Mathematics is one of the scoring subjects where students secure maximum marks in the exam. When it comes to preparing for the annual exam, it is the toughest time when most students struggle to solve problems. So, here at BYJU’S, our expert faculty have formulated RD Sharma Class 8 Maths Solutions, which help students prepare for their exams effortlessly. All the solutions are well designed, keeping in mind the latest CBSE syllabus and exam pattern. Also, students can learn easy tricks and shortcut methods by practising these solutions on a regular basis. The PDFs of this chapter are available here, and students can download them for free from the links provided below.
Chapter 9 – Linear Equation in One Variable contains four exercises, and the RD Sharma Solutions available on this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts covered in this chapter.
- Linear equation and its definitions.
- A solution of a linear equation.
- Solving equations having variable terms on one side and number(s) on the other side.
- Transposition method for solving linear equations in one variable.
- Cross-multiplication method for solving equations.
- Applications of linear equations to practical problems.
RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equation in One Variable
Access Answers to Maths RD Sharma Solutions for Class 8 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
EXERCISE 9.2 PAGE NO: 9.11
Solve each of the following equations and also check your results in each case:
1. (2x+5)/3 = 3x – 10
Solution:
(2x+5)/3 = 3x – 10
Let us simplify,
(2x+5)/3 – 3x = – 10
By taking LCM
(2x + 5 – 9x)/3 = -10
(-7x + 5)/3 = -10
By using cross-multiplication, we get,
-7x + 5 = -30
-7x = -30 – 5
-7x = -35
x = -35/-7
= 5
Let us verify the given equation now,
(2x+5)/3 = 3x – 10
By substituting the value of ‘x’, we get,
(2×5 + 5)/3 = 3(5) – 10
(10+5)/3 = 15-10
15/3 = 5
5 = 5
Hence, the given equation is verified
2. (a-8)/3 = (a-3)/2
Solution:
(a-8)/3 = (a-3)/2
By using cross-multiplication, we get,
(a-8)2 = (a-3)3
2a – 16 = 3a – 9
2a – 3a = -9 + 16
-a = 7
a = -7
Let us verify the given equation now,
(a-8)/3 = (a-3)/2
By substituting the value of ‘a’ we get,
(-7 – 8)/3 = (-7 – 3)/2
-15/3 = -10/2
-5 = -5
Hence, the given equation is verified
3. (7y + 2)/5 = (6y – 5)/11
Solution:
(7y + 2)/5 = (6y – 5)/11
By using cross-multiplication, we get,
(7y + 2)11 = (6y – 5)5
77y + 22 = 30y – 25
77y – 30y = -25 – 22
47y = -47
y = -47/47
y = -1
Let us verify the given equation now,
(7y + 2)/5 = (6y – 5)/11
By substituting the value of ‘y’, we get,
(7(-1) + 2)/5 = (6(-1) – 5)/11
(-7 + 2)/5 = (-6 – 5)/11
-5/5 = -11/11
-1 = -1
Hence, the given equation is verified
4. x – 2x + 2 – 16/3x + 5 = 3 – 7/2x
Solution:
x – 2x + 2 – 16/3x + 5 = 3 – 7/2x
Let us rearrange the equation
x – 2x – 16x/3 + 7x/2 = 3 – 2 – 5
By taking LCM for 2 and 3, which is 6
(6x – 12x – 32x + 21x)/6 = -4
-17x/6 = -4
By cross-multiplying
-17x = -4×6
-17x = -24
x = -24/-17
x = 24/17
Let us verify the given equation now,
x – 2x + 2 – 16/3x + 5 = 3 – 7/2x
By substituting the value of ‘x’, we get,
24/17 – 2(24/17) + 2 – (16/3)(24/17) + 5 = 3 – (7/2)(24/17)
24/17 – 48/17 + 2 – 384/51 + 5 = 3 – 168/34
By taking 51 and 17 as the LCM we get,
(72 – 144 + 102 – 384 + 255)/51 = (102 – 168)/34
-99/51 = -66/34
-33/17 = -33/17
Hence, the given equation is verified
5. 1/2x + 7x – 6 = 7x + 1/4
Solution:
1/2x + 7x – 6 = 7x + 1/4
Let us rearrange the equation
1/2x + 7x – 7x = 1/4 + 6 (by taking LCM)
1/2x = (1+ 24)/4
1/2x = 25/4
