RD Sharma Solutions for Class 8 Maths Exercise 9.3 Chapter 9 Linear Equation in One Variable is provided here. In Exercise 9.3, we will discuss the concept of the cross-multiplication method to solve equations. RD Sharma Solutions, which is designed by the subject experts at BYJU’S, helps students understand the concepts clearly. On regular practice, students can excel in their exams. Students can download the solutions in PDFs easily from the links provided below.
RD Sharma Solutions for Class 8 Maths Exercise 9.3 Chapter 9 Linear Equation in One Variable
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EXERCISE 9.3 PAGE NO: 9.17
Solve the following equations and verify your answer:
1. (2x-3) / (3x+2) = -2/3
Solution:
We have,
(2x-3) / (3x+2) = -2/3
Let us perform cross-multiplication, and we get,
3(2x – 3) = -2(3x + 2)
6x – 9 = -6x – 4
When rearranged,
6x + 6x = 9 – 4
12x = 5
x = 5/12
Now, let us verify the given equation,
(2x-3) / (3x+2) = -2/3
By substituting the value of ‘x’, we get,
(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
Hence, the given equation is verified.
2. (2-y) / (y+7) = 3/5
Solution:
We have,
(2-y) / (y+7) = 3/5
Let us perform cross-multiplication, and we get,
5(2-y) = 3(y+7)
10 – 5y = 3y + 21
When rearranged,
10 – 21 = 3y + 5y
8y = – 11
y = -11/8
Now, let us verify the given equation,
(2-y) / (y+7) = 3/5
By substituting the value of ‘x’, we get,
(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
Hence, the given equation is verified.
3. (5x – 7) / (3x) = 2
Solution:
We have,
(5x – 7) / (3x) = 2
Let us perform cross-multiplication, and we get,
5x – 7 = 2(3x)
5x – 7 = 6x
5x – 6x = 7
-x = 7
x = -7
Now, let us verify the given equation,
(5x – 7) / (3x) = 2
By substituting the value of ‘x’, we get,
(5(-7) – 7) / (3(-7)) = 2
(-35 – 7) / -21 = 2
-42/-21 = 2
2 = 2
Hence, the given equation is verified.
4. (3x+5) / (2x + 7) = 4
Solution:
We have,
(3x+5) / (2x + 7) = 4
Let us perform cross-multiplication, and we get,
3x + 5 = 4(2x+7)
3x + 5 = 8x + 28
3x – 8x = 28 – 5
-5x = 23
x = -23/5
Now, let us verify the given equation,
(3x+5) / (2x + 7) = 4
By substituting the value of ‘x’, we get,
(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
Hence, the given equation is verified.
5. (2y + 5) / (y + 4) = 1
Solution:
We have,
(2y + 5) / (y + 4) = 1
Let us perform cross-multiplication, and we get,
2y + 5 = y + 4
2y – y = 4 – 5
y = -1
Now let us verify the given equation,
(2y + 5) / (y + 4) = 1
By substituting the value of ‘y’, we get,
(2(-1) + 5) / (-1 + 4) = 1
(-2+5) / 3 = 1
3/3 = 1
1 = 1
Hence, the given equation is verified.
6. (2x + 1) / (3x – 2) = 5/9
Solution:
We have,
(2x + 1) / (3x – 2) = 5/9
Let us perform cross-multiplication, and we get,
9(2x + 1) = 5(3x – 2)
18x + 9 = 15x – 10
18x – 15x = -10 – 9
3x = -19
x = -19/3
Now, let us verify the given equation,
(2x + 1) / (3x – 2) = 5/9
By substituting the value of ‘x’, we get,
(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
Hence, the given equation is verified.
7. (1 – 9y) / (19 – 3y) = 5/8
Solution:
We have,
(1 – 9y) / (19 – 3y) = 5/8
Let us perform cross-multiplication, and we get,
8(1- 9y) = 5(19-3y)
8 – 72y = 95 – 15y
8 – 95 = 72y – 15y
57y = -87
y = -87/57
= -29/19
Now, let us verify the given equation,
(1 – 9y) / (19 – 3y) = 5/8
By substituting the value of ‘y’, we get,
(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
Hence, the given equation is verified.
