ML Aggarwal Solutions for Chapter 5 Simultaneous Linear Equation Class 9 Maths ICSE
Exercise 5.1
Solve the following systems of simultaneous linear equations by the substitution method (1 to 4):
1. (i) x + y = 14
x – y = 4
(ii) s – t = 3
s/3 + t/2 = 6
(iii) 2x + 3y = 9
3x + 4y = 5
(iv) 3x – 5y = 4
9x – 2y = 7
Solution
(i) x + y = 14
x – y = 4
It can be written as
x = 4 + y
By substituting the value in the above equation
4 + y + y = 14
By further calculation
2y = 14 – 4 = 10
Dividing by 2
y = 10/2 = 5
So we get
x = 4 + 5 = 9
Hence, x = 9 and y = 5.
(ii) s – t = 3
s/3 + t/2 = 6
By taking LCM
2s + 3t = 6×6 = 36
We know that
s – t = 3 …(1)
2s + 3t = 36 …(2)
So we get
s = 3 + t …(3)
By substituting the value of s in equation (2)
2(3 + t) + 3t = 36
By further calculation
6 + 2t + 3t = 36
So we get
5t = 36 – 6 = 30
By division
t = 30/5 = 6
Substituting t in equation (3)
s = 3 + 6 = 9
Hence, s = 9 and t = 6.
(iii) 2x + 3y = 9 …(1)
3x + 4y = 5 …(2)
Equation (1) can be written as
2x = 9 – 3y
x = (9 – 3y)/2 ...(3)
By substituting the value of x in equation (2)
3×(9 – 3y)/ 2 + 4y = 5
By further calculation
(27 – 9y)/2 + 4y = 5
By taking LCM
27 – 9y + 8y = 10
So we get
-y = – 17
y = 17
Substituting y in equation (3)
x = [9 – (3×17)]/2
By further calculation
x = (9 – 51)/ 2
x = – 21
Hence, x = – 21 and y = 17.
(iv) 3x – 5y = 4 …(1)
9x – 2y = 7 …(2)
Multiply equation (1) by 3
9x – 15y = 12
9x – 2y = 7
By subtracting both the equations
– 13y = 5
y = -5/13
Equation (1) can be written as
3x – 5y = 4
x = (4 + 5y)/3 …(3)
By substituting the value of x in equation (2)
9 [(4 + 5y)/3] – 2y = 7
By further calculation
12 + 15y – 2y = 7
13y = – 5
So we get
y = -5/13
Substituting y in equation (3)
Hence, x = 9/13 and y = – 5/13.
2. (i) a + 3b = 5
7a – 8b = 6
(ii) 5x + 4y – 4 = 0
x – 20 = 12y
Solution
(i) a + 3b = 5 …(1)
7a – 8b = 6 …(2)
Now multiply equation (1) by 7
7a + 21b = 35 …(3)
7a – 8b = 6 …(4)
By subtracting both the equations
29b = 29
So we get
b = 29/29 = 1
Now substituting b = 1 in equation (1)
a + 3 (1) = 5
By further calculation
a + 3 = 5
So we get
a = 5 – 3 = 2
Therefore, a = 2 and b = 1.
(ii) 5x + 4y – 4 = 0
x – 20 = 12y
We can write it as
5x + 4y = 4 …(1)
x – 12y = 20 …(2)
Now multiply equation (2) by 5
5x + 4y = 4 …(3)
5x – 60y = 100
By subtracting both the equations
64y = –96
So we get
y = -96/64 = –3/2
Now substitute the value of y in equation (1)
5x + 4 (-3/2) = 4
By further calculation
5x + 2 (-3) = 4
So we get
5x – 6 = 4
⇒ 5x = 4 + 6 = 10
By division
x = 10/5 = 2
Therefore, x = 2 and y = – 3/2.
3. (i) 2x – 3y/4 = 3
5x – 2y – 7 = 0
(ii) 2x + 3y = 23
5x – 20 = 8y
Solution
(i) 2x – 3y/4 = 3
5x – 2y – 7 = 0
We can write it as
2x/1 – 3y/4 = 3
By taking LCM
(8x – 3y)/ 4 = 3
By cross multiplication
8x – 3y = 12 …(1)
5x – 2y = 7 …(2)
Now multiply equation (1) by 2 and (2) by 3
16x – 6y = 24
15x – 6y = 21
By subtracting both the equations
x = 3
Now substituting the value of x in equation (1)
8×3 – 3y = 12
By further calculation
24 – 3y = 12
⇒ –3y = 12 – 24
So we get
– 3y = – 12
⇒ y = – 12/-3 = 4
Therefore, x = 3 and y = 4.
(ii) 2x + 3y = 23
5x – 20 = 8y
We can write it as
2x + 3y = 23 …(1)
5x – 8y = 20 …(2)
By multiplying equation (1) by 5 and equation (2) by 2
10x + 15y = 115
10x – 16y = 40
By subtracting both the equations
31y = 75
So we get
y = 75/31 = 2 13/31
By substituting the value of y in equation (1)
2x + 3 (75/31) = 23
By further calculation
2x + 225/31 = 23
We can write it as
2x = 23/1 – 225/31
Taking LCM
2x = (713 – 225)/31 = 488/31
So we get
x = 488/ (31×2) = 244/31 = 7 27/31
Therefore, x = 7 27/31 and y = 2 13/31.