By cross-multiplying
4x = 25 × 2
4x = 50
x = 50/4
x = 25/2
Let us verify the given equation now,
1/2x + 7x – 6 = 7x + 1/4
By substituting the value of ‘x’, we get,
(1/2) (25/2) + 7(25/2) – 6 = 7(25/2) + 1/4
25/4 + 175/2 – 6 = 175/2 + 1/4
By taking LCM for 4 and 2 is 4
(25 + 350 – 24)/4 = (350+1)/4
351/4 = 351/4
Hence, the given equation is verified
6. 3/4x + 4x = 7/8 + 6x – 6
Solution:
3/4x + 4x = 7/8 + 6x – 6
Let us rearrange the equation
3/4x + 4x – 6x = 7/8 – 6
By taking 4 and 8 as LCM
(3x + 16x – 24x)/4 = (7 – 48)/8
-5x/4 = -41/8
By cross-multiplying
-5x(8) = -41(4)
-40x = -164
x = -164/-40
= 82/20
= 41/10
Let us verify the given equation now,
3/4x + 4x = 7/8 + 6x – 6
By substituting the value of ‘x’, we get,
(3/4)(41/10) + 4(41/10) = 7/8 + 6(41/10) – 6
123/40 + 164/10 = 7/8 + 246/10 – 6
(123 + 656)/40 = (70 + 1968 – 480)/80
779/40 = 1558/80
779/40 = 779/40
Hence, the given equation is verified
7. 7x/2 – 5x/2 = 20x/3 + 10
Solution:
7x/2 – 5x/2 = 20x/3 + 10
Let us rearrange the equation
7x/2 – 5x/2 – 20x/3 = 10
By taking LCM for 2 and 3 is 6
(21x – 15x – 40x)/6 = 10
-34x/6 = 10
By cross-multiplying
-34x = 60
x = 60/-34
= -30/17
Let us verify the given equation now,
7x/2 – 5x/2 = 20x/3 + 10
By substituting the value of ‘x’, we get,
(7-/2)(-30/17) – (5/2)(-30/17) = (20/3)(-30/17) + 10
-210/34 +150/34 = -600/51 + 10
-30/17 = (-600+510)/51
= -90/51
-30/17 = -30/17
Hence, the given equation is verified
8. (6x+1)/2 + 1 = (7x-3)/3
Solution:
(6x+1)/2 + 1 = (7x-3)/3
(6x + 1 + 2)/2 = (7x – 3)/3
By cross-multiplying
(6x + 3)3 = (7x – 3)2
18x + 9 = 14x – 6
18x – 14x = -6 – 9
4x = -15
x = -15/4
Let us verify the given equation now,
(6x+1)/2 + 1 = (7x-3)/3
By substituting the value of ‘x’, we get,
(6(-15/4) + 1)/2 + 1 = (7(-15/4) – 3)/3
(3(-15/2) + 1)/2 + 1 = (-105/4 -3)/3
(-45/2 + 1)/2 + 1 = (-117/4)/3
(-43/4) + 1 = -117/12
(-43+4)/4 = -39/4
-39/4 = -39/4
Hence, the given equation is verified
9. (3a-2)/3 + (2a+3)/2 = a + 7/6
Solution:
(3a-2)/3 + (2a+3)/2 = a + 7/6
Let us rearrange the equation
(3a-2)/3 + (2a+3)/2 – a = 7/6
By taking LCM for 2 and 3, which is 6
((3a-2)2 + (2a+3)3 – 6a)/6 = 7/6
(6a – 4 + 6a + 9 – 6a)/6 = 7/6
(6a + 5)/6 = 7/6
6a + 5 = 7
6a = 7-5
6a = 2
a = 2/6
a = 1/3
Let us verify the given equation now,
(3a-2)/3 + (2a+3)/2 = a + 7/6
By substituting the value of ‘a’, we get,
(3(1/3)-2)/3 + (2(1/3) + 3)/2 = 1/3 + 7/6
(1-2)/3 + (2/3 + 3)/2 = (2+7)/6
-1/3 + (11/3)/2 = 9/6
-1/3 + 11/6 = 3/2
(-2+11)/6 = 3/2
9/6 = 3/2
3/2 = 3/2
Hence, the given equation is verified
10. x – (x-1)/2 = 1 – (x-2)/3
Solution:
x – (x-1)/2 = 1 – (x-2)/3
Let us rearrange the equation
x – (x-1)/2 + (x-2)/3 = 1
By taking LCM for 2 and 3, which is 6
(6x – (x-1)3 + (x-2)2)/6 = 1
(6x – 3x + 3 + 2x – 4)/6 = 1
(5x – 1)/6 = 1
By cross-multiplying
5x – 1 = 6
5x = 6 + 1
x = 7/5
Let us verify the given equation now,
x – (x-1)/2 = 1 – (x-2)/3
By substituting the value of ‘x’, we get,
7/5 – (7/5 – 1)/2 = 1 – (7/5 – 2)/3
7/5 – (2/5)/2 = 1 – (-3/5)/3
7/5 – 2/10 = 1 + 3/15
(14 – 2)/10 = (15+3)/15
12/10 = 18/15
6/5 = 6/5
Hence, the given equation is verified
11. 3x/4 – (x-1)/2 = (x-2)/3
Solution:
3x/4 – (x-1)/2 = (x-2)/3
Let us rearrange the equation
3x/4 – (x-1)/2 – (x-2)/3 = 0
By taking LCM for 4, 2 and 3, which is 12
(9x – (x-1)6 – (x-2)4)/12 = 0
(9x – 6x + 6 – 4x + 8)/12 = 0
(-x + 14)/12 = 0
By cross-multiplying
-x + 14 = 0
x = 14
Let us verify the given equation now,
3x/4 – (x-1)/2 = (x-2)/3
By substituting the value of ‘x’, we get,
3(14)/4 – (14-1)/2 = (14-2)/3
42/4 – 13/2 = 12/3
(42 – 26)/4 = 4
16/4 = 4
4 = 4
