ML Aggarwal Solutions Class 9 Math 5th Chapter Simultaneous Linear Equations Exercise 5.2
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fifth Chapter Simultaneous Linear Equations Exercise 5.2. APC Solution Class 9 Exercise 5.2.
(1) (i) 3x + 4y = 10
2x – 2y = 2
Solution:
3x + 4y = 10 —- (i)
2x – 2y = 2 —- (ii)
Multiplying equation (ii) by 2 then adding equation (ii) with (i) we get,
3x + 4y = 10
4x – 4y = 4
_________________
7x = 14
Or, x = 14/7
Or, x = 2
Substituting the value of x in equation (i) we get,
3 × 2 + 4y = 10
Or, 4y = 10-6
Or, y = 4/4
Or, 4 = 1
(ii) 2x = 5y + 4
3x – 2y + 16 = 0
Solution:
2x = 5y + 4
Or, 2x – 5y = y —- (i)
3x – 2y + 16 = 0
Or, 3x – 2y = -16
Multiplying equation (i) with 3 & equation (ii) with 2 then substituting equation (ii) from equation (i) we get,
6x – 15y = 12
6x – 4y = -32
(-) (+) (+)
______________
– 11y = 44
Or, y = -4
Substituting the value of y in equation (i) we get,
2x – 5 × -4 = 4
Or, 2x = 4 – 20
Or, x = 16/2
Or, x = -8
(3) (i) 15x – 14y = 117
14x – 15y = 115
Solution:
15x – 14y = 117 —- (i)
14x – 15y = 115
Multiplying equation (i) with 14 & equation (ii) with 15 then substituting equation (ii) from equation (i) we get,
210x – 196y = 1638
210x – 225y = 1725
(-) (+) (+)
___________________
29y = -87
Or, y = – 87/29
Or, y = -3
Substituting the value of y in equation (i) we get,
15x – 14x – 3 = 117
Or, 15x = 117 – 42
Or, x = 75/15
Or, x = 5
(ii) 41x + 53y = 135
53x + 41y = 147
Solution:
41x + 53y = 135 —- (i)
53x + 41y = 147 — (ii)
Adding equation (i) & (ii) we get,
41x + 53y = 135
53x + 41y 147
_______________
94x + 44y = 282
Or, 44 (x + y) = 282
Or, x + y = 282/94
Or, x + y = 3 —- (iii)
Adding equation (iii) & (ii) we get,
x + y = 3
x + y = 1
________
2x = 4
Or, x = 2
Subtracting equation (i) from equation (ii) we get,
53x + 41y = 147
41x + 53y = 135
(-) (-) (-)
________________
12x – 12y = 12
Or, 12 (x – y) = 12
Or, x – y = 1 —– (iv)
Substituting the value of x in equation (iii) we get = 2 + y = 3
Or, y = 1
(4) (i) x/6 = y – 6
3x/4 = 1+ y
Solution:
x/6 = y – 6
Or, x = 6y – 36
Or, x – 6y = -36 — (i)
3x/4 = 1+y
Or, 3x = 4 + 4y
Or, 3x – 4y = 4 — (ii)
Multiplying equation (i) with 3 then subtracting equation (ii) from equation (i) we get,
3x – 18y = -108
3x – 4y = 4
(-) (+) (-)
___________
– 14y = – 112
Or, y = 112/14
Or, y = 8
Substituting the value of in equation (ii) we get,
3x – 4 × 8 = 4
Or, x = 4+32/3
Or, x = 36/3
Or, x = 12
(ii) x – 2/3 y = 8/3
2x/5 – y = 7/5
Solution:
x – 2/3 y = 8/3
Or, 3x – 2y/3 = 8/3
Or, 3x – 2y = 8 —- (i)
2x/5 – y = 7/5
Or, 2x-5y/5 = 7/5
Or, 2x – 5y = 7 — (ii)
Multiplying equation (i) with 2 & (ii) with 3 then subtracting equation (ii from (i) we get,
6x – 4y = 16
6x – 13y = 21
(-) (+) (-)
____________
11y = -5
Or, y = -5/11
Substituting the value of y in equation (i) we get,
3x – 2 × -5/11 = 8
Or, 33x + 10 = 8 × 11
Or, x = 88-10/33
Or, x = 78/33 = 26/11
(5) (i) 9 – (x – 4) = y + 7
2 (x + y) = 4 – 3y
Solution:
Substituting the value of y in equation (i) we get,
x + (- 8/3) = 6
Or, x = 6 +8/3
Or, x = 18+8/3
Or, x = 26/3
(ii) 2x + x-y/6 = 2
x – 2x+y/3 = 1
Solution:
2x + x-y/6 = 2
Or, 12x + x – y = 12
Or, 13x – y = 12 — (i)
x – 2x+y/3 = 1
Or, 3x- 2 – y = 3
Or, x – y = 3 —- (ii)
Subtracting equation (ii) from equation (i) we get,
13x – y =12
x – y = 3
(-) (+) (-)
____________
12x = 9
Or, x = 9/12
Or, x = 3/4
Substituting the value of x in equation (ii) we get,
3/4 – y = 3
Or, y = 3/4 – 3
Or, y = 3-12/4
Or, y = – 9/4
(6) (i) x – 3y = 3x – 1 = 2x – y
Solution:
x – 3y = 3x – 1 = 2x – y
x – 3y = 3x – 1
Or, x – 3x – 3y = -1
Or, -2x – 3y = -1
Or, 2x + 3y = 1 —- (i)
3x – 1 = 2x – y
Or, 3x – 2x + y = 1
Or, x + y = 1 — (ii)
Multiplying equation (ii) with 2 then subtracting equation (ii) from equation (i) we get.