4. (i) mx – ny = m2 + n2
x + y = 2m
(ii) 2x/a + y/b = 2
x/a – y/b = 4
Solution
(i) mx – ny = m2 + n2 …(1)
x + y = 2m …(2)
We can write it as
x = 2m – y …(3)
Now substitute the value of x in (1)
m(2m – y) – ny = m2 + n2
By further calculation
2m2 – my – ny = = m2 + n2
Taking out y as common
m2 – y(m + n) = n2
It can be written as
m2 – n2 – y(m + n) = 0
Expanding using formula
(m – n) (m + n) – y(m + n) = 0
Taking (m + n) as common
(m + n)[(m – n) – y] = 0
So we get
m – n – y = 0
y = m – n
From equation (3)
x = 2m – (m – n)
By further calculation
x = 2m – m + n = m + n
Hence, x = m + n and y = m – n.
(ii) 2x/a + y/b = 2 …(1)
x/a – y/b = 4 …(2)
Adding both the equations
3x/a = 6
So we get
x = 6a/3 = 2a
Substituting x in equation (1)
2 (2a)/ a + y/b = 2
By further calculation
4a/a + y/b = 2
So we get
4 + y/b = 2
y/b = 2 – 4 = –2
Here
y = –2b
Therefore, x = 2a and y = –2b.
5. Solve 2x + y = 35, 3x + 4y = 65. Hence, find the value of x/y.
Solution
It is given that
2x + y = 35 …(1)
3x + 4y = 65 …(2)
Now multiply equation (1) by 4
8x + 4y = 140 …(3)
3x + 4y = 65 …(4)
By subtracting both the equations
5x = 75
⇒ x = 75/5 = 15
Now substituting the value of x in equation (1)
8×15 + 4y = 140
By further calculation
120 + 4y = 140
⇒ 4y = 140 – 120
So we get
4y = 20
⇒ y = 20/4 = 5
Here
x/y = 15/5 = 3
Therefore, x/y = 3.
6. Solve the simultaneous equations 3x – y = 5, 4x – 3y = –1. Hence, find p, if y = px –3.
Solution
It is given that
3x – y = 5 …(1)
4x – 3y = –1 …(2)
Now multiply equation (1) by 3
9x – 3y = 15 …(3)
4x – 3y = –1 …(4)
Subtracting equation (3) and (4)
5x = 16
⇒ x = 16/5
Substitute the value of x in equation (3)
3× 16/5 – y = 5
By further calculation
48/5 – y = 5
⇒ 48/5 – 5 = y
Taking LCM
(48 – 25)/5 = y
So we get
y = 23/5
We know that
y = px – 3
⇒ 23/5 = p× 16/5 – 3
Substitute the value of x and y
23/5 + 3 = 16p/5
Taking LCM
(23 + 15)/5 = 16p/5
By further calculation
38/5 = 16p/5
So we get
16p = 38
⇒ p = 19/8
Therefore, x = 16/5, y = 23/5 and p = 19/8.
Exercise 5.2
Solve the following systems of simultaneous linear equations by the elimination method (1 to 9):
1. (i) 3x + 4y = 10
2x – 2y = 2
(ii) 2x = 5y + 4
3x – 2y + 16 = 0
Solution
(i) 3x + 4y = 10 …(1)
2x – 2y = 2 …(2)
Multiplying equation (1) by 1 and (2) by 2
3x + 4y = 10
4x – 4y = 4
By adding both the equations
7x = 14
By division
x = 14/7 = 2
Substituting the value of x in equation (2)
2×2 – 2y = 2
By further calculation
4 – 2y = 2
So we get
2y = 4 – 2 = 2
⇒ y = 2/2 = 1
Therefore, x = 2 and y = 1.
(ii) 2x = 5y + 4
3x – 2y + 16 = 0
We can write it as
2x – 5y = 4 …(1)
3x – 2y = – 16 …(2)
Now multiply equation (1) by 3 and (2) by 2
6x – 15y = 12 …(3)
6x – 4y = – 32 …(4)
By subtracting both the equations
–11y = 44
⇒ y = -44/11 = – 4
Substitute the value of y in equation (1)
2x – 5(-4) = 4
By further calculation
2x + 20 = 4
So we get
2x = 4 – 20 = –16
⇒ x = –16/2 = –8
Therefore, x = –8 and y = –4.
2. (i) ¾ x – 2/3 y = 1
3/8 x – 1/6 y = 1
(ii) 2x – 3y – 3 = 0
2x/3 + 4y + ½ = 0.
Solution
(i) ¾ x – 2/3 y = 1
3/8 x – 1/6 y = 1
We can write it as
¾ x – 2/3 y = 1
(9x – 8y)/12 = 1
By cross multiplication
9x – 8y = 12 …(1)
3/8 x – 1/6 y = 1
⇒ (9x – 4y)/24 = 1
By cross multiplication
9x – 4y = 24 …(2)
Subtracting equations (1) and (2)
–4y = –12
By division
y = –12/–4 = 3
Substitute the value of y in (1)
9x – 8×3 = 12
By further calculation
9x – 24 = 12
⇒ 9x = 12 + 24 = 36
By division
x = 36/9 = 4
Therefore, x = 4 and y = 3.
(ii) 2x – 3y – 3 = 0
2x/3 + 4y + ½ = 0
We can write it as
2x – 3y – 3 = 0
⇒ 2x – 3y = 3 …(1)
⇒ 2x/3 + 4y + ½ = 0
⇒ 2x/3 + 4y = – ½
Taking LCM
(2x + 12y)/ 3 = – ½
By cross multiplication
2 (2x + 12y) = –1×3
So we get
4x + 24y = – 3 …(2)
Multiply equation (1) by 2
4x – 6y = 6
4x + 24y = –3
By subtracting both the equations
–30y = 9
So we get
y = -9/30 = –3/10
Substitute the value of y in equation (1)
2x – 3(-3/10) = 3
By further calculation
2x + 9/10 = 3
We can write it as
2x = 3 – 9/10
By taking LCM
2x = (30 – 9)/ 10
So we get
2x = 21/10
⇒ x = 21/20
Therefore, x = 21/20 and y = –3/10.
