Class 8 RD Sharma Solutions - Chapter 9 Linear Equation In One Variable - Exercise 9.3 | Set 1

Chapter 9 of RD Sharma's Class 8 Mathematics textbook covers linear equations in one variable. This chapter introduces students to the concept of solving the equations where there is only one variable involved. The Linear equations are fundamental in the algebra and provide the basis for the understanding more complex mathematical concepts.

Linear Equations in One Variable

A linear equation in one variable is an equation of the form ax+b=0, where a and b are constants and x is the variable. The objective is to find the value of the x that satisfies the equation. These equations represent straight lines when graphed and are fundamental for the solving the real-world problems where relationships between the quantities are linear.

Solve the following equations and verify your answer:

Question 1. (2x-3)/(3x+2) = -2/3

Solution:

Given:

(2x-3) / (3x+2) = -2/3

After cross-multiplication we will get,

3(2x – 3) = -2(3x + 2)

6x – 9 = -6x – 4

Now rearrange the above equation

6x + 6x = 9 – 4

12x = 5

x = 5/12

Now verify the given equation by putting x = 5/12

(2x-3) / (3x+2) = -2/3

x = 5/12

(2(5/12) – 3) / (3(5/12) + 2) = -2/3

((5/6)-3) / ((5/4) + 2) = -2/3

((5-18)/6) / ((5+8)/4) = -2/3

(-13/6) / (13/4) = -2/3

(-13/6) × (4/13) = -2/3

-4/6 = -2/3

-2/3 = -2/3

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 2. (2-y)/(y+7) = 3/5

Solution:

Given:

(2-y) / (y+7) = 3/5

After cross-multiplication we will get,

5(2-y) = 3(y+7)

10 – 5y = 3y + 21

Now rearrange the above equation,

10 – 21 = 3y + 5y

8y = – 11

y = -11/8

Now verify the given equation, by putting y = -11/8

(2-y) / (y+7) = 3/5

y = -11/8

(2 – (-11/8)) / ((-11/8) + 7) = 3/5

((16+11)/8) / ((-11+56)/8) = 3/5

(27/8) / (45/8) = 3/5

(27/8) × (8/45) = 3/5

27/45 = 3/5

3/5 = 3/5

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 3. (5x – 7)/(3x) = 2

Solution:

Given:

(5x – 7) / (3x) = 2

After cross-multiplication we will get,

5x – 7 = 2(3x)

5x – 7 = 6x

5x – 6x = 7

-x = 7

x = -7

Now verify the given equation but putting x = -7

(5x – 7) / (3x) = 2

x = -7

(5(-7) – 7) / (3(-7)) = 2

(-35 – 7) / -21 = 2

-42/-21 = 2

2 = 2

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 4. (3x+5)/(2x + 7) = 4

Solution:

Given:

(3x+5) / (2x + 7) = 4

After cross-multiplication we will get,

3x + 5 = 4(2x+7)

3x + 5 = 8x + 28

3x – 8x = 28 – 5

-5x = 23

x = -23/5

Now verify the given equation by putting x =-23/5

(3x+5) / (2x + 7) = 4

x = -23/5

(3(-23/5) + 5) / (2(-23/5) + 7) = 4

(-69/5 + 5) / (-46/5 + 7) = 4

(-69+25)/5 / (-46+35)/5 = 4

-44/5 / -11/5 = 4

-44/5 × 5/-11 = 4

44/11 = 4

4 = 4

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 5. (2y + 5)/(y + 4) = 1

Solution:

Given:

(2y + 5) / (y + 4) = 1

After cross-multiplication we will get,

2y + 5 = y + 4

2y – y = 4 – 5

y = -1

Now verify the given equation by substituting y = -1

(2y + 5) / (y + 4) = 1

y = -1

(2(-1) + 5) / (-1 + 4) = 1

(-2+5) / 3 = 1

3/3 = 1

1 = 1

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 6. (2x + 1)/(3x – 2) = 5/9

Solution:

Given:

(2x + 1) / (3x – 2) = 5/9

After cross-multiplication we will get,

9(2x + 1) = 5(3x – 2)

18x + 9 = 15x – 10

18x – 15x = -10 – 9

3x = -19

x = -19/3

Now verify the given equation by substituting x = -19/3

(2x + 1) / (3x – 2) = 5/9

x = -19/3

(2(-19/3) + 1) / (3(-19/3) – 2) = 5/9

(-38/3 + 1) / (-57/3 – 2) = 5/9

(-38 + 3)/3 / (-57 – 6)/3 = 5/9

-35/3 / -63/3 = 5/9

-35/3 × 3/-63 = 5/9

-35/-63 = 5/9

5/9 = 5/9

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 7. (1 – 9y)/(19 – 3y) = 5/8

Solution:

Given:

(1 – 9y) / (19 – 3y) = 5/8

After cross-multiplication we  willget,

8(1- 9y) = 5(19-3y)

