Frank Solutions for Chapter 8 Simultaneous Linear Equations Class 9 Mathematics ICSE
Exercise 8.1
1. Solve the following simultaneous equations by the substitution method:
(a) 2x + y = 8
3y = 3 + 4x
(b) x + 3y = 5
7x – 8y = 6
(c) 5x + 4y – 23 = 0
x + 9 = 6y
(d) 2x + 3y = 31
5x – 4 = 3y
(e) 7x – 3y = 31
9x – 5y = 41
(f) 13 + 2y = 9x
3y = 7x
(g) 0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
(h) 0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
(i) 3 – (x + 5) = y + 2
2 (x + y) = 10 + 2y
(j) 7 (y + 3) – 2 (x + 2) = 14
4 (y – 2) + 3 (x – 3) = 2
Answer
(a) Given
2x + y = 8 …(1)
3y = 3 + 4x …(2)
Now,
Consider the equation 2x + y = 8
y = 8 – 2x …(3)
Substituting the value of y in equation (2), we get,
3 (8 – 2x) = 3 + 4x
⇒ 24 – 6x = 3 + 4x
⇒ – 6x – 4x = 3 – 24
⇒ – 10x = – 21
We get,
x = (21/10)
Now,
Putting the value of x in equation (3), we get,
y = 8 – 2x
⇒ y = 8 – 2 (21/10)
On further calculation, we get,
y = 8 – (21/5)
⇒ y = {(40 – 21)/5}
We get,
y = (19 / 5)
Therefore, the solution set is {(21/10), (19/5)}
(b) Given
x + 3y = 5 …(1)
7x – 8y = 6 …(2)
Now,
Consider the equation x + 3y = 5
x = 5 – 3y …(3)
Substituting the value of x in equation (2), we get,
7 (5 – 3y) – 8y = 6
⇒ 35 – 21y – 8y = 6
⇒ 35 – 29y = 6
⇒ – 29y = 6 – 35
⇒ – 29y = – 29
We get,
y = 1
Putting the value of y in equation (3), we get,
x = 5 – 3y
⇒ x = 5 – 3(1)
⇒ x = 5 – 3
⇒ x = 2
Therefore, the solution set is (2, 1)
(c) Given
5x + 4y – 23 = 0 …(1)
x + 9 = 6y ….(2)
Now,
Consider the equation (2),
x + 9 = 6y
⇒ x = 6y – 9 …(3)
Substituting the value of x in equation (1), we get,
5 (6y – 9) + 4y – 23 = 0
⇒ 30y – 45 + 4y – 23 = 0
⇒ 34 y – 68 = 0
⇒ 34y = 68
⇒ y = (68/34)
We get,
y = 2
Putting the value of y in equation (3), we get,
x = 6 (2) – 9
⇒ x = 12 – 9
We get,
x = 3
Therefore, the solution set is (3, 2)
(d) Given
2x + 3y = 31 ...(1)
5x – 4 = 3y ...(2)
Now,
Consider the equation (1),
2x + 3y = 31
⇒ 2x = 31 – 3y
⇒ x = {(31 – 3y)/2} …(3)
Substituting the value of x in equation (2), we get,
5{(31 – 3y)/2} – 4 = 3y
⇒ {(155 – 15y)/2} – 4 = 3y
On taking LCM, we get,
{(155 – 15y – 8)/2} = 3y
⇒ 147 – 15y = 6y
We get,
21y = 147
⇒ y = (147/21)
⇒ y = 7
Now,
Putting the value of y in equation (3), we get,
x = [{31 – 3(7)}/2]
⇒ x = (31 – 21)/2
⇒ x = (10/2)
We get,
x = 5
Therefore, the solution set is (5, 7)
(e) Given
7x – 3y = 31 …(1)
9x – 5y = 41 ….(2)
Now,
Consider the equation (1),
7x – 3y = 31
⇒ 7x = 31 + 3y
⇒ x = {(31 + 3y) / 7} ...(3)
Substituting the value of x in equation (2), we get,
{9 (31 + 3y)/7} – 5y = 41
⇒ {(279 + 27y)/7} – 5y = 41
On taking LCM, we get,
(279 + 27y – 35y)/7 = 41
⇒ 279 – 8y = 287
⇒ – 8y = 287 – 279
⇒ – 8y = 8
We get,
y = – 1
Putting the value of y in equation (3), we get,
x = [{31 + 3(-1)}/7]
⇒ x = (31 – 3)/7
⇒ x = (28/7)
We get,
x = 4
Therefore, the solution set is (4, -1)
(f) Given
13 + 2y = 9x …(1)
3y = 7x …(2)
Now,
Consider the equation (2),
3y = 7x
⇒ y = (7x/3) …(3)
Substituting the value of y in equation (1), we get,
13 + 2 (7/3) x = 9x
⇒ 13 + (14/3) x = 9x
⇒ 9x – (14/3) x = 13
On taking LCM, we get,
{(27x – 14x)/3} = 13
⇒ 13x = 39
⇒ x = (39/13)
We get,
x = 3
Putting the value of x in equation (3), we get,
y = (7/3) x
⇒ y = (7/3) 3
We get,
y = 7
