In Exercise 9.1 of RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable, we will deal with equations involving rational numbers, as the coefficients and their solutions can also be rational numbers. We will also solve equations having variable terms on one side and numbers on the other side. For Class 8, RD Sharma Solutions is the best reference material for students as it is based on the CBSE syllabus and exam pattern. The expert faculty at BYJU’S have developed the RD Sharma Solutions for Class 8 Maths in a simple and easily understandable language to help students secure high marks in their exams. Download the solutions in PDFs from the links given below.
RD Sharma Solutions for Class 8 Maths Exercise 9.1 Chapter 9 Linear Equation in One Variable
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 9.1 Chapter 9 Linear Equation in One Variable
EXERCISE 9.1 PAGE NO: 9.5
Solve each of the following equations and also verify your solution:
1. 9 ¼ = y – 1 1/3
Solution:
We have,
9 ¼ = y – 1 1/3
37/4 = y – 4/3
Upon solving, we get,
y = 37/4 + 4/3
By taking LCM for 4 and 3, we get 12
y = (37×3)/12 + (4×4)/12
= 111/12 + 16/12
= (111 + 16)/12
= 127/12
∴ y = 127/12
Verification
RHS = y – 1 1/3
= 127/12 – 4/3
= (127 – 16)/12
= 111/12
= 37/4
= 9 ¼
= LHS
2. 5x/3 + 2/5 = 1
Solution:
We have,
5x/3 + 2/5 = 1
5x/3 = 1 – 2/5 (by taking LCM)
= (5-2)/5
By using cross-multiplication, we get,
5x/3 = 3/5
5x = (3×3)/5
x = 9/(5×5)
= 9/25
∴ x = 9/25
Verification
LHS = 5x/3 + 2/5
= 5/3 × 9/25 + 2/5
= 3/5 + 2/5
= (3 + 2)/5
= 5/5
= 1
= RHS
3. x/2 + x/3 + x/4 = 13
Solution:
We have,
x/2 + x/3 + x/4 = 13
let us take LCM for 2, 3 and 4, which is 12
(x×6)/12 + (x×4)/12 + (x×3)/12 = 13
6x/12 + 4x/12 + 3x/12 = 13
(6x+4x+3x)/12 = 13
13x/12 = 13
By using cross-multiplication, we get,
13x = 12×13
x = 156/13
= 12
∴ x = 12
Verification:
LHS = x/2 + x/3 + x/4
= 12/2 + 12/3 + 12/4
= 6 + 4 + 3
= 13
= RHS
4. x/2 + x/8 = 1/8
Solution:
We have,
x/2 + x/8 = 1/8
let us take LCM for 2 and 8, which is 8
(x×4)/8 + (x×1)/8 = 1/8
4x/8 + x/8 = 1/8
5x/8 = 1/8
By using cross-multiplication, we get,
5x = 8/8
5x = 1
x = 1/5
∴ x = 1/5
Verification:
LHS = x/2 + x/8
= (1/5)/2 + (1/5)/8
= 1/10 + 1/40
= (4 + 1)/40
= 5/40
= 1/8
= RHS
5. 2x/3 – 3x/8 = 7/12
Solution:
We have,
2x/3 – 3x/8 = 7/12
By taking LCM for 3 and 8, we get 24
(2x×8)/24 – (3x×3)/24 = 7/12
16x/24 – 9x/24 = 7/12
(16x-9x)/24 = 7/12
7x/24 = 7/12
By using cross-multiplication, we get,
7x×12 = 7×24
x = (7×24)/(7×12)
= 24/12
= 2
∴ x = 2
Verification:
LHS = 2x/3 – 3x/8
= 2(2)/3 – 3(2)/8
= 4/3 – 6/8
= 4/3 – 3/4
= (16 – 9)/ 12
= 7/12
= RHS
6. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0
Solution:
We have,