3. (i) 15x – 14y = 117
14x – 15y = 115
(ii) 41x + 53y = 135
53x + 41y = 147.
Solution
(i) 15x – 14y = 117 …(1)
14x – 15y = 115 …(2)
Now multiply equation (1) by 14 and (2) by 15
210x – 196y = 1638 …(3)
210x – 225y = 1725 …(4)
By subtracting both the equations
29y = –87
So we get
y = -87/29 = –3
Substitute the value of y in equation (1)
15x – 14(-3) = 117
By further calculation
15x + 42 = 117
So we get
15x = 117 – 42 = 75
By division
x = 75/15 = 5
Therefore, x = 5 and y = – 3.
(ii) 41x + 53y = 135 …(1)
53x + 41y = 147 …(2)
Now multiply equation (1) by 53 and (2) by 41
2173x + 2809y = 7155 …(3)
2173x + 1681y = 6027 …(4)
By subtracting both the equations
1128y = 1128
So we get
y = 1128/1128 = 1
Substitute the value of y in equation (1)
41x + 53×1 = 135
By further calculation
41x + 53 = 135
So we get
41x = 135 – 53 = 82
By division
x = 82/41 = 2
Therefore, x = 2 and y = 1.
4. (i) x/6 = y – 6
3x/4 = 1 + y
(ii) x – 2/3 y = 8/3
2x/5 – y = 7/5.
Solution
(i) x/6 = y – 6
3x/4 = 1 + y
We can write it as
x = 6 (y – 6)
⇒ x = 6y – 36
⇒ x – 6y = –36 …(1)
⇒ 3x/4 = 1 + y
By cross multiplication
3x = 4 (1 + y)
So we get
3x = 4 + 4y
⇒ 3x – 4y = 4 ...(2)
Multiply equation (1) by 3
3x – 18y = – 108
⇒ 3x – 4y = 4
Subtracting both the equations
– 14y = – 112
So we get
y = – 112/-14 = 8
Substitute the value of y in equation (1)
x – 6×8 = – 36
By further calculation
x – 48 = – 36
⇒ x = – 36 + 48
⇒ x = 12
Therefore, x = 12 and y = 8.
(ii) x – 2/3 y = 8/3
⇒ 2x/5 – y = 7/5
We can write it as
x – 2/3 y = 8/3
Taking LCM
(3x – 2y)/3 = 8/3
By cross multiplication
3x – 2y = 8/3 ×3 = 8
⇒ 3x – 2y = 8 …(1)
⇒ 2x/5 – y = 7/5
Taking LCM
(2x – 5y)/ 5 = 7/5
By cross multiplication
2x – 5y = 7/5 ×5 = 7
⇒ 2x – 5y = 7 ...(2)
Multiply equation (1) by 2 and (2) by 3
6x – 4y = 16 …(3)
6x – 15y = 21 …(4)
Subtracting both the equations
11y = –5
⇒ y = –5/11
Substitute the value of y in equation (1)
3x – 2(-5/11) = 8
By further calculation
3x + 10/11 = 8
We can write it as
3x = 8 – 10/11
Taking LCM
3x = (88 – 10)/11 = 78/11
By cross multiplication
x = 78/(11×3) = 26/11
Therefore, x = 26/11 and y = –5/11.
5. (i) 9 – (x – 4) = y + 7
2 (x + y) = 4 – 3y
(ii) 2x + (x – y)/ 6 = 2
x – (2x + y)/3 = 1.
Solution
(i) 9 – (x – 4) = y + 7
⇒ 2(x + y) = 4 – 3y
We can write it as
9 – (x – 4) = y + 7
⇒ 9 – x + 4 = y + 7
By further calculation
13 – x = y + 7
⇒ – x – y = 7 – 13 = –6
⇒ x + y = 6 …(1)
⇒ 2(x + y) = 4 – 3y
⇒ 2x + 2y = 4 – 3y
By further calculation
2x + 2y + 3y = 4
So we get
2x + 5y = 4 …(2)
Now multiply equation (1) by 5 and (2) by 1
5x + 5y = 30
⇒ 2x + 5y = 4
By subtracting both the equations
3x = 26
So we get
x = 26/3
Substitute the value of x in (1)
26/3 + y = 6
We can write it as
y = 6 – 26/3
Taking LCM
y = (18 – 26)/3
So we get
y = – 8/3
Therefore, x = 26/3 and y = – 8/3.
(ii) 2x + (x – y)/6 = 2
⇒ x – (2x + y)/3 = 1
⇒ 2x + (x – y)/6 = 2
Multiply by 6
12x + x – y = 12
By further calculation
13x – y = 12 ...(2)
⇒ x – (2x + y)/3 = 1
Multiply by 3
3x – 2x – y = 3
By further calculation
x – y = 3 …(2)
So we get
x = 3 + y …(3)
Substitute the value of x in (1)
13 (3 + y) – y = 12
By further calculation
39 + 13y – y = 12
So we get
12y = 12 – 39 = – 27
By division
y = –27/12 = – 9/4
Substitute the value of y in (3)
x = 3 + y
⇒ x = 3 + (-9)/4
By further calculation
x = 3 – 9/4
Taking LCM
x = (12 – 9)/ 4
⇒ x = ¾
Therefore, x = ¾ and y = –9/4.
6. x – 3y = 3x – 1 = 2x – y.