Therefore, the solution set is (3, 7)
(g) Given
0.5x + 0.7y = 0.74 ...(1)
0.3x + 0.5y = 0.5 ….(2)
Now,
Consider equation 0.5x + 0.7y = 0.74
0.5x = 0.74 – 0.7y
⇒ x = {(0.74 – 0.7y)/0.5} ...(3)
Substituting the value of x in equation (2), we get,
0.3 {(0.74 – 0.7y)/0.5} + 0.5y = 0.5
⇒ {(0.222 – 0.21y)/0.5} + 0.5y = 0.5
On taking LCM, we get,
{(0.222 – 0.21y + 0.25y) / 0.5} = 0.5
⇒ 0.222 + 0.04y = 0.25
⇒ 0.04y = 0.028
⇒ y = (0.028/0.04)
⇒ y = (28/40)
⇒ y = (7/10)
We get,
y = 0.7
Putting the value of y in equation (3), we get,
x = [{0.74 – 0.7 (0.7)}/0.5]
⇒ x = (0.74 – 0.49)/0.5
⇒ x = (0.25/0.5)
⇒ x = (25/50)
⇒ x = (1/2)
We get,
x = 0.5
Therefore, the solution set is (0.5, 0.7)
(h) Given
0.4x + 0.3y = 1.7 ...(1)
0.7x – 0.2y = 0.8 …(2)
Multiplying both the equations by 10, we get,
4x + 3y = 17 ...(3)
7x – 2y = 8 ...(4)
Now,
Consider the equation,
4x + 3y = 17
⇒ 4x = 17 – 3y
⇒ x = (17 – 3y)/4 ….(5)
Substituting the value of x in equation (4)
7 {(17 – 3y)/4} – 2y = 8
⇒ {119 – 21y)/4} – 2y = 8
On taking LCM, we get,
{(119 – 21y – 8y)/4 = 8
⇒ 119 – 29y = 32
⇒ -29y = 32 – 119
⇒ -29y = – 87
⇒ y = (-87/–29)
We get,
y = 3
Putting the value of y in equation (5), we get,
x = [{17 – 3(3)}/4]
⇒ x = (17 – 9)/4
⇒ x = (8/4)
We get,
x = 2
Therefore, the solution set is (2, 3)
(i) Given
3 – (x + 5) = y + 2 ...(1)
2 (x + y) = 10 + 2y …(2)
Now,
Consider equation (1)
3 – (x + 5) = y + 2
⇒ 3 – x – 5 = y + 2
⇒ – x – 2 = y + 2
⇒ x + y = – 4
⇒ x = – 4 – y ….(3)
Now,
Consider equation
2 (x + y) = 10 + 2y
⇒ 2x + 2y = 10 + 2y
⇒ 2x = 10
We get,
x = 5
Substituting the value of x in equation (3), we get,
5 = – 4 – y
⇒ y = – 4 – 5
⇒ y = – 9
Therefore, the solution set is (5, – 9)
(j) Given
7 (y + 3) – 2 (x + 2) = 14 ….(1)
4 (y – 2) + 3 (x – 3) = 2 …(2)
Consider the equation (1)
7 (y + 3) – 2 (x + 2) = 14
⇒ 7y + 21 – 2x – 4 = 14
⇒ – 2x + 7y = 14 + 4 – 21
⇒ – 2x + 7y = -3
⇒ 2x – 7y = 3
⇒ 2x = 7y + 3
We get,
x = {(7y + 3)/2} ….(3)
Now,
Consider the equation,
4 (y – 2) + 3(x – 3) = 2
⇒ 4y – 8 + 3x – 9 = 2
⇒ 3x + 4y = 19
Substituting the value of x, we get,
3 {(7y + 3)/2} + 4y = 19 from (3)
⇒ (21y + 9)/2} + 4y = 19
On taking LCM, we get,
(21y + 9 + 8y)/2 = 19
⇒ 29y + 9 = 38
⇒ 29y = 29
We get,
y = 1
Substituting the value of y in equation (3), we get,
x = {(7y + 3)/2}
⇒ x = {7(1) + 3}/2
We get,
x = (10/2)
⇒ x = 5
Therefore, the solution set is (5, 1)
2. Solve the following simultaneous equations:
(i) 6x + 3y = 7xy
3x + 9y = 11xy
(ii) 8v – 3u = 5uv
6v – 5u = – 2uv
(iii) 3 (2u + v) = 7uv
3 (u + 3v) = 11uv
(iv) 2 (3u – v) = 5uv
2 (u + 3v) = 5uv
Answer:
(i) 6x + 3y = 7xy
3x + 9y = 11xy
Dividing both sides of each equation by xy, we get,
(6/y) + (3/x) = 7 ...(1)
(3/y) + (9/x) = 11 ...(2)
Multiplying equation (2) by 2, we get,
(6/y) + (18/x) = 22 …(3)
Subtracting (1) from (3), we get,
(15/x) = 15
⇒ x = 1
Hence,
(3/y) + 9 = 11
⇒ (3/y) = 11 – 9
⇒ (3/y) = 2
We get,
y = (3/2)
Therefore, the solution set is {1, (3/2)}
(ii) 8v – 3u = 5uv
6v – 5u = – 2uv
Dividing both sides of each equation by uv, we get,
(8/u) – (3/v) = 5 …(1)
(6/u) – (5/v) = – 2 …(2)
Multiplying equation (1) by 3 and equation (2) by 4,
We get,
(24/u) – (9/v) = 15 …(3)
(24/u) – (20/v) = – 8 …(4)
Subtracting (4) from (3), we get,
