ML Aggarwal Solutions Class 9 Math 5th Chapter Simultaneous Linear Equations Exercise 5.1
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fifth Chapter Simultaneous Linear Equations Exercise 5.1. APC Solution Class 9 Exercise 5.1.
(1) (i) x + y = 14
x – y = 4
Solution:
x + y = 14 —– (i),
x – y = 4
Or, x = 4 + y
Substituting the value of x in equation (i) we get,
x + y = 14
Or, 4 + y = 14
Or, 2y = 14 – 4
Or, y = 10/2
Or, y = 5
Substituting the value of y in equation (i) we get,
x + y = 14
Or, x + 5 = 14
Or, x = 9
(ii) s – t = 3
s/3 + t/2 = 6
Solution:
s – t = 3 —– (i), s/3 + t/2 = 6 —- (ii)
Or, s = 3 + t
Substituting the value of s in equation (ii) we get,
s/3 + t/2 = 6
Or, 3+t/3 + t/2 = 6
Or, 6+2t+3t/6 = 6
Or, 5t = 36 – 6
Or, t = 30/5 = 6
Substituting the value of t in equation (i) we get,
s – t = 3
Or, s – 6 = 3
Or, s = 9
(2) (i) a + 3b = 5
7a – 8b = 6
Solution:
(i) a + 3b = 5 —- (i),
7a – 8b = 6 —- (ii)
Or, a = 5 – 3b
Substituting the value of a in equation (ii) we get,
7a – 8b = 6
Or, 7 (5 – 3b) – 8b = 6
Or, 35 – 21b – 8b = 6
Or, – 2ab = – 29
Or, b = 1
Substituting the value of b in equation (i) we get,
a + 3 × 1 = 5
Or, a = 2
(ii) 5x + 4y – 4 = 0
x – 20 = 12y
Solution:
5x + 4y – 4 = 0
Or, 5x + 4y = 4 —- (i)
x – 20 = 12y
Or, x – 12y = 20 —- (ii)
Or, 5x = 4 – 4y
Or, x = 4-4y/5
Substituting the value of x in equation (ii) we get,
4-4y/5 – 12y = 20
Or, 4-44-60y/5 = 20
Or, – 64y = 100 – 4
Or, y = – 96/64
Or, y = – 3/2
Substituting the value of y in equation (i) we get,
5x + 4 × -3/2 = 4
Or, x = 4+6/5
Or, x = 2
(3) (i) 2x – 3y/4 = 3
5x – 2y – 7 = 0
Solution:
2x – 3y/4 = 3
= 8x – 3y = 3 × 4
Or, 8x – 3y = 12 — (i)
Or, 8x = 12+3y/8
5x – 2y – 7 = 0
Or, 5x – 2y = 7 —- (ii)
Substituting the value of x in equation (ii) we get,
5 × 12+3y/8 – 2y = 7
Or, 60+15y-16y/8 = 7
Or, -y = 56 – 60
Or, -y = -4
Or, y = 4
Substituting the value of y in equation (ii) we get,
5x – 2 × 4 = 7
Or, 5x = 7+8
Or, x = 15/5
Or, x = 3
(ii) 2x + 3y = 23
5x – 20 = 8y
Solution:
2x + 3y = 23 — (i),
Or, x = 23-3y/2
5x – 20 = 8y
Or, 5x – 8y = 20 —- (ii)
Substituting the value of x in equation (ii) we get,
5 × 23-3y/2 – 8y = 20
Or, 115-15y-16y/2 = 20
Or, -31y = 40-115
Or, y = 75/31
(4) (i) mx – my = m2 + n2
x + y = 2m
(ii) 2x/a + y/b = 2
x/a – y/b = 4
Solution:
mx – ny = m2 + n2 — (i)
x + y = 2m —- (ii)
Or, x = 2m – y
Substituting the value of x in equation (i) we get,
m (2m – y) – ny = m2 + n2
Or 2m2 – my – ny = m2 + n2
Or, -my – ny = m2 + n2 – 2m2
Or –y (m + n) = n2 – m2
Or, -y = (n + m) (n-n)/m+n
Or, y = m – n
Substituting the value of y in equation (ii) we get,
x + m – n = 2m
Or, x = 2m – m + n
Or, x = m + n
(ii) 2x/a + y/b = 2 —- (i) x/a – y/b = 4 —- (ii)
Or x/a = 4b+y/b
Substituting the value of x/a in equation (i) we get,
2 4b+y/b + y/b = 2
Or, 8b+2y+y/b = 2
Or, 8b + 3y = 2b
Or, 3y = 2b – 8b
Or, y = -6b/3
Or, y = -2b
Substituting the value of y in equation (ii) we get,
x/a – -2b/b = y
Or, x/a = 4 – 2
Or, x = 2a
(5) Solve 2x + y = 35, 3x + 4y = 65, Hence, find the value of x/y.
Solution:
2x + y = 35 —- (i)
3x + 4y = 65 —- (ii)
Or, y = 35 – 2x
Substituting the value of y in equation (ii) we get,
3x + 4 (35 – 2x) = 65
Or, 3x + 140 – 8x = 65
Or, -5x = 65 – 140
Or, x = 75/5
Or, x = 15
Substituting the value of x in equation (i) we get,
2 × 15 + y = 35
Or, y = 35 – 30
Or, y = 5
∴ Value of x/y = 15/5 = 3
(6) Solve the simultaneous equations 3x – y = 5, 4x – 3y = -1. Hence, find p, if y = px -3
Solution:
3x – y = 5 —– (i), 4x – 3y = -1 —- (ii)
Or, 3x = 5 + y
Or, x = 5+y/3
Substituting the value of x in equation (ii) we get,
4 × 5+y/3 – 3y = -1
Or, 20+4y+9y/3 = -1
Or, -5y = -3-20
Or, y = 23/5
Substituting the value of y in equation (i) we get,
3x – 23/5 = 5
Or, 15x – 23 = 5 × 5
Or, x = 25+23/15
Or, x = 48/15
Or, x = 16/5
Given, y = px – 3
Or, 23/5 = p × 16/5 – 3
Or, 23/5 = 16p-15/5
Or, p = 23+15/16
Or, P = 38/16
Or, p = 19/8