Solution
It is given that
x – 3y = 3x – 1 = 2x – y
Here,
x- 3y = 3x – 1
⇒ x – 3x – 3y = –1
By further calculation
⇒ –2x – 3y = –1
⇒ 2x + 3y = 1 …(1)
3x – 1 = 2x – y
⇒ 3x – 2x + y = 1
By further simplification
⇒ x + y = 1 …(2)
Multiply equation (2) by 2 and subtract from equation (1)
2x + 3y = 1
2x + 2y = 2
So we get
y = –1
Substitute the value of y in equation (1)
2x + 3(-1) = 1
So we get
2x – 3 = 1
⇒ 2x = 1 + 3 = 4
By division
⇒ x = 4/2 = 2
Therefore, x = 2 and y = –1.
7. (i) 4x + (x – y)/8 = 17
2y + x – (5y + 2)/3 = 2
(ii) (x + 1)/2 + (y – 1)/3 = 8
(x – 1)/3 + (y + 1)/ 2 = 9.
Solution
(i) 4x + (x–y)/8 = 17
2y + x – (5y+2)/3 = 2
We can write it as
4x + (x–y)/8 = 17
⇒ (32 + x – y)/8 = 17
By further calculation
⇒ (33x – y)/8 = 17
By cross multiplication
⇒ 33x – y = 136 …(1)
2y + x – (5y + 2)/3 = 2
Taking LCM
⇒ [3(2y + x) – 5(5y + 2)]/3 = 2
By further calculation
6y + 3x – 5y – 2 = 2×3
So we get
y + 3x – 2 = 6
⇒ 3x + y = 6 + 2
⇒ 3x + y = 8 …(2)
By adding both the equations
36x = 144
By division
x = 144/36 = 4
Substitute the value of x in equation (1)
33×4 – y = 136
By further calculation
132 – y = 136
⇒ –y = 136 – 132
So we get
–y = 4
⇒ y = –4
Therefore, x = 4 and y = –4.
(ii) (x + 1)/2 + (y – 1)/3 = 8
(x – 1)/3 + (y + 1)/2 = 9
We can write it as
(x + 1)/2 + (y – 1)/3 = 8
Taking LCM
⇒ (3x + 3 + 2y – 2)/6 = 8
By further calculation
3x + 2y + 1 = 48
So we get
3x + 2y = 47 …(1)
(x – 1)/3 + (y + 1)/2 = 9
Taking LCM
⇒ (2x – 2 + 3y + 3)/6 = 9
By further calculation
⇒ 2x + 3y + 1 = 54
So we get
⇒ 2x + 3y = 53 …(2)
By adding equation (1) and (2)
5x + 5y = 100
Dividing by 5
⇒ x + y = 20 ...(3)
By subtracting equation (1) and (2)
x – y = –6 …(4)
Now add equation (3) and (4)
2x = 14
⇒ x = 14/2 = 7
Subtracting equation (4) and (3)
2y = 26
⇒ y = 26/2 = 13
Therefore, x = 7 and y = 13.
8. (i) 3/x + 4y = 7
5/x + 6y = 13
(ii) 5x – 9 = 1/y
x + 1/y = 3.
Solution
(i) 3/x + 4y = 7 …(1)
5/x + 6y = 13 …(2)
Substitute 1/x = a in equation (1) and (2)
3a + 4y = 7 …(3)
5a + 6y = 13 …(4)
Multiply equation (3) by 5 and (4) by 3
15a + 20y = 35
15a + 18y = 39
Subtracting both the equations
2y = -4
So we get
⇒ y = –4/2 = –2
Substitute the value of y in equation (3)
3a + 4 (-2) = 7
By further calculation
3a – 8 = 7
⇒ 3a = 7 + 8 = 15
So we get
3a = 15
⇒ a = 15/3 = 5
Here x = 1/a = 1/5
Therefore, x = 1/5 and y = –2.
(ii) 5x – 9 = 1/y …(1)
x + 1/y = 3 …(2)
Substitute 1/y = b in (1) and (2)
5x – 9 = b
⇒ 5x – b = 9 …(3)
⇒ x + b = 3 …(4)
By adding equation (3) and (4)
5x – b = 9 …(3)
x + b = 3 …(4)
So we get
6x = 12
By division
⇒ x = 12/6 = 2
Substitute the value of x in equation (4)
2 + b = 3
⇒ b = 3 – 2
⇒ b = 1
Here 1/y = b
b = 1/y
⇒ y = 1
Therefore, x = 2 and y = 1.
9. (i) px + qy = p – q
qx – py = p + q
(ii) x/a – y/b = 0
ax + by = a2 + b2.
Solution
(i) px + qy = p – q …(1)
qx – py = p + q …(2)
Now multiply equation (1) by p and (2) by q
p2x + pqy = p2 – pq
q2x – pqy = pq + q2
By adding both the equations
(p2 + q2) x = p2 + q2
By further calculation
x = (p2+q2)/(p2+q2) = 1
From equation (1)
p×1 + qy = p – q
By further calculation
⇒ p – qy = p – q
So we get
⇒ qy = p – q – p = –q
Here
⇒ y = -q/q = – 1
Therefore, x = 1 and y = –1.
(ii) x/a – y/b = 0
ax + by = a2 + b2
We can write it as
x/a – y/b = 0
Taking LCM
(bx – ay)/ab = 0
By cross multiplication
⇒ bx – ay = 0 ...(1)
ax + by = a2 + b2 …(2)
Multiply equation (1) by b and equation (2) by a
b2x – aby = 0
a2x + aby = a2 + ab2
By adding both the equations
(a2 + b2)x = a2+ ab2 = a (a2 + b2)
So we get
⇒ x = a(a2 + b2)/(a2+b2) = a
From equation (2)
b×a – ay = 0
By further calculation
ab – ay = 0
⇒ ay = ab
So we get
y = ab/a = b
Therefore, x = a and y = b.