(11/v) = 23
⇒ v = (11/23)
Hence,
(6/u) – (5/11) × 23 = – 2
⇒ (6/u) – (115/11) = – 2
⇒ (6/u) = – 2 + (115/11)
⇒ (6/u) = (-22 + 115)/11
⇒ (6/u) = (93/11)
⇒ u = (6 × 11)/93
We get,
u = (22/31)
Therefore, the solution set is {(22/31), (11/23)}
(iii) 3 (2u + v) = 7uv
3 (u + 3v) = 11uv
Dividing by uv, we get,
(6/v) + (3/u) = 7 …(1)
Consider,
3 (u + 3v) = 11uv
⇒ 3u + 9v = 11uv
Dividing by uv, we get,
(3/v) + (9/u) = 11 ...(2)
Multiplying (1) by 3, we get,
(18/v) + (9/u) = 21 ...(3)
Subtracting (2) from (3), we get,
(15/v) = 10
⇒ v = (15/10)
⇒ v = (3/2)
Hence,
(3/u) = 7 – (6/v) from (1)
⇒ (3/u) = 7 – {(6 × (2/3)}
⇒ (3/u) = 7 – 4
⇒ (3/u) = 3
We get,
u = 1
Therefore, the solution set is {1, (3/2)}
(iv) 2 (3u – v) = 5uv
⇒ 2 (u + 3v) = 5uv
⇒ 2 (3u – v) = 5uv
⇒ 6u – 2v = 5uv
Dividing by uv, we get,
(6/v) – (2/u) = 5 ….(1)
⇒ 2 (u + 3v) = 5uv
⇒ 2u + 6v = 5uv
Dividing by uv, we get,
(2/v) + (6/u) = 5 …..(2)
Multiplying equation (1) by 3, we get,
(18/v) – (6/u) = 15 …(3)
Adding equation (2) and (3), we get,
(20/v) = 20
⇒ v = 1
Hence,
(6/u) = 5 – (2/1)
⇒ (6/u) = 3
⇒ u = (6/3)
We get,
u = 2
Therefore, the solution set is (2, 1)
3. Solve the following simultaneous equations:
(a) 13a – 11b = 70
11a – 13b = 74
(b) 41x + 53y = 135
53x + 41y = 147
(c) 65x – 33y = 97
33x – 65y = 1
(d) 103a + 51b = 617
97a + 49b = 583
Answer
(a) The given equations are,
13a – 11b = 70 ...(1)
11a – 13b = 74 ...(2)
Multiplying equation (1) by 13 and equation (2) by 11, we get,
169a – 143b = 910 ...(3)
121a – 143b = 814 …(4)
Subtracting equation (4) from equation (3), we get,
48a = 96
We get,
a = 2
Substituting the value of a in equation (1), we get,
13(2) – 11b = 70
⇒ 26 – 11b = 70
⇒ – 11b = 70 – 26
⇒ – 11b = 44
We get,
b = – 4
Therefore, the solution set is (2, – 4)
(b) The given equations are,
41x + 53y = 135 ...(1)
53x + 41y = 147 …(2)
Multiplying equation (1) by 53 and equation (2) by 41, we get,
2173x + 2809y = 7155 …(3)
2173x + 1681y = 6027 ...(4)
Subtracting equation (4) from (3) we get,
1128y = 1128
⇒ y = 1
Substituting the value of y in equation (1), we get,
41x + 53(1) = 135
⇒ 41x + 53 = 135
⇒ 41x = 135 – 53
⇒ 41x = 82
We get,
x = 2
Therefore, the solution set is (2, 1)
(c) The given equations are,
65x – 33y = 97 ...(1)
33x – 65y =1 ...(2)
Multiplying equation (1) by 33 and equation (2) by 65, we get,
2145x – 1089y = 3201 ...(3)
2145x – 4225y = 65 ...(4)
Subtracting equation (4) from equation (3), we get,
3136y = 3136
⇒ y = 1
Substituting the value of y in equation (2), we get,
33x – 65 (1) = 1
⇒ 33x – 65 = 1
⇒ 33x = 1 + 65
⇒ 33x = 66
We get,
x = 2
Therefore, the solution set is (2, 1)
(d) The given equations are
103a + 51b = 617 …(1)
97a + 49b = 583 …(2)
Subtracting equation (2) from (1), we get,
6a + 2b = 34
Dividing throughout by 2, we get,
3a + b = 17 …(3)
Adding both the equations, we get,
200a + 100b = 1200
Dividing throughout by 100, we get,
2a + b = 12 ...(4)
Subtracting equation (4) from equation (3), we get,
a = 5
Substituting the value of a in equation (3), we get,
3(5) + b = 17
⇒ 15 + b = 17
⇒ b = 17 – 15
⇒ b = 2
Therefore, the solution set is (5, 2)
4. Solve the following pairs of equations:
(a) (3/5) x – (2/3) y + 1 = 0
(1/3) y + (2/5) x = 4
(b) (x/3) + (y/4) = 11