8. 2x / (3x + 1) = 1
Solution:
We have,
2x / (3x + 1) = 1
Let us perform cross-multiplication, and we get,
2x = 1(3x + 1)
2x = 3x + 1
2x – 3x = 1
-x = 1
x = -1
Now, let us verify the given equation,
2x / (3x + 1) = 1
By substituting the value of ‘x’, we get,
2(-1) / (3(-1) + 1) = 1
-2 /(-3+1) = 1
-2/-2 = 1
1 = 1
Hence, the given equation is verified.
9. y – (7 – 8y)/9y – (3 + 4y) = 2/3
Solution:
We have,
y – (7 – 8y)/9y – (3 + 4y) = 2/3
(y – 7 + 8y) / (9y – 3 – 4y) = 2/3
(-7 + 9y) / (5y – 3) = 2/3
Let us perform cross-multiplication, and we get,
3(-7 + 9y) = 2(5y – 3)
-21 + 27y = 10y – 6
27y – 10y = 21 – 6
17y = 15
y = 15/17
Now, let us verify the given equation,
y – (7 – 8y)/9y – (3 + 4y) = 2/3
By substituting the value of ‘y’, we get,
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
Hence, the given equation is verified.
10. 6/ 2x – (3 – 4x) = 2/3
Solution:
We have,
6/ 2x – (3 – 4x) = 2/3
6/(2x – 3 + 4x) = 2/3
6/(6x – 3) = 2/3
Let us perform cross-multiplication, and we get,
3(6) = 2(6x – 3)
18 = 12x – 6
12x = 18 + 6
12x = 24
x = 24/12
= 2
Now, let us verify the given equation,
6/ 2x – (3 – 4x) = 2/3
6/(6x – 3) = 2/3
By substituting the value of ‘x’, we get,
6/ (6(2) – 3) = 2/3
6/(12-3) = 2/3
6/9 = 2/3
2/3 = 2/3
Hence, the given equation is verified.
11. 2/3x – 3/2x = 1/12
Solution:
We have,
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
By cross-multiplying, we get,
12(-5) = 1 (6x)
-60 = 6x
x = -60/6
= -10
Now, let us verify the given equation,
2/3x – 3/2x = 1/12
By substituting the value of ‘x’, we get,
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
Hence, the given equation is verified.
12. (3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)
Solution:
We have,
(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
By cross-multiplying, we 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, let us verify the given equation,
(3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)
By substituting the value of ‘x’, we get,
(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
Hence, the given equation is verified.
13. (7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
Solution:
We have,
(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
(7x – 2) / (5x – 1) – (7x +3)/(5x + 4) = 0
By taking LCM as (5x – 1) (5x + 4)
((7x-2) (5x+4) – (7x+3)(5x-1)) / (5x – 1) (5x + 4) = 0
By cross-multiplying, we get,
(7x-2) (5x+4) – (7x+3)(5x-1) = 0
Upon simplification,
35x2 + 28x – 10x – 8 – 35x2 + 7x – 15x + 3 = 0
10x – 5 = 0
10x = 5
x = 5/10
= 1/2
Now, let us verify the given equation,
(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
By substituting the value of ‘x’, we get,
(7(1/2) – 2) / (5(1/2) – 1) = (7(1/2) + 3) /(5(1/2) + 4)
(7/2 – 2) / (5/2 – 1) = (7/2 + 3) / (5/2 + 4)
((7-4)/2) / ((5-2)/2) = ((7+6)/2) / ((5+8)/2)
(3/2) / (3/2) = (13/2) / (13/2)
1 = 1
Hence, the given equation is verified.