10. Solve 2x + y = 23, 4x – y = 19. Hence, find the values of x – 3y and 5y – 2x.
Solution
It is given that
2x + y = 23 …(1)
4x – y = 19 …(2)
Adding both the equations
6x = 42
⇒ x = 42/6 = 7
Substitute the value of x in equation (1)
2×7 + y = 23
By further calculation
14 + y = 23
So we get
y = 23 – 14 = 9
Therefore, x = 7 and y = 9.
x – 3y = 7 – 3×9 = 7 – 27 = –20
5y – 2x = 5 × 9 – 2×7 = 45 – 14 = 31
11. The expression ax + by has value 7 when x = 2, y = 1. When x = –1, y = 1, it has value 1, find a and b.
Solution
It is given that
ax + by = 7 when x = 2 and y = 1
Substituting the values
a(2) + b(1) = 7
⇒ 2a + b = 7 …(1)
Here
ax + by = 1 when x = –1 and y = 1
Substituting the values
a(-1) + b(1) = 1
⇒ –a + b = 1 …(2)
By subtracting both the equations
–3a = –6
So we get
a = – 6/–3 = 2
Substituting the value of a in equation (1)
2×2 + b = 7
By further calculation
⇒ 4 + b = 7
⇒ b = 7 – 4 = 3
Therefore, a = 2 and b = 3.
12. Can the following equations hold simultaneously?
3x – 7y = 7
11x + 5y = 87
5x + 4y = 43.
If so, find x and y.
Solution
3x – 7y = 7 …(1)
11x + 5y = 87 …(2)
5x + 4y = 43 …(3)
Now multiply equation (1) by 5 and (2) by 7
15x – 35y = 35
77x + 35y = 609
By adding both the equations
92x = 644
By division
x = 644/92 = 7
Substitute the value of x in equation (1)
3×7 – 7y = 7
By further calculation
21 – 7y = 7
So we get
– 7y = 7 – 21 = – 14
⇒ y = – 14/–7 = 2
Therefore, x = 7 and y = 2.
If x = 7 and y – 2 satisfy the equation (3) then we can say that the equations hold simultaneously
Substitute the value of x and y in equation (3)
5x + 4y = 43
By further calculation
⇒ 5×7 + 4×2 = 43
So we get
35 + 8 = 43
43 = 43 which is true.
Therefore, the equations hold simultaneously.
Exercise 5.3
1. Solve the following systems of simultaneous linear equations by cross-multiplication method:
(i) 3x + 2y = 4
8x + 5y = 9
(ii) 3x – 7y + 10 = 0
y – 2x = 3
Solution
(i) 3x + 2y = 4
8x + 5y = 9
We can write it as
3x + 2y – 4 = 0
8x + 4y – 9 = 0
By cross multiplication method
x/(-18 + 20) = y/(-32 + 27) = 1/(15 – 16)
By further calculation
x/2 = y/-5 = 1/-1
So we get
x/2 = –1
⇒ x = –2
⇒ y = –5(-1) = 5
Therefore, x = – 2 and y = 5.
(ii) 3x – 7y + 10 = 0
y – 2x = 3
We can write it as
3x – 7y + 10 = 0
y – 2x – 3 = 0
By cross multiplication method
x/(21 – 10) = y/(-20 + 9) = 1/(3 – 14)
By further calculation
x/11 = y/-11 = 1/-11
So we get
x/11 = 1/-11
⇒ x = – 1
Similarly
y/-11 = 1/-11
⇒ y = 1
Therefore, x = –1 and y = 1.
2. Solve the following pairs of linear equations by cross-multiplication method:
(i) x – y = a + b
ax + by = a2 – b2
(ii) 2bx + ay = 2ab
bx – ay = 4ab.
Solution
(i) x – y = a + b
ax + by = a2 – b2
We can write it as
x – y – (a + b) = 0
ax + by – (a2 – b2) = 0
By cross multiplication method
x/[a2 – b2 + b (a + b)] = y/[-a(a+b) + a2 – b2] = 1/(b + a)
By further calculation
⇒ x/(a2 – b2 + ab + b2) = y/(-a2 – ab + a2 – b2) = 1/(a + b)
So we get
⇒ x/[a(a + b)] = y/[-b(a + b)] = 1/(a + b)
⇒ x = a (a + b)/(a + b) = a
⇒ y = [-b(a + b)]/(a + b) = – b
Therefore, x = a and y = – b.
(ii) 2bx + ay = 2ab
bx – ay = 4ab
We can write it as
2bx + ay – 2an = 0
bx – ay – 4ab = 0
By cross multiplication method
x/(-4a2b – 2a2b) = y/(-2ab2 + 8ab2) = 1/(-2ab – ab)
By further calculation
x/-6a2b = y/6ab2= 1/-3ab
So we get
⇒ x = -6a2b/-3ab = 2a
⇒ y = 6ab2/-3ab = –2b
Therefore, x = 2a and b = –2b.
Exercise 5.4
Solve the following pairs of linear equations (1 to 5):
1. (i) 2/x + 2/3y = 1/6
2/x – 1/y = 1
(ii) 3/2x + 2/3y = 5
5/x – 3/y = 1.
Solution
(i) 2/x + 2/3y = 1/6 …(1)
2/x – 1/y = 1 …(2)
By subtracting both the equations
5/3y = -5/6
By cross multiplication
–15y = 30
By division
y = 30/ -15 = – 2
Substitute the value of y in equation (1)
2/x + 2/(3×(-2)) = 1/6
By further calculation
2/x – 1/3 = 1/6
So we get
2/x = 1/6 + 1/3
Taking LCM
2/x = (1+2)/6 = 3/6
By cross multiplication
x = (2×6)/3 = 12/3 = 4
Therefore, x = 4 and y = –2.