(5x/6) – (y/3) = – 7
(c) (3/2x) + (2/3y) = 5
(5/x) – (3/y) = 1
(d) (3/x) – (1/y) = – 9
(2/x) + (3/y) = 5
(e) y – x = 0.8
[13/{2(x + y)}] = 1
(f) (2/x) + (3/y) = (9/xy)
(4/x) + (9/y) = (21/xy)
Where x ≠ 0, y ≠ 0
(g) (x + y)/xy = 2
(x – y)/xy = 6
(h) {2/(x + 1)} – {1/(y – 1)} = (1/2)
{1/(x + 1)} + (2/(y – 1)} = (5/2)
(i) {6/(x + y)} = {7/(x – y)} + 3
1/{2(x + y)} = 1/{3 (x – y)}
Where x + y ≠ 0 and x – y ≠ 0
(j) {5/(x + y)} – {2/(x – y)} = -1
{15/(x + y)} + {7/(x – y)} = 10
(k) {2/(3x + 2y)} + {3/(3x – 2y)} = (17/5)
{5/(3x + 2y)} + {1/(3x – 2y)} = 2
(l) {xy/(x + y)} = (6/5)
{xy/(y – x)} = 6
where x + y ≠ 0 and y – x ≠ 0
Answer
(a) Given equations are,
(3/5) x – (2/3) y + 1 = 0
⇒ (1/3) y + (2/5) x = 4
⇒ (3/5) x – (2/3) y + 1 = 0
On taking LCM, we get,
9x – 10y + 15 = 0
⇒ 9x – 10y = – 15 ….(1)
(1/3) y + (2/5) x = 4
On taking LCM, we get,
5y + 6x = 60
⇒ 6x + 5y = 60 …(2)
Multiplying equation (2) by 2, we get,
12x + 10y = 120 …(3)
Adding equations (1) and (3), we get,
21x = 105
⇒ x = 5
Substituting the value of x in equation (2), we get,
6 (5) + 5y = 60
⇒ 30 + 5y = 60
⇒ 5y = 60 – 30
⇒ 5y = 30
⇒ y = 6
Therefore, the solution set is (5, 6)
(b) Given equations are,
(x/3) + (y/4) = 11
⇒ (5x/6) – (y/3) = – 7
⇒ (x/3) + (y/4) = 11
On taking LCM, we get,
4x + 3y = 132 ….(1)
⇒ (5x/6) – (y/3) = – 7
On taking LCM, we get,
5x – 2y = – 42 ...(2)
Multiplying equation (1) by 2 and equation (2) by 3, we get,
8x + 6y = 264 ….(3)
15x – 6y = – 126 ...(4)
Adding equations (3) and (4), we get,
23x = 138
⇒ x = 6
Substituting the value of x in equation (1), we get,
4(6) + 3y = 132
⇒ 24 + 3y = 132
⇒ 3y = 132 – 24
⇒ 3y = 108
We get,
y = 36
Therefore, the solution set is (6, 36)
(c) Given equations are,
(3/2x) + (2/3y) = 5
⇒ (5/x) – (3/y) = 1
Let (1/x) = a and (1/y) = b
Then,
We have
(3/2)a + (2/3) b = 5
On taking LCM, we get,
9a + 4b = 30 ….(1)
5a – 3b = 1 …(2)
Multiplying equation (1) by 3 and equation (2) by 4, we get,
27a + 12b = 90 ...(3)
20a – 12b = 4 ….(4)
Adding equations (3) and (4), we get,
47a = 94
⇒ a = 2
Hence,
(1/x) = 2
⇒ x = (1/2)
Substituting the value of a in (1), we get,
9 (2) + 4b = 30
⇒ 18 + 4b = 30
⇒ 4b = 30 – 18
⇒ 4b = 12
We get,
b = 3
⇒ (1/y) = 3
⇒ y = (1/3)
Therefore, the solution set is {(1/2), (1/3)}
(d) Given equations are,
(3/x) – (1/y) = – 9
⇒ (2/x) + (3/y) = 5
Let (1/x) = a and (1/y) = b
Then,
We have,
3a – b = – 9 ….(1)
2a + 3b = 5 ….(2)
Multiplying equation (1) by 3, we get,
9a – 3b = – 27 ...(3)
Adding equations (2) and (3), we get,
11a = – 22
⇒ a = – 2
⇒ (1/x) = – 2
⇒ x = (- 1/2)
Substituting the value of a in equation (1), we get,
3 (-2) – b = – 9
⇒ – 6 – b = – 9
⇒ b = – 6 + 9
⇒ b = 3
(1/y) = 3
⇒ y = (1/3)
Therefore, the solution set is {(- 1/2), (1/3)}
(e) Given equations are,
y – x = 0.8
⇒ [13/{2 (x + y)}] = 1
⇒ y – x = 0.8
⇒ – x + y = 0.8 …(1)
And [1/{2 (x + y)}] = 1
On cross multiplication, we get,
13 = 2 (x + y)
⇒ 13 = 2x + 2y
⇒ 2x + 2y = 13 …(2)
Multiplying equation (1) by 2, we get,
– 2x + 2y = 1.6 ...(3)
Adding equations (2) and (3) we get,
4y = 14.6
⇒ y = 3.65
Substituting the value of y in equation (1), we get,
– x + 3.65 = 0.8
⇒ – x = – 2.85
⇒ x = 2.85
Therefore, the solution set is (2.85, 3.65)
(f) Given equations are,
(2/x) + (3/y) = (9/xy)
