x 
x = 3 
Question 3. 
 = -7 
Solution: 
3x + 2 = -7 (x – 6) (by cross multiplying) 
3x + 2 = -7x + 42 
3x + 7x = 42 – 2 
10x = 40 
x = 4 
Question 4. 
3a – 4 = 2(4 – a) 
Solution: 
3a – 4 = 8 – 2a 
3a + 2a = 8+4 
5a = 12 
a = 2.4 
Question 5. 
3(b – 4) = 2(4 – b) 
Page 2


14. Linear Equations in one Variable 
(With Problems Based on Linear equations) 
EXERCISE 14(A) 
Solve the following equations: 
Question 1. 
20 = 6 + 2x 
Solution: 
20 = 6 + 2x 
20 – 6 = 2x 
14 = 2x 
7 = x 
x = 7 
Question 2. 
15 + x = 5x + 3 
Solution: 
15 – 3 = 5x – x 
12 = 4x 
3 = x 
x = 3 
Question 3. 
 = -7 
Solution: 
3x + 2 = -7 (x – 6) (by cross multiplying) 
3x + 2 = -7x + 42 
3x + 7x = 42 – 2 
10x = 40 
x = 4 
Question 4. 
3a – 4 = 2(4 – a) 
Solution: 
3a – 4 = 8 – 2a 
3a + 2a = 8+4 
5a = 12 
a = 2.4 
Question 5. 
3(b – 4) = 2(4 – b) 
Solution: 
 
Question 6. 
 
Solution: 
 
Question 7. 
 
Solution: 
 
Question 8. 
5(8x + 3) = 9(4x + 7) 
Page 3


14. Linear Equations in one Variable 
(With Problems Based on Linear equations) 
EXERCISE 14(A) 
Solve the following equations: 
Question 1. 
20 = 6 + 2x 
Solution: 
20 = 6 + 2x 
20 – 6 = 2x 
14 = 2x 
7 = x 
x = 7 
Question 2. 
15 + x = 5x + 3 
Solution: 
15 – 3 = 5x – x 
12 = 4x 
3 = x 
x = 3 
Question 3. 
 = -7 
Solution: 
3x + 2 = -7 (x – 6) (by cross multiplying) 
3x + 2 = -7x + 42 
3x + 7x = 42 – 2 
10x = 40 
x = 4 
Question 4. 
3a – 4 = 2(4 – a) 
Solution: 
3a – 4 = 8 – 2a 
3a + 2a = 8+4 
5a = 12 
a = 2.4 
Question 5. 
3(b – 4) = 2(4 – b) 
Solution: 
 
Question 6. 
 
Solution: 
 
Question 7. 
 
Solution: 
 
Question 8. 
5(8x + 3) = 9(4x + 7) 
Solution: 
 
Question 9. 
3(x +1) = 12 + 4(x – 1) 
Solution: 
3(x + 1) = 12 + 4(x – 1) 
3x + 3 = 12 + 4x – 4 
3x – 4x = 12 – 4 – 3 
-x = 5 
x = -5 
Question 10. 
 
Solution: 
 
Question 11. 
 
Page 4


14. Linear Equations in one Variable 
(With Problems Based on Linear equations) 
EXERCISE 14(A) 
Solve the following equations: 
Question 1. 
20 = 6 + 2x 
Solution: 
20 = 6 + 2x 
20 – 6 = 2x 
14 = 2x 
7 = x 
x = 7 
Question 2. 
15 + x = 5x + 3 
Solution: 
15 – 3 = 5x – x 
12 = 4x 
3 = x 
x = 3 
Question 3. 
 = -7 
Solution: 
3x + 2 = -7 (x – 6) (by cross multiplying) 
3x + 2 = -7x + 42 
3x + 7x = 42 – 2 
10x = 40 
x = 4 
Question 4. 
3a – 4 = 2(4 – a) 
Solution: 
3a – 4 = 8 – 2a 
3a + 2a = 8+4 
5a = 12 
a = 2.4 
Question 5. 
3(b – 4) = 2(4 – b) 
Solution: 
 
Question 6. 
 
Solution: 
 
Question 7. 
 
Solution: 
 
Question 8. 
5(8x + 3) = 9(4x + 7) 
Solution: 
 
Question 9. 
3(x +1) = 12 + 4(x – 1) 
Solution: 
3(x + 1) = 12 + 4(x – 1) 
3x + 3 = 12 + 4x – 4 
3x – 4x = 12 – 4 – 3 
-x = 5 
x = -5 
Question 10. 
 
Solution: 
 
Question 11. 
 
Solution: 
 
Question 12. 
 
Solution: 
 
 
Page 5


14. Linear Equations in one Variable 
(With Problems Based on Linear equations) 
EXERCISE 14(A) 
Solve the following equations: 
Question 1. 
20 = 6 + 2x 
Solution: 
20 = 6 + 2x 
20 – 6 = 2x 
14 = 2x 
7 = x 
x = 7 
Question 2. 
15 + x = 5x + 3 
Solution: 
15 – 3 = 5x – x 
12 = 4x 
3 = x 
x = 3 
Question 3. 
 = -7 
Solution: 
3x + 2 = -7 (x – 6) (by cross multiplying) 
3x + 2 = -7x + 42 
3x + 7x = 42 – 2 
10x = 40 
x = 4 
Question 4. 
3a – 4 = 2(4 – a) 
Solution: 
3a – 4 = 8 – 2a 
3a + 2a = 8+4 
5a = 12 
a = 2.4 
Question 5. 
3(b – 4) = 2(4 – b) 
Solution: 
 
Question 6. 
 
Solution: 
 
Question 7. 
 
Solution: 
 
Question 8. 
5(8x + 3) = 9(4x + 7) 
Solution: 
 
Question 9. 
3(x +1) = 12 + 4(x – 1) 
Solution: 
3(x + 1) = 12 + 4(x – 1) 
3x + 3 = 12 + 4x – 4 
3x – 4x = 12 – 4 – 3 
-x = 5 
x = -5 
Question 10. 
 
Solution: 
 
Question 11. 
 
Solution: 
 
Question 12. 
 
Solution: 
 
 
Question 13. 
 
Solution: 
 
Question 14. 
 
Solution: 
 
 
Question 15.