(ii) 3/2x + 2/3y = 5 …(1)
5/x – 3/y = 1 …(2)
Multiply equation (1) by 1 and (2) by 2/9
3/2x + 2/3y = 5
10/9x – 2/3y = 2/9
By adding both the equations
(3/2 + 10/9)1/x = 5 + 2/9
Taking LCM
⇒ (27 + 20)/18 × 1/x = (45+2)/9
By further calculation
⇒ 47/18x = 47/9
By cross multiplication
⇒ x = (47×9)/(47×18) = ½
Substitute the value of x in equation (2)
5/½ – 3/y = 1
By further calculation
10 – 3/y = 1
⇒ 3/y = 10 – 1 = 9
So we get
y = 3/9 = 1/3
Therefore, x = ½ and y = 1/3
2. (i) (7x – 2y)/ xy = 5
(8x + 7y)/ xy = 15
(ii) 99x + 101y = 499xy
101x + 99y = 501xy.
Solution
(i) (7x – 2y)/xy = 5
(8x + 7y)/xy = 15
We can write it as
7x/xy – 2y/xy = 5
8x/xy + 7y/xy = 15
By further simplification
7/y – 2/x = 5 …(1)
8/y + 7/x = 15 …(2)
Now multiply equation (1) by 7 and (2) by 2
49/y – 14/x = 35
16/y + 14/x = 30
By adding both the equations
65/y = 65
So we get
⇒ y = 65/65 = 1
Substitute the value of y in equation (1)
7/1 – 2/x = 5
By further calculation
⇒ 2/x = 7–5 = 2
So we get
⇒ x = 2/2 = 1
Therefore, x = 1 and y = 1.
(ii) 99x + 101y = 499xy
101x + 99y = 501xy
Now divide each term by xy
99 x/xy + 101 y/xy = 499 xy/xy
101 y/xy + 99 x/xy = 501 xy/xy
By further calculation
99/y + 101/x = 499 …(1)
101/y + 99/x = 501 …(2)
By adding both the equations
200/y + 200/x = 1000
Divide by 200
1/y + 1/x = 5 ...(3)
Subtracting both the equations
-2/y + 2/x = – 2
Divide by 2
⇒ -1/y + 1/x = – 1 …(4)
By adding equation (3) and (4)
2/x = 4
So we get
x = 2/4 = ½
By subtracting equation (3) and (4)
2/y = 6
So we get
⇒ y = 2/6 = 1/3
Therefore, x = ½ and y = 1/3 if x ≠ 0, y ≠ 0.
3. (i) 3x + 14y = 5xy
21y – x = 2xy
(ii) 3x + 5y = 4xy
2y – x = xy.
Solution
(i) 3x + 14y = 5xy
21y – x = 2xy
Now dividing each equation by xy of x ≠ 0, y ≠ 0
3x/xy + 14y/xy = 5xy/xy
By further calculation
3/y = 14/x = 5 …(1)
(ii) 3x + 5y = 4xy
2y – x = xy
We can write it as
3x + 5y = 4xy
– x + 2y = xy
Divide each equation by xy if x≠ 0 and y ≠ 0
3x/xy + 5y/xy = 4xy/xy
So we get
3/y + 5/x = 4 …(1)
-x/xy + 2y/xy = xy/xy
So we get
-1/y + 2/x = 1 …(2)
Now multiply equation (1) by 1 and (2) by 3
3/y + 5/x = 4
-3/y + 6/x = 3
By adding both the equations
11/x = 7
So we get
⇒ x = 11/7
Substitute the value of x in equation (2)
-1/y + 2/11/7 = 1
By further calculation
-1/y + (2×7)/11 = 1
⇒ -1/y + 14/11 = 1
We can write it as
⇒ -1/y = 1 – 14/11
Taking LCM
⇒ -1/y = (11 – 14)/11
So we get
⇒ -1/y = -3/11
By cross multiplication
⇒ -3y = –11
⇒ y = –11/-3 = 11/3
Therefore, x = 11/7 and y = 11/3.
4. (i) 20/ (x + 1) + 4/ (y – 1) = 5
10/ (x + 1) – 4/ (y – 1) = 1
(ii) 3/ (x + y) + 2/ (x – y) = 3
2/ (x + y) + 3/ (x – y) = 11/3.
Solution
(i) 20/(x+1) + 4/(y–1) = 5 …(1)
10/(x+1) – 4/(y–1) = 1 …(2)
Add equation (1) and (2)
30/(x + 1) = 6
By cross multiplication
⇒ 30 = 6(x+1)
By further calculation
⇒ 30/6 = x + 1
⇒ 5 = x + 1
So we get
⇒ x = 5 – 1 = 4
Substitute the value of x in equation (1)
20/(x+1) + 4/(y–1) = 5
⇒ 20/(4+1) + 4/(y–1) = 5
By further calculation
20/5 + 4/(y–1) = 5
⇒ 4 + 4/(y–1) = 5
We can write it as
⇒ 4/(y–1) = 5 – 4 = 1
⇒ 4/(y–1) = 1
By cross multiplication
⇒ 4 = 1(y–1)
So we get
⇒ 4 = y – 1
⇒ y = 4 + 1 = 5
Therefore, x = 4 and y = 5.