⇒ (4/x) + (9/y) = (21/xy)
Let (1/x) = a and (1/y) = b
Then,
We have,
2a + 3b = 9ab ….(1)
4a + 9b = 21ab ...(2)
Multiplying equation (1) by 2, we get,
4a + 6b = 18ab ….(3)
Subtracting equation (3) from equation (2), we get,
3b = 3ab
⇒ a = 1
⇒ (1/x) = 1
⇒ x = 1
Substituting the value of a in equation (1), we get,
2(1) + 3b = 9(1) b
⇒ 2 + 3b = 9b
⇒ 6b = 2
⇒ b = (1/3)
⇒(1/y) = (1/3)
y = 3
Therefore, the solution set is (1, 3)
(g) Given equations are,
(x + y)/xy = 2
⇒ (x – y)/xy = 6
⇒ (x + y)/xy = 2
⇒ x + y = 2xy ...(1)
(x – y)/xy = 6
⇒ x – y = 6xy ….(2)
Adding equations (1) and (2), we get,
2x = 8xy
⇒ y = (1/4)
Substituting the value of y in equation (1), we get,
x + (1/4) = 2x (1/4)
⇒ {(4x + 1)/4} = (x/2)
On cross multiplication, we get,
8x + 2 = 4x
⇒ 4x = – 2
⇒ x = (- 1/2)
Therefore, the solution set is {(- 1/2), (1/4)}
(h) Given equations are,
{2/(x + 1)} – {1/(y – 1)} = (1/2)
⇒ {1/(x + 1)} + (2/(y – 1)} = (5/2)
Let {1/(x + 1)} = a and {1/(y – 1)} = b
Then,
We have,
2a – b = (1/2) ...(1)
a + 2b = (5/2) ...(2)
Multiplying equation (1) by 2, we get,
4a – 2b = 1 …(3)
Adding equations (2) and (3), we get,
5a = (7/2)
⇒ a = (7/10)
⇒ {1/(x + 1)} = (7/10)
On cross multiplication, we get,
10 = 7x + 7
⇒ 7x = 3
⇒ x = (3/7)
Substituting the value of a in equation (3), we get,
4(7/10) – 2b = 1
⇒ (14/5) – 2b = 1
⇒ 2b = (14/5) – 1
⇒ 2b = (9/5)
⇒ b = (9/10)
{1/(y – 1)} = (9/10)
On cross multiplication, we get,
10 = 9y – 9
⇒ 9y = 19
⇒ y = (19/9)
Therefore, the solution set is {(3/7), (19/9)}
(i) Given equations are,
{6/(x + y)} = {7/(x – y)} + 3
⇒ 1/{2 (x + y)} = 1/{3 (x – y)}
Let {1/(x + y)} = a and {1/(x – y)} = b
Then,
We have,
6a = 7b + 3
⇒ 6a – 7b = 3 …(1)
And (1/2) a = (1/3) b
3a = 2b
Multiplying by 2, we get,
6a = 4b ...(2)
Substituting the value of 6a in equation (1), we get,
4b – 7b = 3
⇒ – 3b = 3
⇒ b = -1
⇒ 6a = – 4
⇒ a = (- 2/3)
x + y = (- 3/2) and x – y = -1
Adding both these equations, we get,
2x = (-5/2)
⇒ x = (- 5/4)
⇒ (-5/4) – y = – 1
⇒ y = (- 5/4) + 1
⇒ y = (-1/4)
Therefore, the solution set is {(- 5/4), (-1/4)}
(j) Given equations are,
{5/(x + y)} – {2/(x – y)} = -1
{15/(x + y)} + {7/(x – y)} = 10
Let {1/(x + y)} = a and {1/(x – y)} = b
Then,
We have,
5a – 2b = -1 …(1)
15a + 7b = 10 ...(2)
Multiplying equation (1) by 3, we get,
15a – 6b = – 3 ….(3)
Subtracting equation (3) from equation (2), we get,
13b = 13
⇒ b = 1
Substituting the value of b in equation (1), we get,
5a – 2(1) = – 1
⇒ 5a = – 1 + 2
⇒ 5a = 1
⇒ a = (1/5)
x + y = 5 and x – y = 1
Adding these two equations, we get,
2x = 6
⇒ x = 3
⇒ 3 + y = 5
⇒ y = 5 – 3
⇒ y = 2
Therefore, the solution set is (3, 2)
(k) Given equations are,
{2/(3x + 2y)} + {3/(3x – 2y)} = (17/5)
⇒ {5/(3x + 2y)} + {1/(3x – 2y)} = 2
Let {1/(3x + 2y)} = a and {1/(3x – 2y)} = b
Then,
We have,
2a + 3b = (17/5) ….(1)
5a + b = 2 ….(2)
Multiplying equation (1) by 5 and equation (2) by 2, we get,
10a + 15b = 17 ...(3)
10a + 2b = 4 ...(4)
Subtracting equation (4) from equation (3), we get,
13b = 13
⇒ b = 1
Substituting the value of b in equation (2), we get,
5a + 1 = 2
⇒ 5a = 1
⇒ a = (1/5)
3x + 2y = 5 and 3x – 2y = 1
Adding these two equations, we get,
6x = 6
⇒ x = 1
⇒ 3(1) + 2y = 5
⇒ 2y = 5 – 3
⇒ 2y = 2
⇒ y = 1