Hence, the given equation is verified
12. 5x/3 – (x-1)/4 = (x-3)/5
Solution:
5x/3 – (x-1)/4 = (x-3)/5
Let us rearrange the equation
5x/3 – (x-1)/4 – (x-3)/5 = 0
By taking LCM for 3, 4 and 5, which is 60
((5x×20) – (x-1)15 – (x-3)12)/60 = 0
(100x – 15x + 15 -12x + 36)/60 = 0
(73x + 51)/60 = 0
By cross-multiplying
73x + 51 = 0
x = -51/73
Let us verify the given equation now,
5x/3 – (x-1)/4 = (x-3)/5
By substituting the value of ‘x’, we get,
(20x – (x-1)3)/12 = (-51/73 – 3)/5
(20x – 3x + 3)/12 = (-270/73)/5
(17x + 3)/12 = -270/365
(17(-51/73) + 3)/12 = -54/73
(-867/73 + 3)/12 = -54/73
((-867 + 219)/73)/12 = -54/73
(-648)/876 = -54/73
-54/73 = -54/73
Hence, the given equation is verified
13. (3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14
Solution:
(3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14
Let us rearrange the equation
(3x+1)/16 + (2x-3)/7 – (x+3)/8 – (3x-1)/14 = 0
By taking LCM for 16, 7, 8 and 14, which is 112
((3x+1)7 + (2x-3)16 – (x+3)14 – (3x-1)8)/112 = 0
(21x + 7 + 32x – 48 – 14x – 42 – 24x + 8)/112 = 0
(21x + 32x – 14x – 24x + 7 – 48 – 42 + 8)/112 = 0
(15x – 75)/112 = 0
By cross-multiplying
15x – 75 = 0
15x = 75
x = 75/15
= 5
Let us verify the given equation now,
(3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14
By substituting the value of ‘x’, we get,
(3(5)+1)/16 + (2(5)-3)/7 = (5+3)/8 + (3(5)-1)/14
(15+1)/16 + (10-3)/7 = 8/8 + (15-1)/14
16/16 + 7/7 = 8/8 + 14/14
1 + 1 = 1 + 1
2 = 2
Hence, the given equation is verified
14. (1-2x)/7 – (2-3x)/8 = 3/2 + x/4
Solution:
(1-2x)/7 – (2-3x)/8 = 3/2 + x/4
Let us rearrange the equation
(1-2x)/7 – (2-3x)/8 – x/4 = 3/2
By taking LCM for 7, 8 and 4, which is 56
((1-2x)8 – (2-3x)7 – 14x)/56 = 3/2
(8 – 16x – 14 + 21x – 14x)/56 = 3/2
(-9x – 6)/56 = 3/2
By cross-multiplying
2(-9x-6) = 3(56)
-18x – 12 = 168
-18x = 168+12
-18x = 180
x = 180/-18
x = -10
Let us verify the given equation now,
(1-2x)/7 – (2-3x)/8 = 3/2 + x/4
By substituting the value of ‘x’, we get,
(1-2(-10))/7 – (2-3(-10))/8 = 3/2 + (-10)/4
(1+20)/7 – (2+30)/8 = 3/2 – 5/2
21/7 – 32/8 = 3/2 – 5/2
3 – 4 = -2/2
-1 = -1
Hence, the given equation is verified
15. (9x+7)/2 – (x – (x-2)/7) = 36
Solution:
(9x+7)/2 – (x – (x-2)/7) = 36
Let us simplify the given equation into a simple form
(9x+7)/2 – (7x-x+2)/7 = 36
(9x+7)/2 – (6x+2)/7 = 36
By taking LCM for 2 and 7 is 14
(7(9x+7) – 2(6x+2))/14 = 36
(63x+49 – 12x – 4)/14 = 36
(51x + 45)/14 = 36
By cross-multiplying
51x + 45 = 36(14)
51x + 45 = 504
51x = 504-45
51x = 459
x = 459/51
= 9
Let us verify the given equation now,
(9x+7)/2 – (x – (x-2)/7) = 36
(9x+7)/2 – (6x+2)/7 = 36
By substituting the value of ‘x’, we get,
(9(9)+7)/2 – (6(9)+2)/7 = 36
(81+7)/2 – (54+2)/7 = 36
88/2 – 56/7 = 36
44 – 8 = 36
36 = 36
Hence, the given equation is verified
16. 0.18(5x – 4) = 0.5x + 0.8
Solution:
0.18(5x – 4) = 0.5x + 0.8
Let us rearrange the equation
0.18(5x – 4) – 0.5x = 0.8
0.90x – 0.72 – 0.5x = 0.8
0.90x – 0.5x = 0.8 + 0.72
0.40x = 1.52
x = 1.52/0.40
= 3.8
Let us verify the given equation now,
0.18(5x – 4) = 0.5x + 0.8
By substituting the value of ‘x’, we get,
0.18(5(3.8)-4) = 0.5(3.8) + 0.8
0.18(19-4) = 1.9 + 0.8
2.7 = 2.7
Hence, the given equation is verified
17. 2/3x – 3/2x = 1/12
Solution:
2/3x – 3/2x = 1/12
By taking LCM for 3x and 2x, which is 6x
((2×2) – (3×3))/6x = 1/12
(4-9)/6x = 1/12
-5/6x = 1/12
By cross-multiplying
6x = -60
x = -60/6
= -10
Let us verify the given equation now,
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
((-2×2) + (3×3))/60 = 1/12
(-4+9)/60 = 1/12
5/60 = 1/12