(ii) 3/(x+y) + 2/(x–y) = 3 …(1)
2/(x+y) + 3/(x–y) = 11/3 …(2)
Multiply equation (1) by 3 and (2) by 2
9/(x+y) + 6/(x–y) = 9 …(3)
4/(x+y) + 6/(x–y) = 22/3 …(4)
Subtracting both the equations
5/(x+y) = 9 – 22/3
Taking LCM
⇒ 5/(x+y) = 5/3
By cross multiplication
⇒ 5×3 = 5(x + y)
By further calculation
⇒ (5×3)/5 = x + y
⇒ x + y = (3×1)/3
⇒ x + y = 3 ...(5)
Substitute equation (5) in (1)
3/3 + 2/(x–y) = 3
By further calculation
1 + 2/(x–y) = 3
⇒ 2/(x–y) = 3 – 1 = 2
So we get
⇒ 2/2 = x – y
Here
1 = x – y …(6)
We can write it as
x – y = 1
x + y = 3
By adding both the equations
2x = 4
⇒ x = 4/2 = 2
Substitute x = 2 in equation (5)
2 + y = 3
⇒ y = 3 – 2 = 1
Therefore, x = 2 and y = 1.
5. (i) 1/2(2x+3y) + 12/7(3x–2y) = ½
7/(2x+3y) + 4/(3x–2y) = 2
(ii) 1/2(x+2y) + 5/3(3x–2y) = –3/2
5/ 4(x 2y) – 3/ 5(3x–2y) = 61/60.
Solution
(i) 1/ 2(2x+3y) + 12/ 7(3x–2y) = ½
7/ (2x+3y) + 4/ (3x–2y) = 2
Consider 2x + 3y = a and 3x – 2y = b
We can write it as
1/2a + 12/7b = ½
7/a + 4/b = 2
Now multiply equation (1) by 7 and (2) by ½
7/2a + 12/b = 7/2
7/2a + 2/b = 1
Subtracting both the equations
10/b = 5/2
So we get
b = (10×2)/5 = 4
Substitute the value of b in equation (2)
7/a + 4/4 = 2
7/a + 1 = 2
So we get
⇒ 7/a = 2 – 1 = 1
⇒ a = 7
Here
2x + 3y = 7 …(3)
3x – 2y = 4 …(4)
Multiply equation (3) by 2 and (4) by 3
4x + 6y = 14
9x – 6y = 12
So we get
13x = 26
⇒ x = 26/13 = 2
Substitute the value of x in (3)
2×2 + 3y = 7
By further calculation
⇒ 4 + 3y = 7
So we get
3y = 7 – 4 = 3
⇒ y = 3/3 = 1
Therefore, x = 2 and y = 1.
(ii) 1/ 2(x + 2y) + 5/ 3(3x – 2y) = – 3/2
5/ 4(x + 2y) – 3/ 5 (3x – 2y) = 61/60
Consider x + 2y = a and 3x – 2y = b
1/2a + 5/3b = –3/2 …(1)
5/4a – 3/5b = 61/60 …(2)
Now, multiply equation (1) by 5/2 and (2) by (1)
5/4a + 25/6b = –15/4
5/4a – 3/5b = 61/60
Subtracting both the equations
25/6b + 3/5b = –15/4 – 61/60
Taking LCM
(125 + 18)/ 30b = (-225 – 61)/ 60
By further calculation
143/30b = –286/60
By cross multiplication
30b×(-286) = 60×143
So we get
⇒ b = (60×143)/(30×-286) = –1
Substitute the value of b in equation (1)
1/2a + 5/(3×–1) = -3/2
By further calculation
1/2a – 5/4 = –3/2
We can write it as
1/2a = –3/2 + 5/3
Taking LCM
1/2a = (-9+ 10)/ 6 = 1/6
So we get
a = 6/2 = 3
Here
x + 2y = 3 …(3)
3x – 2y = – 1 …(4)
Adding both the equations
4x = 2
⇒ x = 2/4 = ½
Substitute the value of x in equation (3)
½ + 2y = 3
By further calculation
2y = 3 – ½
Taking LCM
2y = 5/2
⇒ y = 5/(2×2) = 5/4
Therefore, x = ½ and y = 5/4.
Chapter Test
Solve the following simultaneous linear equations (1 to 4):
1.(i) 2x – (¾)y = 3,
5x – 2y = 7
Solution
8x-3y = 12 …(i)
5x-2y = 7 …(ii)
Multiply (i) by 5 and (ii) by 8, we get
40x-15y = 60 ...(iii)
40x -16y = 56 ...(iv)
Subtract (iv) from (iii), we get
y = 4
Substitute y in (i)
8x- 3×4 = 12
⇒ 8x = 12+12
⇒ 8x = 24
⇒ x = 24/8
⇒ x = 3
Hence, x = 3 and y = 4.
(ii) 2(x-4) = 9y+2
x – 6y = 2
Solution
2(x-4) = 9y+2
⇒ 2x-8 = 9y+2
⇒ 2x-9y = 2+8
⇒ 2x-9y = 10 …(i)
⇒ x-6y = 2 …(ii)
Multiply (ii) by 2, we get
2x -12y = 4 …(iii)
Subtract (iii) from (i), we get
2x-9y - (2x -12y) = 10 -4
⇒ 2x - 9y -2x +12y = 6
⇒ 0+3y = 6
⇒ 3y = 6
⇒ y = 6/3
⇒ y = 2
Substitute the value of y in (i)
2x- 9×2 = 10
⇒ 2x-18 = 10
⇒ 2x = 10+18
⇒ 2x = 28
⇒ x = 28/2
⇒ x = 14
Hence, x = 14 and y = 2.
2. (i) 97x+53y = 177
53x+97y = 573
Solution
Given equations are as follows.