Therefore, the solution set is (1, 1)
(l) Given equations are,
{xy/(x + y)} = (6/5)
⇒ {xy/(y – x)} = 6
⇒ {xy/(x + y)} = (6/5)
⇒ (x + y)/xy = (5/6)
⇒ (1/y) + (1/x) = (5/6) ...(1)
⇒ {xy /(y – x)} = 6
⇒ (y – x)/xy = (1/6)
⇒ (1/x) – (1/y) = (1/6) ….(2)
Adding equations (1) and (2), we get,
(2/x) = 1
⇒ x = 2
⇒ (1/y) + (1/2) = (5/6)
⇒ (1/y) = (5/6) – (1/2)
⇒ (1/y) = (5 – 3)/6
⇒ (1/y) = (2/6)
⇒ (1/y) = (1/3)
⇒ y = 3
Therefore, the solution set is (2, 3)
5. If 2x + y = 23 and 4x – y = 19; find the value of x – 3y and 5y – 2x
Answer
Given
2x + y = 23 …..(1)
4x – y = 19 …(2)
Adding equations (1) and (2), we get,
6x = 42
⇒ x = 7
Substitute the value of x in equation (1), we get,
2x + y = 23
⇒ 2 (7) + y = 23
⇒ 14 + y = 23
⇒ y = 23 – 14
⇒ y = 9
Hence,
x – 3y = 7 – 3 (9) = 7 – 27 = – 20
⇒ 5y – 2x = 5 (9) – 2 (7)
⇒ y = 45 – 14
⇒ y = 31
6. If 10y = 7x – 4 and 12x + 18y = 1; find the value of 4x + 6y and 8y – x.
Answer
7. Solve: 4x + 6/y = 15 and 6x – 8/y = 14. Hence, find a if y = ax – 2.
Answer
8. Solve: 3/x – 2/y = 0 and 2/x + 5/y = 19, Hence, find a if y = ax + 3.
Answer
9. Can the following equations hold simultaneously?
7y – 3x = 7
5y – 11x = 87
5x + 4y = 43
If yes, find the value of x and y.
Answer
10. If the following three equations hold simultaneously for x and y, find the value of ‘m’.
2x + 3y + 6 = 0
4x – 3y – 8 = 0
x + my – 1 = 0
Answer
Exercise 8.2
1. Draw the graphs of the following linear equations:
(a) x = 3
(b) y + 5 = 0
(c) 3x + 2y – 6 = 0
(d) 5x – 5y = 8
Answer
(a) The graph of x = 3 is shown below
y + 5 = 0
i.e, y = – 5
The graph is shown below
(c) 3x + 2y – 6 = 0
3x + 2y = 6
2y = 6 – 3x
We get,
y = (6 – 3x) / 2
Corresponding values of x and y can be tabulated as follows:
X |
0 |
1 |
-1 |
Y |
3 |
1.5 |
4.5 |
Plotting the points (0, 3), (1, 1.5) and (-1, 4.5),
We get the graph as shown below
5x – 5y = 8
⇒ 5x = 8 – 5y
⇒ x = (8 – 5y)/5
Corresponding values of x and y can be tabulated as follows:
X |
1.6 |
0.6 |
2.6 |
Y |
0 |
1 |
-1 |
Plotting the points (1.6, 0), (0.6, 1) and (2.6, – 1),
We get the graph as shown below
2. Draw the graph for each of the following equation: Also, find the coordinates of the points where the graph of the equation meets the coordinate axes:
(a) (1/2) x + (1/3) y = 1
(b) {(3x + 14)/2} = {(y – 10)/5}
Answer
(a) Given
(1/2) x + (1/3) y = 1
⇒ 3x + 2y = 6
⇒ 2y = 6 – 3x
We get,
y = {(6 – 3x) / 2}
Corresponding values of x and y can be tabulated as follows:
X |
0 |
1 |
-1 |
Y |
3 |
1.5 |
4.5 |
Plotting the points (0, 3), (1, 1.5) and (-1, 4.5),
We get the graph as shown below
Therefore, the graph of the equation meets the X – axis at (2, 0) and Y- axis at (0, 3)
(b) Given
{(3x + 14)/2} = {(y – 10)/5}
On cross multiplication, we get,
15x + 70 = 2y – 20
⇒ 15x – 2y = – 20 – 70
⇒ 15x – 2y = – 90
⇒ 2y = 90 + 15x
⇒ y = {(90 + 15x) / 2}
Corresponding values of x and y can be tabulated as below
X |
– 5 |
– 2 |
– 1 |
Y |
7.5 |
30 |
37.5 |
Plotting the points (-5, 7.5), (-2, 30) and (-1, 37.5),
We get the graph as shown below:
Therefore, the graph of the equation meets the X-axis at (-6, 0) and Y-axis at (0, 45)
3. Draw the graph of the equation 4x – 3y + 12 = 0
Also, find the area of the triangle formed by the line drawn and the coordinate axes.