1/12 = 1/12
Hence, the given equation is verified
18. 4x/9 + 1/3 + 13x/108 = (8x+19)/18
Solution:
4x/9 + 1/3 + 13x/108 = (8x+19)/18
Let us rearrange the equation
4x/9 + 13x/108 – (8x+19)/18 = -1/3
By taking LCM for 9, 108 and 18, which is 108
((4x×12) + 13x×1 – (8x+19)6)/108 = -1/3
(48x + 13x – 48x – 114)/108 = -1/3
(13x – 114)/108 = -1/3
By cross-multiplying
(13x – 114)3 = -108
39x – 342 = -108
39x = -108 + 342
39x = 234
x = 234/39
= 6
Let us verify the given equation now,
4x/9 + 1/3 + 13x/108 = (8x+19)/18
By substituting the value of ‘x’, we get,
4(6)/9 + 1/3 + 13(6)/108 = (8(6)+19)/18
24/9 + 1/3 + 78/108 = 67/18
8/3 + 1/3 + 13/18 = 67/18
((8×6) + (1×6) + (13×1))/18 = 67/18
(48 + 6 + 13)/18 = 67/18
67/18 = 67/18
Hence, the given equation is verified
19. (45-2x)/15 – (4x+10)/5 = (15-14x)/9
Solution:
(45-2x)/15 – (4x+10)/5 = (15-14x)/9
By rearranging
(45-2x)/15 – (4x+10)/5 – (15-14x)/9 = 0
By taking LCM for 15, 5 and 9, which is 45
((45-2x)3 – (4x+10)9 – (15-14x)5)/45 = 0
(135 – 6x – 36x – 90 – 75 + 70x)/45 = 0
(28x – 30)/45 = 0
By cross-multiplying
28x – 30 = 0
28x = 30
x = 30/28
= 15/14
Let us verify the given equation now,
(45-2x)/15 – (4x+10)/5 = (15-14x)/9
By substituting the value of ‘x’, we get,
(45-2(15/14))/15 – (4(15/14) + 10)/5 = (15 – 14(15/14))/9
(45- 15/7)/15 – (30/7 + 10)/5 = (15-15)/9
300/105 – 100/35 = 0
(300-300)/105 = 0
0 = 0
Hence, the given equation is verified
20. 5(7x+5)/3 – 23/3 = 13 – (4x-2)/3
Solution:
5(7x+5)/3 – 23/3 = 13 – (4x-2)/3
By rearranging
(35x + 25)/3 + (4x – 2)/3 = 13 + 23/3
(35x + 25 + 4x – 2)/3 = (39+23)/3
(39x + 23)/3 = 62/3
By cross-multiplying
(39x + 23)3 = 62(3)
39x + 23 = 62
39x = 62 – 23
39x = 39
x = 1
Let us verify the given equation now,
5(7x+5)/3 – 23/3 = 13 – (4x-2)/3
By substituting the value of ‘x’, we get,
(35x + 25)/3 – 23/3 = 13 – (4x-2)/3
(35+25)/3 – 23/3 = 13 – (4-2)/3
60/3 – 23/3 = 13 – 2/3
(60-23)/3 = (39-2)/3
37/3 = 37/3
Hence, the given equation is verified
21. (7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3
Solution:
(7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3
Upon expansion
(7x-1)/4 – (4x-1+x)/6 = 10/3
(7x-1)/4 – (5x-1)/6 = 10/3
By taking LCM for 4 and 6, we get 24
((7x-1)6 – (5x-1)4)/24 = 10/3
(42x – 6 – 20x + 4)/24 = 10/3
(22x – 2)/24 = 10/3
By cross-multiplying
22x – 2 = 10(8)
22x – 2 = 80
22x = 80+2
22x = 82
x = 82/22
= 41/11
Let us verify the given equation now,
(7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3
By substituting the value of ‘x’, we get,
(7x-1)/4 – (5x-1)/6 = 10/3
(7(41/11)-1)/4 – (5(41/11)-1)/6 = 10/3
(287/11 – 1)/4 – (205/11 – 1)/6 = 10/3
(287-11)/44 – (205-11)/66 = 10/3
276/44 – 194/66 = 10/3
69/11 – 97/33 = 10/3
((69×3) – (97×1))/33 = 10/3
(207 – 97)/33 = 10/3
110/33 = 10/3
10/3 = 10/3
Hence, the given equation is verified
22. 0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1
Solution:
0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1
Let us simplify
(0.5/0.35)(x – 0.4) – (0.6/0.42)(x – 2.71) = x + 6.1
(x – 0.4)/0.7 – (x – 2.71)/0.7 = x + 6.1
(x – 0.4 – x + 2.71)/0.7 = x + 6.1
-0.4 + 2.71 = 0.7(x + 6.1)
0.7x = 2.71 – 0.4 – 4.27
= -1.96
x = -1.96/0.7
= -2.8
Let us verify the given equation now,
0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1
By substituting the value of ‘x’, we get,
0.5(-2.8 – 0.4)/0.35 – 0.6(-2.8 – 2.71)/0.42 = -2.8 + 6.1
-1.6/0.35 + 3.306/0.42 = 3.3
-4.571 + 7.871 = 3.3
3.3 = 3.3
Hence, the given equation is verified
23. 6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2
Solution:
6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2
By rearranging
6.5x + (19.5x – 32.5)/2 – 6.5x – (13x – 26)/2 = 13
(19.5x – 32.5)/2 – (13x – 26)/2 = 13
(19.5x – 32.5 – 13x + 26)/2 = 13
(6.5x – 6.5)/2 = 13
6.5x – 6.5 = 13×2
6.5x – 6.5 = 26
6.5x = 26+6.5
6.5x = 32.5
x = 32.5/6.5
= 5
Let us verify the given equation now,
6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2
By substituting the value of ‘x’, we get,
6.5(5) + (19.5(5) – 32.5)/2 = 6.5(5) + 13 + (13(5) – 26)/2
32.5 + (97.5 – 32.5)/2 = 32.5 + 13 + (65 – 26)/2
32.5 + 65/2 = 45.5 + 39/2
(65 + 65)/2 = (91+39)/2
130/2 = 130/2
65 = 65
Hence, the given equation is verified
24. (3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Solution:
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Let us simplify
9x2 + 6x – 24x – 16 – 8x2 – 4x + 22x + 11 = x2 + 7x – 3x – 21
9x2 + 6x – 24x – 16 – 8x2 – 4x + 22x + 11 – x2 – 7x + 3x + 21 = 0
9x2 – 8x2 – x2 + 6x – 24x – 4x + 22x – 7x + 3x – 16 + 21 + 11 = 0
-4x + 16 = 0
-4x = -16
x = 4
Let us verify the given equation now,
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
By substituting the value of ‘x’, we get,
(3(4) – 8) (3(4) + 2) – (4(4) – 11) (2(4) + 1) = (4 – 3) (4 + 7)
(12-8) (12+2) – (16-11) (8+1) = 1(11)
4 (14) – 5(9) = 11
56 – 45 = 11
11 = 11
Hence, the given equation is verified
25. [(2x+3) + (x+5)]2 + [(2x+3) – (x+5)]2 = 10x2 + 92
Solution:
[(2x+3) + (x+5)]2 + [(2x+3) – (x+5)]2 = 10x2 + 92Let us simplify the given equation
[3x + 8]2 + [x – 2]2 = 10x2 + 92By using the formula (a+b)2
9x2 + 48x + 64 + x2 – 4x + 4 = 10x2 + 92
By rearranging
9x2 – 10x2 + x2 + 48x – 4x = 92 – 64 – 4
44x = 24
x = 24/44
= 6/11
Let us verify the given equation now,
[(2x+3) + (x+5)]2 + [(2x+3) – (x+5)]2 = 10x2 + 92By substituting the value of ‘x’, we get,
[2(6/11) + 3 + (6/11) + 5]2 + [2(6/11) + 3 – (6/11) – 5]2 = 10(6/11)2 + 92 [(12/11 + 3) + (6/11 + 5)]2 + [(12/11 + 3) – (6/11 + 5)]2 = 10(6/11)2 + 92 [(12+33)/11 + (6+55)/11]2 + [(12+33)/11- (6+55)/11]2 = 10(6/11)2 + 92 [(45/11)+ (61/11)]2 + [(45/11) – (61/11)]2 = 360/121 + 92(106/11)2 + (-16/11)2 = (360 + 11132)/121
11236/121 + 256/121 = 11492/121
11492/121 = 11492/121
Hence, the given equation is verified
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 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, 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, 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 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, 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, 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, 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, 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, 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, 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
EXERCISE 9.4 PAGE NO: 9.29
1. Four-fifth of a number is more than three-fourths of the number by 4. Find the number.
Solution:
Let us consider the number as ‘x’
So, Three-fourth of the number is 3x/4
Fourth-fifth of the number is 4x/5
4x/5 – 3x/4 = 4
By taking LCM of 5 and 4, we get 20
(16x – 15x)/20 = 4
By cross-multiplying, we get,
16x – 15x = 4(20)
x = 80
∴ The number is 80.
2. The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution:
Let the two consecutive numbers be x and (x – 1)
So now,
x2 – (x-1)2 = 31
By using the formula (a-b)2 = a2 + b2 – 2ab
x2 – (x2 – 2x + 1) = 31
x2 – x2 + 2x – 1 = 31
2x – 1 = 31
2x = 31+1
2x = 32
x = 32/2
= 16
Two consecutive numbers are, x and (x-1) : 16 and (16-1) =15
∴ The two consecutive numbers are 16 and 15.