14. ((x+1)/(x+2))2 = (x+2) / (x + 4)
Solution:
We have,
((x+1)/(x+2))2 = (x+2) / (x + 4)
(x+1)2 / (x+2)2 – (x+2) / (x + 4) = 0
By taking LCM as (x+2)2 (x+4)
((x+1)2 (x+4) – (x+2) (x+2)2) / (x+2)2 (x+4) = 0
By cross-multiplying, we get,
(x+1)2 (x+4) – (x+2) (x+2)2 = 0
Let us expand the equation
(x2 + 2x + 1) (x + 4) – (x + 2) (x2 + 4x + 4) = 0
x3 + 2x2 + x + 4x2 + 8x + 4 – (x3 + 4x2 + 4x + 2x2 + 8x + 8) = 0
x3 + 2x2 + x + 4x2 + 8x + 4 – x3 – 4x2 – 4x – 2x2 – 8x – 8 = 0
-3x – 4 = 0
x = -4/3
Now let us verify the given equation,
((x+1)/(x+2))2 = (x+2) / (x + 4)
By substituting the value of ‘x’, we get,
(x+1)2 / (x+2)2 = (x+2) / (x + 4)
(-4/3 + 1)2 / (-4/3 + 2)2 = (-4/3 + 2) / (-4/3 + 4)
((-4+3)/3)2 / ((-4+6)/3)2 = ((-4+6)/3) / ((-4+12)/3)
(-1/3)2 / (2/3)2 = (2/3) / (8/3)
1/9 / 4/9 = 2/3 / 8/3
1/4 = 2/8
1/4 = 1/4
Hence, the given equation is verified.
15. ((x+1)/(x-4))2 = (x+8)/(x-2)
Solution:
We have,
((x+1)/(x-4))2 = (x+8)/(x-2)
(x+1)2 / (x-4)2 – (x+8) / (x-2) = 0
By taking LCM as (x-4)2 (x-2)
((x+1)2 (x-2) – (x+8) (x-4)2) / (x-4)2 (x-2) = 0
By cross-multiplying, we get,
(x+1)2 (x-2) – (x+8) (x-4)2 = 0
Upon expansion, we get,
(x2 + 2x + 1) (x-2) – ((x+8) (x2 – 8x + 16)) = 0
x3 + 2x2 + x – 2x2 – 4x – 2 – (x3 – 8x2 + 16x + 8x2 – 64x + 128) = 0
x3 + 2x2 + x – 2x2 – 4x – 2 – x3 + 8x2 – 16x – 8x2 + 64x – 128 = 0
45x – 130 = 0
x = 130/45
= 26/9
Now, let us verify the given equation,
((x+1)/(x-4))2 = (x+8)/(x-2)
(x+1)2 / (x-4)2 = (x+8) / (x-2)
By substituting the value of ‘x’, we get,
(26/9 + 1)2 / (26/9 – 4)2 = (26/9 + 8) / (26/9 – 2)
((26+9)/9)2 / ((26-36)/9)2 = ((26+72)/9) / ((26-18)/9)
(35/9)2 / (-10/9)2 = (98/9) / (8/9)
(35/-10)2 = (98/8)
(7/2)2 = 49/4
49/4 = 49/4
Hence, the given equation is verified.
16. (9x-7)/(3x+5) = (3x-4)/(x+6)
Solution:
We have,
(9x-7)/(3x+5) = (3x-4)/(x+6)
(9x-7)/(3x+5) – (3x-4)/(x+6) = 0
By taking LCM as (3x+5) (x+6)
((9x-7) (x+6) – (3x-4) (3x+5)) / (3x+5) (x+6) = 0
By cross-multiplying, we get,
(9x-7) (x+6) – (3x-4) (3x+5) = 0
Upon expansion, we get,
9x2 + 54x – 7x – 42 – (9x2 + 15x – 12x – 20) = 0
44x – 22 = 0
44x = 22
x = 22/44
= 2/4
= 1/2
Now, let us verify the given equation,
(9x-7)/(3x+5) = (3x-4)/(x+6)
By substituting the value of ‘x’, we get,
(9(1/2) – 7) / (3(1/2) + 5) = (3(1/2) – 4) / ((1/2) + 6)
(9/2 – 7) / (3/2 + 5) = (3/2 – 4) / (1/2 + 6)
((9-14)/2) / ((3+10)/2) = ((3-8)/2) / ((1+12)/2)
-5/2 / 13/2 = -5/2 / 13/2
-5/13 = -5/13
Hence, the given equation is verified.