97x+53y = 177 …(i)
53x+97y = 573 …(ii)
Multiply (i) by 53 and (ii) by 97
53(97x+53y) = 53×177
⇒ 5141x + 2809y = 9381 …(iii)
97(53x+97y) = 97×573
⇒ 5141x+9409y = 55581 …(iv)
Subtract (iv) from (iii)
(5141x+2809y) - (5141x+9409y) = 9381-55581
⇒ 5141x + 2809y - 5141x - 9409y = -46200
⇒ 0x -6600y = -46200
⇒ -6600y = -46200
⇒ y = -46200/-6600
⇒ y = 7
Substitute the value of y in (i)
97x + 53×7 = 177
⇒ 97x+371 = 177
⇒ 97x = 177-371
⇒ 97x = –194
⇒ x = -194/97
⇒ x = -2
Hence, x = -2 and y = 7.
(ii) x+y = 5.5
x-y = 0.9
Solution
x+y = 5.5 …(i)
x-y = 0.9 …(ii)
Adding (i) and (ii), we get
2x = 5.5+0.9
⇒ 2x = 6.4
⇒ x = 6.4/2
⇒ x = 3.2
Substitute value of x in (i)
3.2+y = 5.5
⇒ y = 5.5-3.2
⇒ y = 2.3
Hence, x = 3.2 and y = 2.3.
3. (i) x+y = 7xy
2x-3y+xy = 0
Solution
x + y = 7xy …(i)
2x - 3y + xy = 0 ...(ii)
Divide (i) by xy, we get
Divide (ii) by xy, we get
Multiplying (iii) by 3, we get
Adding (v) and (iv), we get
Substitute value of y in (iv)
Hence x = 1/3 and y = 1/4.
(ii) 30/(x-y) + 44/(x+y) = 10
40/(x-y) + 55/(x+y) = 13
Solution
Multiply (i) by 4 and (ii) by 3, we get
Subtracting (iv) from (iii), we get
Substitute (v) in (i), we get
Now solve for (v) and (vi)
x+y = 11
x-y = 5
Add (v) and (vi)
2x = 16
⇒ x = 16/2 = 8
Substitute x in (v)
8+y = 11
⇒ y = 11-8
⇒ y = 3
Hence, x = 8 and y = 3.
4. (i) ax+by = a-b
bx-ay = a+b
Solution
ax + by = a-b …(i)
bx - ay = a+b …(ii)
Multiplying (i) by a and (ii) by b, we get
a(ax+by) = a(a-b)
⇒ a2x +aby = a2-ab …(iii)
b(bx-ay) = b(a+b)
⇒ b2x -aby = ab+b2 …(iv)
Adding (iii) and (iv)
a2x +aby + b2x -aby = a2-ab + ab+b2
⇒ (a2+b2)x = (a2+b2)
⇒ x = (a2+b2)/(a2+b2)
⇒ x = 1
Substitute the value of x in (i), we get
a×1 + by = a-b
⇒ a + by = a-b
⇒ by = -b
⇒ y = -b/b
⇒ y = -1
Hence, x = 1 and y = -1.
(ii) 3x + 2y = 2xy
Solution
3x + 2y = 2xy …(i)
Divide (i) by xy
Multiply (ii) by 2, we get
Subtract (iii) from (iv)
—————
Substitute y in (iii)
(3/3) + (2/x) = 2
⇒ 1+(2/x) = 2
⇒ (2/x) = 1
⇒ x = 2
Hence, x = 2 and y = 3.
5. Solve
2x -(3/y) =9
3x + (7/y) = 2.
Hence find the value of k if x = ky + 5.
Solution
2x -(3/y) = 9 …(i)
3x + (7/y) = 2 …(ii)
Multiply (i) by 3 and (ii) by 2, we get
6x- (9/y) = 27 ...(iii)
6x+ (14/y) = 4 …(iv)
Subtracting (iv) from (iii), we get
-23/y = 23
⇒ y = 23/-23
⇒ y = -1
Substitute y in (i)
2x-(3/-1) = 9
⇒ 2x+3 = 9
⇒ 2x = 9-3
⇒ 2x = 6
⇒ x = 6/2
x = 3
Hence, x = 3 and y = -1.
Given x = ky+5
Substitute x and y in above equation.
3 = k×-1+5
⇒ 3 = -k+5
⇒ k = 5-3
⇒ k = 2
Hence the value of k is 2.
6. Solve,
Hence find the value of 2x2-y2.
Solution
Let (x+y) = a
Multiply (iii) by 5
Subtracting (ii) from (iv)
Substitute x in (iii)
(1/a) -1/(2×3) = 1/30
⇒ (1/a) – (1/6) = 1/30
⇒ 1/a = (1/30)+(1/6)
⇒ 1/a = (1+5)/30
⇒ 1/a = 6/30
⇒ a = 30/6
⇒ a = 5
Substitute a in x+y = a
3+y = 5
⇒ y = 5-3
⇒ y = 2
Hence, x = 3, y = 2.
2x2-y2 = 2×32 - 22
= 2×9 - 4
= 18-4
= 14
Hence, the value of 2x2-y2 is 14.
7. Can x, y be found to satisfy the following equations simultaneously ?
If so, find them.
Solution
Multiply (i) by 5 and (ii) by 2, we get
Subtract (v) from (iv)
31/x = 95-2
⇒ 31/x = 93
⇒ x = 31/93
⇒ x = 1/3
Substitute x in (i)
(2/y)+ 5÷(1/3) = 19
⇒ (2/y)+ 5×3 = 19
⇒ (2/y) = 19-15
⇒ (2/y) = 4
⇒ y = 2/4
⇒ y = 1/2
Substitute x and y in (iii)
3×(1/3) + 8×(1/2) = 5
⇒ 1+4 = 5
The value of x and y satisfies (iii).
Hence the given equations are simultaneous.