3. Find a number whose double is 45 greater than its half.
Solution:
Let us consider the number as ‘x’
So,
2x – x/2 = 45
(4x-x)/2 = 45
By cross-multiplying, we get,
3x = 90
x = 90/3
= 30
∴ The number is 30.
4. Find a number such that when 5 is subtracted from 5 times that number, the result is 4, more than twice the number.
Solution:
Let us consider the number as ‘x’
Then, five times the number will be 5x
And, two times, the number will be 2x
So,
5x – 5 = 2x + 4
5x – 2x = 5 + 4
3x = 9
x = 9/3
x = 3
∴ The number is 3.
5. A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.
Solution:
Let us consider the number as ‘x’
So,
x/5 + 5 = x/4 – 5
x/5 – x/4 = -5 – 5
By taking LCM for 5 and 4, which is 20
(4x-5x)/20 = -10
By cross-multiplying, we get,
4x – 5x = -10(20)
-x = -200
x = 200
∴ The number is 200.
6. A number consists of two digits whose sum is 9. If 27 is subtracted from the number the digits are reversed. Find the number.
Solution:
We know that one of the digits be ‘x’
The other digit is 9-x
So, the two digit number is 10(9-x) + x
The number obtained after interchanging the digits is 10x + (9-x)
10(9-x) + x – 27 = 10x + (9-x)
Upon simplification,
90 – 10x + x – 27 = 10x + 9 – x
-10x + x – 10x + x = 9 – 90 + 27
-18x = -54
x = 54/18
= 9/3
= 3
The two-digit number is 10(9-x) + x
Substituting the value of x, we get,
10(9-x) + x
10(9 – 3) + 3
10(6) + 3
60+3
63
∴ The number is 63.
7. Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.
Solution:
Let one of the numbers be ‘x’
The other number is 184 – x
So, One-third of one part may exceed one-seventh of another part by 8.
x/3 – (184-x)/7 = 8
LCM for 3 and 7 is 21
(7x – 552 + 3x)/21 = 8
By cross-multiplying, we get,
(7x – 552 + 3x) = 8(21)
10x – 552 = 168
10x = 168 + 552
10x = 720
x = 720/10
= 72
∴ One of the numbers is 72, and the other number is 184 – x => 184 – 72 = 112.
8. The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 2/3. What is the original fraction equal to?
Solution:
Let us consider the denominator as x and numerator as (x-6)
By using the formula,
Fraction = numerator/denominator = (x-6)/x
(x – 6 + 3)/x = 2/3
(x – 3)/x = 2/3
By cross-multiplying
3(x-3) = 2x
3x – 9 = 2x
3x – 2x = 9
x = 9
∴ The denominator is x = 9, numerator is (x-6) = (9-6) = 3
And the fraction = numerator/denominator = (x-6)/x = 3/9 = 1/3
9. A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50. Find the number of notes of each type.
Solution:
Let the number of 10Rs notes be x
Number of 20Rs notes be 50 – x
Amount due to 10Rs notes = 10 × x = 10x
Amount due to 20Rs notes = 20 × (50 – x) = 1000 – 20x
So the total amount is Rs 800
10x + 1000 – 20x = 800
-10x = 800 – 1000
-10x = -200
x = -200/-10
= 20
∴ The number of 10Rs notes is 20
Number of 20Rs notes are 50 – 20 = 30
10. Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty- five paise coins as she has fifty-paise coins. How many coins of each kind does she have?
Solution:
Let the number of fifty paise coins be x
The number of twenty-five paise coins be 2x
Amount due to fifty paise coins = (50×x)/100 = 0.50x
Amount due to twenty five paise coins = (25×2x)/100 = 0.50x
So the total amount is Rs 9
0.50x + 0.50x = 9
1x = 9
x = 9
∴ The number of fifty paise coins is x = 9
Number of twenty-five paise coins, 2x = 2×9 = 18
11. Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times Ashima’s age. How old were they two years ago?
Solution:
Let the present age of Ashima be ‘x’ years
The present age of Sunita is 2x years
Ashima’s new age = (x – 6) years
Sunita’s new age = (2x + 4) years
So, (2x + 4) = 4 (x – 6)
2x + 4 = 4x – 24
2x – 4x = -24 – 4
-2x = -28
x = -28/-2
= 14
∴ The age of Ashima is x years = 14 years
Age of Sunita is 2x years = 2(14) = 28 years
Two years ago, age of Ashima is 14 – 2 = 12 years, age of Sunita = 28 – 2 = 26 years
12. The ages of Sonu and Monu are in the ratio 7:5 ten years hence, the ratio of their ages will be 9:7. Find their present ages.
Solution:
Let the present age of Sonu be 7x years
The present age of Monu is 5x years
Sonu’s age after 10 years = (7x + 10) years
Monu’s age after 10 years = (5x + 10) years
So,
(7x + 10) / (5x + 10) = 9/7
by using cross-multiplication, we get,
7(7x + 10) = 9(5x + 10)
49x + 70 = 45x + 90
49x – 45x = 90 – 70
4x = 20
x = 20/4
= 5
∴ Present age of Sonu is 7x = 7(5) = 35years
Present age of Monu is 5x = 5(5) = 25years
13. Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.