17. (x+2)/(x+5) = x/(x+6)
Solution:
We have,
(x+2)/(x+5) = x/(x+6)
(x+2)/(x+5) – x/(x+6) = 0
By taking LCM as (x+5) (x+6)
((x+2) (x+6) – x(x+5)) / (x+5) (x+6) = 0
By cross-multiplying, we get,
(x+2) (x+6) – x(x+5) = 0
Upon expansion,
x2 + 8x + 12 – x2 – 5x = 0
3x + 12 = 0
3x = -12
x = -12/3
= -4
Now let us verify the given equation,
(x+2)/(x+5) = x/(x+6)
By substituting the value of ‘x’, we get,
(-4 + 2) / (-4 + 5) = -4 / (-4 + 6)
-2/1 = -4 / (2)
-2 = -2
Hence, the given equation is verified.
18. 2x – (7-5x) / 9x – (3+4x) = 7/6
Solution:
We have,
2x – (7-5x) / 9x – (3+4x) = 7/6
(2x – 7 + 5x) / (9x – 3 – 4x) = 7/6
(7x – 7) / (5x – 3) = 7/6
By cross-multiplying, we get,
6(7x – 7) = 7(5x – 3)
42x – 42 = 35x – 21
42x – 35x = -21 + 42
7x = 21
x = 21/7
= 3
Now, let us verify the given equation,
2x – (7-5x) / 9x – (3+4x) = 7/6
(7x – 7) / (5x – 3) = 7/6
By substituting the value of ‘x’, we get,
(7(3) -7) / (5(3) – 3) = 7/6
(21-7) / (15-3) = 7/6
14/12 = 7/6
7/6 = 7/6
Hence, the given equation is verified.
19. (15(2-x) – 5(x+6)) / (1-3x) = 10
Solution:
We have,
15(2-x) – 5(x+6) / (1-3x) = 10
(30-15x) – (5x + 30) / (1-3x) = 10
By cross-multiplying, we get,
(30-15x) – (5x + 30) = 10(1- 3x)
30- 15x – 5x – 30 = 10 – 30x
30- 15x – 5x – 30 + 30x = 10
10x = 10
x = 10/10
= 1
Now let us verify the given equation,
(15(2-x) – 5(x+6)) / (1-3x) = 10
By substituting the value of ‘x’, we get,
(15(2-1) – 5(1+6)) / (1- 3) = 10
(15 – 5(7))/-2 = 10
(15-35)/-2 = 10
-20/-2 = 10
10 = 10
Hence, the given equation is verified.
20. (x+3)/(x-3) + (x+2)/(x-2) = 2
Solution:
We have,
(x+3)/(x-3) + (x+2)/(x-2) = 2
By taking LCM as (x-3) (x-2)
((x+3)(x-2) + (x+2) (x-3)) / (x-3) (x-2) = 2
By cross-multiplying, we get,
(x+3)(x-2) + (x+2) (x-3) = 2 ((x-3) (x-2))
Upon expansion,
x2 + 3x – 2x – 6 + x2 – 3x + 2x – 6 = 2(x2 – 3x – 2x + 6)
2x2 – 12 = 2x2 – 10x + 12
2x2 – 2x2 + 10x = 12 + 12
10x = 24
x = 24/10
= 12/5
Now, let us verify the given equation,
(x+3)/(x-3) + (x+2)/(x-2) = 2
By substituting the value of ‘x’, we get,
(12/5 + 3)/(12/5 – 3) + (12/5 + 2)/(12/5 – 2) = 2
((12+15)/5)/((12-15)/5) + ((12+10)/5)/((12-10)/5) = 2
(27/5)/(-3/5) + (22/5)/(2/5) = 2
-27/3 + 22/2 = 2
((-27×2) + (22×3))/6 = 2
(-54 + 66)/6 = 2
12/6 = 2
2 = 2
Hence, the given equation is verified.