Answer
Given equation is,
4x – 3y + 12 = 0
⇒ 4x = 3y – 12
⇒ x = (3y – 12)/4
Corresponding values of x and y can be tabulated as follows:
X |
-3 |
– 1.5 |
0 |
Y |
0 |
2 |
4 |
Plotting the points (-3, 0), (-1.5, 2) and (0, 4)
We get the graph as shown below:
Area of △OAB = (1/2) × OB × OA
Area of △OAB = (1/2) × 3 × 4
We get,
Area of △OAB = 6 sq. units
4. Draw the graph of the equation
y = 5x – 4 Find graphically
(a) the value of x, when y = 1
(b) the value of y, when x = – 2
Answer
Given equation is,
y = 5x – 4
Corresponding values of x and y can be tabulated as follows:
X |
0 |
2 |
-1 |
Y |
– 4 |
6 |
-9 |
Plotting the points (0, -4), (2, 6) and (-1, -9),
We get the graph as shown below:
X |
1 |
2 |
3 |
P |
Y |
1 |
q |
-5 |
7 |
Find graphically the values of ‘p’ and ‘q’.
Answer
The graph is shown below:
From the graph, we find that,
p = -1 and q = -2
6. A straight line passes through the points (2, 5) and (-4, -7). Plot these points on a graph paper and draw the straight line passes through these points. If points (a, - 1) and (-5, b) lie on the line drawn, find the value of a and b.
Answer
The graph is as follows:
7. Solve the following equations graphically:
(i) x + 3y = 8
3x = 2 + 2y
(ii) 2x + 4y = 7
3x + 8y = 10
(iii) 2x – y = 9
5x + 2y = 27
(iv) x + 4y + 9 = 0
3y = 5x – 1
(v) x = 4
3x/3 – y = 5
(vi) 3y = 5 – x
2x = y + 3
(vii) x – 2y = 2
x/2 – y = 1
(viii) 2x – 6y + 10 = 0
3x – 9y + 25 = 0
(ix) 2 + 3y/x = 6/x
6x/y – 5 = 4/y
(x) x + 2y – 7 = 0
2x – y – 4 = 0
Answer
(i)
(ii)8. (i) Find graphically the vertices of the triangle, whose sides have the equations 2y – x = 8, 5y – x = 14 and y = 2x – 1.
(ii) Find graphically the vertices of the triangle, whose sides have are given by 3y = x + 18, x + 7y = 22 and y + 3x = 26.
Answer
(i)
9. Solve the following system of linear equations graphically:
4x – 5y – 20 = 0
3x + 3y – 15 = 0
Determine the vertices of the triangle formed by the lines, represented by the above equations and the y –axis.
Answer
10. Solve the following system of equations graphically
x – y + 1 = 0
4x + 3y = 24
Answer
11. Draw the graphs of the following equations:
3x + 2y + 6 = 0
3x + 8y – 12 = 0
Also, determine the co-ordinates of the vertices of the triangle formed by these lines and x-axis.
Answer
12. (a) Solve the following system of equations graphically:
2x = 23 – 3y
5x = 20 + 8y
(b) Solve the following system of equations graphically:
6x – 3y + 2 = 7x + 1
5x + 1 = 4x – y + 2
Also, find the area of the triangle formed by these lines and x-axis in each graph.
Answer
(a)
(b)Exercise 8.3
1. The length of a rectangle is twice its width. If its perimeter is 30 units, find its dimensions.
Answer
Let the length of a rectangle = x units and
Breadth of a rectangle = y units
As per the given condition, we have,
x = 2y
⇒ x – 2y = 0 …(1)
Also,
Perimeter of a rectangle = 30 units
2 (x + y) = 30
⇒ x + y = 15 ...(2)
Subtracting equation (2) from equation (1), we get,
-3y = – 15
⇒ y = 5
Substituting the value of y in equation (1), we get,
x – 2y = 0
⇒ x – 2(5) = 0
⇒ x – 10 = 0
We get,
x = 10
Therefore, the length and breadth of a rectangle are 10 units and 5 units respectively.
2. The difference of two numbers is 3, and the sum of three times the larger one and twice the smaller one is 19. Find the two numbers.
Answer
Let the larger number be x and the smaller number be y
According to the given information,
We have,
x – y = 3
⇒ x = 3 + y ...(1)
Also,
3x + 2y = 19
⇒ 3 (3 + y) + 2y = 19 From equation (1)
⇒ 9 + 3y + 2y = 19
⇒ 5y = 19 – 9
⇒ 5y = 10
We get,
y = 2
x = 3 + 2
⇒ x = 5
Therefore, the required numbers are 5 and 2 respectively.
3. If a number is thrice the other and their sum is 68, find the numbers.
Answer
Let the two numbers be x and y respectively
Then,
As per the given statement, we have,
x = 3y ….(1)
And,
x + y = 68
⇒ 3y + y = 68
⇒ 4y = 68
⇒ y = 17
⇒ x = 3y = 3 × 17 = 51
Therefore, the required two numbers are 51 and 17 respectively.
4. The sum of four times the first number and three times the second number is 15. The difference of three times the first number and twice the second number is 7. Find the numbers.
Answer
Let the two numbers be x and y respectively
Then,
We have,
4x + 3y = 15 ….(1)
3x – 2y = 7 …(2)
Multiplying equation (1) by (2) and equation (2) by 3,
We get,
8x + 6y = 30 …(3)
9x – 6y = 21 ….(4)
Adding equations (3) and (4),
We get,
17x = 51
⇒ x = (51/17)
⇒ x = 3
Substituting the value of x in equation (1), we get,
4x + 3y = 15
⇒ 4 (3) + 3y = 15
⇒ 12 + 3y = 15
⇒ 3y = 15 – 12
⇒ 3y = 3
We get,
y = 1
Therefore, the two required numbers are 3 and 1 respectively.