Solution:
Let the age of the son five years ago be x years
The age of man five years ago be 7x years
After five years, the son’s age is x + 5 years
After five years father’s age is 7x + 5 years
So, since five years, the relation in their ages are
7x + 5 + 5 = 3(x + 5 + 5)
7x + 10 = 3x + 15 + 15
7x + 10 = 3x + 30
7x – 3x = 30 – 10
4x = 20
x = 5
∴ Present father’s age is 7x + 5 = 7(5) + 5 = 35 + 5 = 40years
Present son’s age is x + 5 = 5 + 5 = 10years
14. I am currently 5 times as old as my son. In 6 years time, I will be three times as old as he will be then. What are our ages now?
Solution:
Let the present son’s age be x years
Present father’s age be 5x years
Son’s age after 6 years = (x + 6) years
Fathers’ age after 6 years = (5x + 6) years
So,
5x + 6 = 3(x + 6)
5x + 6 = 3x + 18
5x – 3x = 18 – 6
2x = 12
x = 12/2
= 6
∴ present son’s age is x = 6years
Present father’s age is 5x = 5(6) = 30years
15. I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?
Solution:
Let the number of five rupee notes be x
The number of ten rupee notes be (x + 10)
Amount due to five rupee notes = 5 × x = 5x
Amount due to ten rupee notes = 10 (x + 10) = 10x + 100
The total amount = Rs 1000
5x + 10x +100 = 1000
15x = 900
x = 900/15
= 60
∴ the number of five rupee notes is x = 60
The number of ten rupee notes is x + 10 = 60+10 = 70
16. At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two-fifths drank fruit juice, and just three did not drink anything. How many guests were in all?
Solution:
Let the number of guests be x
The given details are the number of guests who drank colas are x/4
The number of guests who drank squash is x/3
The number of guests who drank fruit juice is 2x/5
The number of guests who did not drink anything was 3
x/4 + x/3 + 2x/5 + 3 = x
By taking LCM for 4, 3 and 5, we get 60
(15x+20x+24x-60x)/60 = -3
By cross-multiplying, we get,
(15x+20x+24x-60x) = -3(60)
-x = -180
x = 180
∴ The total number of guests in all was 180
17. There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question, one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?
Solution:
Let the number of correct answers be x
The number of questions answered wrong is (180 – x)
Total score when answered right = 4x
Marks deducted when answered wrong = 1(180 – x) = 180 – x
So,
4x – (180 – x) = 450
4x – 180 + x = 450
5x = 450 + 180
5x = 630
x = 630/5
= 126
∴ 126 questions he answered correctly.
18. A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day he works, and he will be fined Rs 5 for each day he is absent. If he receives Rs 745 in all, how many days he remained absent?
Solution:
Let us consider the number of absent days as x
So, the number of present days is (20 – x)
The wage for one day of work = Rs 60
Fine for absent day = Rs 5
So,
60(20 – x) – 5x = 745
1200 – 60x – 5x = 744
-65x = 744-1200
-65x = -456
x = -456/-65
= 7
∴ For 7 days, the labourer was absent.
19. Ravish has three boxes whose total weight is 60 ½ Kg. Box B weighs 3 ½ kg more than box A, and box C weighs 5 1/3 kg more than box B. Find the weight of box A.
Solution:
The given details are the total weight of three boxes is 60 ½ kg = 121/2 kg
Let the weight of box A be x kg
Weight of box B be x + 7/2 kg
Weight of box C be x + 7/2 + 16/3 kg
So,
x + x + 7/2 + x + 7/2 + 16/3 = 121/2
3x = 121/2 – 7/2 – 7/2 – 16/3
By taking LCM for 2 and 3 is 6
3x = (363 – 21 – 21 – 32)/6
3x = 289/6
x = 289/18
∴ The weight of box A is 289/18 kg
20. The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number.
Solution:
Le the denominator be x and the numerator be (x – 3)
By using the formula
Fraction = numerator/denominator
= (x – 3)/x
So, when the numerator is increased by 2 and Denominator is increased by 5, then the fraction is ½
(x – 3 + 2)/(x + 5) = 1/2
(x – 1)/(x + 5) = 1/2
By using cross-multiplication, we get
2(x – 1) = x + 5
2x – 2 = x + 5
2x – x = 2 + 5
x = 7
∴ Denominator is x = 7, numerator is (x – 3) = 7 – 3 = 4
And the fraction = numerator/denominator = 4/7
21. In a rational number, twice the numerator is 2 more than the denominator, if 3 is added to each, the numerator and the denominator. The new fraction is 2/3. Find the original number.