21. ((x+2) (2x-3) – 2x2 + 6)/(x-5) = 2
Solution:
We have,
((x+2) (2x-3) – 2x2 + 6)/(x-5) = 2
By cross-multiplying, we get,
(x+2) (2x-3) – 2x2 + 6) = 2(x-5)
2x2 – 3x + 4x – 6 – 2x2 + 6 = 2x – 10
x = 2x – 10
x – 2x = -10
-x = -10
x = 10
Now let us verify the given equation,
((x+2) (2x-3) – 2x2 + 6)/(x-5) = 2
By substituting the value of ‘x’, we get,
((10+2) (2(10) – 3) – 2(10)2 + 6)/ (10-5) = 2
(12(17) – 200 + 6)/5 = 2
(204 – 194)/5 = 2
10/5 = 2
2 = 2
Hence, the given equation is verified
22. (x2 – (x+1) (x+2))/(5x+1) = 6
Solution:
We have,
(x2 – (x+1) (x+2))/(5x+1) = 6
By cross-multiplying, we get,
(x2 – (x+1) (x+2)) = 6(5x+1)
x2 – x2 – 2x – x – 2 = 30x + 6
-3x – 2 = 30x + 6
30x + 3x = -2 – 6
33x = -8
x = -8/33
Now, let us verify the given equation,
(x2 – (x+1) (x+2))/(5x+1) = 6
By substituting the value of ‘x’, we get,
((-8/33)2 – ((-8/33)+1) (-8/33 + 2))/(5(-8/33)+1) = 6
(64/1089 – ((-8+33)/33) ((-8+66)/33)) / (-40+33)/33) = 6
(64/1089 – (25/33) (58/33)) / (-7/33) = 6
(64/1089 – 1450/1089) / (-7/33) = 6
((64-1450)/1089 / (-7/33)) = 6
-1386/1089 × 33/-7 = 6
1386 × 33 / 1089 × -7 = 6
6 = 6
Hence, the given equation is verified.
23. ((2x+3) – (5x-7))/(6x+11) = -8/3
Solution:
We have,
((2x+3) – (5x-7))/(6x+11) = -8/3
By cross-multiplying, we get,
3((2x+3) – (5x-7)) = -8(6x+11)
3(2x + 3 – 5x + 7) = -48x – 88
3(-3x + 10) = -48x – 88
-9x + 30 = -48x – 88
-9x + 48x = -88 – 30
39x = -118
x = -118/39
Now, let us verify the given equation,
((2x+3) – (5x-7))/(6x+11) = -8/3
By substituting the value of ‘x’, we get,
((2(-118/39) + 3) – (5(-118/39) – 7)) / (6(-118/39) + 11) = -8/3
((-336/39 + 3) – (-590/39 – 7)) / (-708/39 + 11) = -8/3
(((-336+117)/39) – ((-590-273)/39)) / ((-708+429)/39) = -8/3
(-219+863)/39 / (-279)/39 = -8/3
644/-279 = -8/3
-8/3 = -8/3
Hence, the given equation is verified.
24. Find the positive value of x for which the given equation is satisfied:
(i) (x2 – 9)/(5+x2) = -5/9
Solution:
We have,
(x2 – 9)/(5+x2) = -5/9
By cross-multiplying, we get,
9(x2 – 9) = -5(5+x2)
9x2 – 81 = -25 – 5x2
9x2 + 5x2 = -25 + 81
14x2 = 56
x2 = 56/14
x2 = 4
x = √4
= 2
(ii) (y2 + 4)/(3y2 + 7) = 1/2
Solution:
We have,
(y2 + 4)/(3y2 + 7) = 1/2
By cross-multiplying, we get,
2(y2 + 4) = 1(3y2 + 7)
2y2 + 8 = 3y2 + 7
3y2 – 2y2 = 7 – 8
y2 = -1
y = √-1
= 1
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