5. In a two-digit number, the sum of the digits is 7. The difference of the number obtained by reversing the digits and the number itself is 9. Find the number.
Answer
Let digit at ten’s place be x and the digit at unit’s place by y
Then,
The number is 10x + y
Number obtained by reversing the digits = 10y + x
According to the given information, we have,
x + y = 7 ...(1)
And,
(10y + x) – (10x + y) = 9
⇒ 10y + x – 10x – y = 9
⇒ 9y – 9x = 9
Taking 9 as common, we get,
9 (y – x) = 9
⇒ y – x = 1 ...(2)
Adding equations (1) and (2), we get,
2y = 8
⇒ y = 4
Substitute the value of y in equation (1),
We get,
x + y = 7
⇒ x + 4 = 7
⇒ x = 7 – 4
We get,
x = 3
Therefore,
Required number = 10x + y
= 10 (3) + 4
= 30 + 4
= 34
Hence, the required number is 34.
6. The sum of a two-digit number and the number obtained by reversing the digits is 110 and the difference of two digits is 2. Find the number.
Answer
7. Seven more than a 2-digit number is equal to two less than the number obtained by reversing the digits. The sum of the digits is 5. Find the number.
Answer
8. If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it becomes 4/5. Find the fraction.
Answer
9. The ratio of two numbers is 2/5. If 4 is added in first and 32 is subtracted from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.
Answer
10. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes ½. Find the fraction.
Answer
11. If 1 added to the denominator of a fraction, the fraction becomes ½. If 1 is added to the numerator of the fraction, the fraction becomes 1. Find the fraction.
Answer
12. The age of the father is seven times the age of the son. Ten years later, the age of the father will be thrice the age of the son. Find their present ages.
Answer
13. The present ages of Kapil and Karuna are in the ratio 2 : 3. Six years later, the ratio will be 5 : 7. Find their present ages.
Answer
14. A father’s age is three times the age of his child. After 12 years, twice the age of father will be 36 more than thrice the age of his child. Find his present age.
Answer
15. In a triangle, the sum of two angles is equal to the third angle. If the difference between these two angles is 20°, determine all the angles.
Answer
16. In a Î”ABC, ∠A = x°, ∠B = (2x - 30°). ∠C = y° and also, ∠A + ∠B = one right angle. Find the angles. Also, state the type of this triangle.
Answer
17. A two-digit number is such that the ten’s digit exceeds thrice the unit’s digit by 3 and the number obtained by interchanging the digits is 2 more than twice the sum of the digits. Find the number.
Answer
18. Anil and Sunita have incomes in the ratio 3 : 5. If they spend in the ratio 1: 3, each saves T 5000. Find the income of each.
Answer
19. The ratio of passed and failed students in an examination was 3 : 1. Had 30 less appeared and 10 less failed, the ratio of passes to failures would have been 13: 4. Find the number of students who appeared for the examination.
Answer
20. An eraser costs Rs 1.50 less than a sharpener. Also, the cost of 4 erasers and 3 sharpeners is Rs 29. Taking x and y as the costs (in Rs) of an eraser and a sharpener respectively, write two equations for the above statements and find the value of x and y.
Answer
21. A person goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the person in still water and the speed of this stream.
Answer
22. A boat goes 18 km upstream in 3 hours and 24km downstream in 2 hours. Find the speed of the boat in still water and the speed of the stream.
Answer
23. Salman and Kirti start at the same time from two places 28 km apart. If they walk in the same direction, Salman overtakes Kirti in 28 hours but if they walk in the opposite directions, they meet in 4 hours. Find their speeds (in km/h).
Answer
24. A solution containing 12% alcohol is to be mixed with a solution containing 4% alcohol to make 20 gallons of solution containing 9% alcohol. How much of each solution should be used ?
Answer
25. 9 pens and 5 pencils cost Rs 32, and 7 pens and 8 pencils cost Rs 29. Find the unit price for each pen and pencil.
Answer
26. Sunil and Kafeel both have some oranges. If Sunil gives 2 oranges to Kafeel, then Kafeel will have thrice as many as Sunil. And if Kafeel gives 2 oranges to Sunil, then they will have the same numbers of oranges. How many oranges does each have?
Answer
27. Samidha and Shreya have pocket money Rs x. and Rs y. respectively at the beginning of a week. They both spend money throughout the week. At the end of the week, Samidha spends Rs 500 and is left with as much money as Shreya had in the the beginning of the week. Shreya spends Rs 500 and is left with 3/5 of what Samidha had in the beginning of the week. Find their pocket money.
Answer
28. Two mobiles S1 and S2 are sold for Rs 10, 490 marking 4% profit on S1 and 6% on S2. If the two mobiles are sold for Rs 10,510, a profit of 6% is made on S1 and 4% on S2. Find the cost price of both the mobiles.
Answer
29. A and B can build a wall in 6.2/3 days. If A’s one day work is 1.1/4 of one day work of B, find in 4 how many days A and B alone can build the wall.
Answer