Solution:
Le the numerator be x and the denominator be (2x – 2)
By using the formula
Fraction = numerator/denominator
= x / (2x – 2)
So, the numerator and denominator are increased by 3, then the fraction is 2/3
(x + 3)/(2x – 2 + 3) = 2/3
(x + 3)/(2x + 1) = 2/3
By cross-multiplying, we get,
3(x + 3) = 2(2x + 1)
3x + 9 = 4x + 2
3x – 4x = 2 – 9
-x = -7
x = 7
∴ The numerator is x = 7, denominator is (2x – 2) = (2(7) – 2) = 14-2 = 12
And the fraction is numerator/denominator = 7/12
22. The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.
Solution:
Let the speed of one train be x km/hr.
The speed of the other train be (x + 5) km/hr.
The total distance between the two stations = 340 km
By using the formula
Distance = speed × time
So, the distance covered by one train in 2 hrs. Will be x×2 = 2x km
Distance covered by the other train in 2 hrs. Will be 2(x + 5) = (2x + 10) km
The distance between the trains is 30 km
2x + 2x + 10 + 30 = 340
4x + 40 = 340
4x = 340 – 40
4x = 300
x = 300/4
= 75
∴ The speed of one train is x = 75 km/hr.
Speed of other train is (x + 5) = 75 + 5 = 80 km/hr.
23. A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream is 1 km/hr., find the speed of the steamer in still water and the distance between the ports.
Solution:
Let the speed of the steamer be x km/hr.
Speed of stream = 1 km/hr.
Downstream speed = (x + 1) km/hr.
Upstream speed = (x – 1) km/hr.
By using the formula
Distance = speed × time
= (x + 1) × 9 and
= (x – 1) × 10
9x + 9 = 10x – 10
9x – 10x = -10 -9
-x = -19
x = 19 km/hr.
∴ The speed of the steamer in still water is 19 km/hr.
Distance between the ports is 9(x + 1) = 9(19+1) = 9(20) = 180 km.
24. Bhagwanti inherited Rs 12000.00. She invested part of it at 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00 How much did she invest at each rate?
Solution:
At a rate of 10%, let the investment be Rs x
At the rate of 12%, the investment will be Rs (12000 – x)
At 10% of rate the annual income will be x × (10/100) = 10x/100
At 12% of rate, the annual income will be (12000 – x) × 12/100 = (144000 – 12x)/100
Total investment = 1280
So, 10x/100 + (144000 – 12x)/100 = 1280
(10x + 144000 – 12x)/100 = 1280
(144000 – 2x)/100 = 1280
By cross-multiplying, we get,
144000 – 2x = 1280(100)
-2x = 128000 – 144000
-2x = -16000
x = -16000/-2
= 8000
∴ At 10% of rate, she invested Rs 8000, and at 12% of the rate she invested Rs (12000 – x) = Rs (12000 – 8000) = Rs 4000
25. The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm2 more than that of the given rectangle. Find the length and breadth of the given rectangle.
Solution:
Let the breadth of the rectangle be x meter
Length of the rectangle be (x + 9) meter
Area of the rectangle length×breadth = x(x +9) m2
When length and breadth increased by 3cm, then,
New length = x + 9 + 3 = x + 12
New breadth = x + 3
So, the area is
(x + 12) (x + 3) = x (x + 9) + 84
x2 + 15x + 36 = x2 + 9x + 84
15x – 9x = 84 – 36
6x = 48
x = 48/6
= 8
∴ The length of the rectangle (x + 9) = (8 + 9) = 17cm, and the breadth of the rectangle is 8cm.
26. The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?
Solution:
Let the age of Anup be x years
So the age of Anup’s father will be (100 – x) years
The age of Anuj is (100-x)/5 years
So, When Anup is as old as his father after (100 – 2x) years,
Then Anuj’s age = present age of his father (Anup) + 8
Present age of Anuj + 100 – 2x = Present age of Anup + 8
(100 – x)/5 + (100 – 2x) = x + 8
(100-x)/5 – 3x = 8 – 100
(100 – x – 15x)/5 = -92
By cross-multiplying, we get,
100 – 16x = -460
-16x = -460 – 100
-16x = -560
x = -560/-16
= 35
∴ The present age of Anup is 35 years then, the age of Anup’s father will be (100-x) = 100-35 = 65 years
The age of Anuj is (100-x)/5 = (100 – 35)/5 = 65/5 = 13 years
27. A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?
Solution:
Let the amount lady had be Rs x
Amount spent for hankies and given to beggar is x/2 + 1
Remaining amount is x – (x/2 + 1) = x/2 – 1 = (x-2)/2
Amount spent for lunch (x-2)/2×1/2 = (x-2)/4
The amount given as a tip is Rs 2
Remaining amount after lunch = (x-2)/2 – (x-2)/4 – 2 = (2x – 4 – x + 2 – 8)/4 = (x – 10)/4
Amounts spent for books =1/2 × (x-10)/4 = (x-10)/8
The bus fare is Rs 3
Amount left = (x-10)/4 – (x-10)/8 – 3 = (2x – 20 – x + 10 – 24)/8 = (x-34)/8
So from the question, we know that the amount left = Rs 1
(x-34)/8 = 1
By cross-multiplying, we get,
x – 34 = 8
x = 8 + 34
= 42
∴ the lady started with Rs. 42
Comments