2x + 3y = 1
2x + 2y = 2
(-) (-) (-)
_____________
y = -1
Substituting the value of y in equation (ii) we get,
x – 1 = 1
Or, x = 2
(7) (i) 4x + x-y/8 = 17
2y + x – 5y+2/3 = 2
Solution:
4x + x-y/8 = 17
Or, 32x + x – y = 17×8
Or, 33x – y = 136 —- (i)
2y + x – 5y+2/3 = 2
Or, 6y + 3x – 5y – 2 = 2 × 3
Or, 3x + y = 6+2
Or, 3x + y = 8 —- (ii)
Adding equation & (ii) we get,
33x – y = 136
3x + 4 = 8
__________
35x = 144
Or, x = 144/35
Or, x = 4
Substituting the value of x in equation (ii) we get,
3 × 4 + 4 = 8
Or, y = 8 – 12
Or, y = -4
(ii) x+1/2 + y-1/3 = 8
x-1/3 + y+1/2 = 9
Solution:
x+1/2 + y-1/3 = 8
Or, 3x+3+2y-2/6 = 8
Or, 3x + 2y = 48 – 1
Or, 3x + 2y = 47 —- (i)
x-1/3 + y+1/2 = 9
Or, 2x-2+3y+3/6 = 9
Or, 2x + 3y = 54 – 1
Or, 2x + 3y = 53 —– (ii)
Adding equation (i) & (ii) we get,
3x + 2y = 47
2x + 3y = 53
___________
5x + 5y = 100
Or, 5 = (x + y) = 100
Or, x + y = 20 —- (iii)
Subtracting equation (ii) from equation (ii) we get,
3x + 2y = 47
2x + 3y = 53
(-) (-) (-)
_____________
x – y = -6 —- (iv)
Substituting the value of x in equation (iii) we get,
7 + y = 20
Or, y = 13
Adding equation (iii) & (iv) we get,
x + y = 20
x – y = -6
__________
2x = 14
Or, x = 7
(8) (i) 3/x + 4y = 7
5/x + 6y = 13
Solution:
3/x + 4y = 7 — (i)
5/x + 6y = 13 —– (ii)
Multiplying equation (i) with 3 & equation (ii) with 2 then subtracting equation (ii) from equation (i) we get,
9/x + 12y = 21
(9) (i) px + qy = p – q
qx – py = p + q
Solution:
px + qy = p – q —- (i)
qx – py = p + q —- (ii)
Multiplying equation p & equation (ii) with q then then we get,
p2x + pqy = p (p – q)
pq2x – pqy = q (p + q)
____________________
p2x + q2x = p2 – pq + pq + q2
Or, x (p2 + q2) = p2 + q2
Or, x = 1
Substituting the value of x in equation (i) we get,
p × 1 + qy = p – q
Or, qy = p – q – p
Or, y = -1
(ii) x/a – y/b = 0
ax + by = a2 + b2
Solution:
x/a – y/b = 0, ax + by = a2 + b2 —- (ii)
Or, bx – ay = 0 — (ii)
Multiplying equation (i) with a & equation (ii) with b then adding then we get,
a2x + ab/y = a (a2 + b2)
b2x – aby = 0
___________________
a2x + b2x = a (a2 + b2)
Or, x (a2 + b2) = a (a2 + b2)
Or, x = a
Substituting the value of x in equation [i] we get,
b × a – ay = 0
Or, -ay = – ab
Or, y = b
(10) Solve 2x + y = 23, 4x – y = 19, Hence, find the values of x – 3y and 5y – 2x.
Solution:
2x + y = 23 —- (i), 4x – y = 19 —- (ii)
Adding equation (i) & (ii) we get,
2x + y = 23
4x – y = 19
____________
6x = 42
Or, x = 7
Now, x – 3y = 7 – 3 × 9
= 7 – 27
= – 20
5y – 2x = 5 × 9 – 2 × 7
= 45 – 14
= 31
Substituting the value of x in equation (i) we get,
2 × 7 + y = 23
Or, y = 23 – 14
Or, y = 9
(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:
ax + by = 7 when x = 2, y = 1
∴ a×2 + b × 1 = 7
Or, 2a + b = 7 —- (i)
Also, ax + by = 1 when x = -1
y = 1
∴ a× – 1 + b ×y = 1
Or, – a + b = 1 —- (ii)
Subtracting equation (ii) from equation (i) we get,
2a + b = 7
– a + b = 1
____________
3a = 6
Or, a = 2
Substituting the value of a in equation (ii) we get,
– 2 + b = 1
Or, 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 —- (i)
11x + 5y = 87 — (ii)
5x + 4y = 43 —- (iii)
Multiplying equation with 5 & equation (ii) with 7 then adding them we get,
15x – 35y = 35
77x + 35y = 609
_______________
92x = 644
Or, x = 644/92
Or, x = 7
Substituting the value of x in equation (i) we get,
3 × 7 – 7y = 7
Or, -7y = 7 – 21
Or, -7y = -14
Or, y = 2
Now, putting the value of x & y in equation 3 on the L.H.S we get,
5 × 7 + 4 × 2
= 35 + 8
= 43
= R.H.S
∴ The following equation hold simultaneously when,
x = 7, y = 2