(x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0
Upon expansion, we get,
x2 + 5x + 6 + x2 – 5x +6 – 2x2 – 2x =0
-2x + 12 = 0
By dividing the equation using -2, we get,
x – 6 = 0
x = 6
∴ x = 6
Verification:
LHS = (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1)
= (6 + 2) (6 + 3) + (6 – 3) (6 – 2) – 2(6) (6 + 1)
= (8) (9) + (3) (4) – 12(7)
= 72 + 12 – 84
= 84 – 84
= 0
= RHS
7. x/2 – 4/5 + x/5 + 3x/10 = 1/5
Solution:
We have,
x/2 – 4/5 + x/5 + 3x/10 = 1/5
upon solving, we get,
x/2 + x/5 + 3x/10 = 1/5 + 4/5
by taking LCM for 2, 5 and 10, which is 10
(x×5)/10 + (x×2)/10 + (3x×1)/10 = 5/5
5x/10 + 2x/10 + 3x/10 = 1
(5x+2x+3x)/10 = 1
10x/10 = 1
x = 1
∴ x = 1
Verification:
LHS = x/2 – 4/5 + x/5 + 3x/10
= ½ – 4/5 + 1/5 + 3(1)/10
= (5 – 8 + 2 + 3)/10
= (10 – 8)/10
= 2/10
= 1/5
= RHS
8. 7/x + 35 = 1/10
Solution:
We have,
7/x + 35 = 1/10
7/x = 1/10 – 35
= ((1×1) – (35×10))/10
= (1 – 350)/10
7/x = -349/10
By using cross-multiplication, we get,
x = -70/349
∴ x = -70/349
Verification:
LHS = 7/x + 35
= 7/(-70/349) + 35
= (-7 × 349)/70 + 35
= -349/10 + 35
= (-349 + 350)/ 10
= 1/10
= RHS
9. (2x-1)/3 – (6x-2)/5 = 1/3
Solution:
We have,
(2x-1)/3 – (6x-2)/5 = 1/3
By taking LCM for 3 and 5, which is 15
((2x-1)×5)/15 – ((6x-2)×3)/15 = 1/3
(10x – 5)/15 – (18x – 6)/15 = 1/3
(10x – 5 – 18x + 6)/15 = 1/3
(-8x + 1)/15 = 1/3
By using cross-multiplication, we get,
(-8x + 1)3 = 15
-24x + 3 = 15
-24x = 15 – 3
-24x = 12
x = -12/24
= -1/2
∴ x = -1/2
Verification:
LHS = (2x – 1)/3 – (6x – 2)/5
= [2(-1/2) – 1]/3 – [6(-1/2) – 2]/5
= (- 1 – 1)/3 – (-3 – 2)/5
= – 2/3 – (-5/5)
= -2/3 + 1
= (-2 + 3)/3
= 1/3
RHS
10. 13(y – 4) – 3(y – 9) – 5(y + 4) = 0
Solution:
We have,
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
Upon expansion, we get,
13y – 52 – 3y + 27 – 5y – 20 = 0
13y – 3y – 5y = 52 – 27 + 20
5y = 45
y = 45/5
= 9
∴ y = 9
Verification:
LHS = 13(y – 4) – 3 (y – 9) – 5 (y + 4)
= 13 (9 – 4) – 3 (9 – 9) – 5 (9 + 4)
= 13 (5) – 3 (0) – 5 (13)
= 65 – 0 – 65
= 0
= RHS
11. 2/3(x – 5) – 1/4(x – 2) = 9/2
Solution:
We have,
2/3(x – 5) – 1/4(x – 2) = 9/2
Upon expansion, we get,
2x/3 – 10/3 – x/4 + 2/4 = 9/2
2x/3 – 10/3 – x/4 + 1/2 = 9/2
2x/3 – x/4 = 9/2 + 10/3 – 1/2
By taking LCM for (3 and 4 is 12) (2 and 3 is 6)
(2x×4)/12 – (x×3)/12 = (9×3)/6 + (10×2)/6 – (1×3)/6
8x/12 – 3x/12 = 27/6 + 20/6 – 3/6
(8x-3x)/12 = (27+20-3)6
5x/12 = 44/6
By using cross-multiplication, we get,
5x×6 = 44×12
30x = 528
x = 528/30
= 264/15
= 88/5
Verification:
LHS = 2/3 (x – 5) – ¼ (x – 2)
= 2/3 [(88/5) – 5] – ¼ [(88/5) – 2]
= 2/3 [(88 – 25)/5] – ¼ [(88 – 10)/5]
= 2/3 × 63/5 – ¼ × 78/5
= 42/5 – 39/10
= (84 – 39)/10
= 45/10
= 9/2
= RHS
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