8 – 72y = 95 – 15y

8 – 95 = 72y – 15y

57y = -87

y = -87/57

y = -29/19

Now verify the given equation by substituting y = -29/19

(1 – 9y) / (19 – 3y) = 5/8

y = -29/19

(1 – 9(-29/19)) / (19 – 3(-29/19)) = 5/8

(19+261)/19 / (361+87)/19 = 5/8

280/19 × 19/448 = 5/8

280/ 448 = 5/8

5/8 = 5/8

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 8. 2x/(3x + 1) = 1

Solution:

Given:

2x / (3x + 1) = 1

After cross-multiplication we will get,

2x = 1(3x + 1)

2x = 3x + 1

2x – 3x = 1

-x = 1

x = -1

Now verify the given equation by substituting x = -1

2x / (3x + 1) = 1

x = -1

2(-1) / (3(-1) + 1) = 1

-2 /(-3+1) = 1

-2/-2 = 1

1 = 1

Here, L.H.S. = R.H.S.,

Thus the given equation is verified.

Question 9. y – (7 – 8y)/9y – (3 + 4y) = 2/3

Solution:

Given:

y – (7 – 8y)/9y – (3 + 4y) = 2/3

(y – 7 + 8y) / (9y – 3 – 4y) = 2/3

(-7 + 9y) / (5y – 3) = 2/3

After cross-multiplication we will get,

3(-7 + 9y) = 2(5y – 3)

-21 + 27y = 10y – 6

27y – 10y = 21 – 6

17y = 15

y = 15/17

Now verify the given equation by substituting y = 15/17

y – (7 – 8y)/9y – (3 + 4y) = 2/3

y = 15/17

15/17 – (7-8(15/17))/ 9(15/17) – (3 + 4(15/17)) = 2/3

15/17 – (7 – 120/17) / 135/17 – (3 + 60/17) = 2/3

15/17 – ((119-120)/17) / 135/17 – ((51+60)/17) = 2/3

15/17 – (-1/17) / 135/17 – (111/17) = 2/23

((15 + 1)/17) / ((135-111)/17) = 2/3

16/17 / 24/17 = 2/3

16/24 = 2/3

2/3 = 2/3

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Question 10. 6/2x – (3 – 4x) = 2/3

Solution:

Given:

6/ 2x – (3 – 4x) = 2/3

6/(2x – 3 + 4x) = 2/3

6/(6x – 3) = 2/3

After cross-multiplication we will get,

3(6) = 2(6x – 3)

18 = 12x – 6

12x = 18 + 6

12x = 24

x = 24/12

x = 2

Now verify the given equation by substituting x = 2

6/ 2x – (3 – 4x) = 2/3

6/(6x – 3) = 2/3

x = 2

6/ (6(2) – 3) = 2/3

6/(12-3) = 2/3

6/9 = 2/3

2/3 = 2/3

Here, L.H.S. = R.H.S.,

Thus the given equation is verified.

Question 11. 2/3x – 3/2x = 1/12

Solution:

Given:

2/3x – 3/2x = 1/12

By taking LCM for 2 and 3 which is 6

4-9/6x = 1/12

-5/6x = 1/12

After cross-multiplying we will get,

12(-5) = 1 (6x)

-60 = 6x

x = -60/6

x = -10

Now verify the given equation by substituting x = -10

2/3x – 3/2x = 1/12

x = -10

2/3(-10) – 3/2(-10) = 1/12

2/-30 – 3/-20 = 1/12

-4+6/60 = 1/12

5/60 = 1/12

1/12 = 1/12

Here, L.H.S. = R.H.S., 

Thus, the given equation is verified.

Question 12. (3x + 5)/(4x + 2) = (3x + 4)/(4x + 7)

Solution:

Given:

(3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)

(3x + 5)/ (4x + 2) – (3x + 4)/(4x + 7) = 0

By taking LCM as (4x + 2) (4x + 7)

((3x + 5) (4x + 7) – (3x + 4) (4x + 2)) / (4x + 2) (4x + 7) = 0

After cross-multiplying we will get,

(3x + 5) (4x + 7) – (3x + 4) (4x + 2) = 0

(3x + 5) (4x + 7) – (3x + 4) (4x + 2) = 0

12x² + 21x + 20x + 35 – 12x² – 6x – 16x – 8 = 0

19x + 35 – 8 = 0

19x = -27

x = -27/19

Now verify the given equation by substituting, x = -27/19

(3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)

x = -27/19

(3(-27/19) +5) / (4(-27/19) + 2) = (3(-27/19) + 4) / (4(-27/19) + 7)

(-81/19 + 5) / (-108/19 + 2) = (-81/19 + 4) / (-108/19 + 7)

((-81+95)/19) / ((-108+38)/19) = ((-81+76)/19) / ((-108+133)/19)

14/19 / -70/19 = -5/19 / 25/19

-14/70 = -5/25

-1/5 = -1/5

Here, L.H.S. = R.H.S.,

Thus, the given equation is verified.

Chapter 9 Linear Equation In One Variable - Exercise 9.3 | Set 2

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