Selina Concise Class 9 Maths Chapter 6 Simultaneous Equations Exercise 6A Solutions
EXERCISE – 6A
(Q1) Solve the points of liner (Simultaneous) equations by the method of elimination by substitution.
(1) 8x + 5y = 9
3x + 2y = 4
Solution:
Given equations are –
8x + 5y = 9 —- (i)
3x + 2y = 4 —- (ii)
From equation (ii) =>
3x + 2y = 4
3x = 4 – 2y
x = 4-2y/3
Substitute x = 4-2y/3 in equation (i), we get,
8x + 5y = 9
8 × (4-2y/3) + 5y = 9
32-16y/3 + 5y = 9
32-16y+15y/3 = 9
32 – 16y + 15y = 9 × 3
32 – y = 27
-y = 27 – 32
-y = -5
y = 5
Substitute y = 5 in equation (ii), we get,
3x + 2y = 4
3×x + 2×5 = 4
3x + 10 = 4
3x = 4 – 10
3x = -6
x = -6/3
x = -2
(2) 2x – 3y = 7
5x + y = 9
Solution:
Given equations are,
2x – 3y = 7 —- (i)
5x + y = 9 —– (ii)
From equation (ii) =>
5x + y = 9
5x = 9 – y
x = 9-y/5
substitute x = 9-y/5 in equation (i), we get,
2 × (9-y/5) – 3y = 7
18 – 2y/5 – 3y = 7
18 – 2y – 15y/5 = 7
18 – 2y – 15y = 7×5
18 – 17y = 35
-17y = 35 – 18
-17y = 17
y = 17/-17
y = -1
Substitute y = -1 in equation (i), we get,
2x – 3×(-1) = 7
2x + 3 = 7
2x = 7 – 3
2x = 4
x = 4/2
x = 2
(3) 2x + 3y = 8
2x = 2 + 3y
Solution:
Given equations are –
2x + 3y = 8 —- (i)
2x = 2 + 3y — (ii)
From equation (ii) =>
2x = 2 + 3y
x = 2+3y/2
Substitute x = 2+3y/2 in equation (i),
We get, 2x + 3y = 8
=> 2 × (2+3y/2) + 3y = 8
=> 4+6y/2 + 3y = 8
=> 4 + 6y + 6y/2 = 8
=> 4 + 6y + 6y = 8 × 2
=> 4 + 12y = 16
12y = 16 – 4
12y = 12
y = 12/12
y = 1
Substitute y = 1 in equation (i),
2x + 3 × 1 = 8
2x + 3 = 8
2x = 8 – 3
2x = 5
x = 5/2
x = 2.5
∴ The value of x = 2.5 and y = 1.
(4) 0.2x + 0.1y = 25
2 (x – 2) – 1.6y = 116
Solution:
Given equations are –
0.2x + 0.1y = 25 — (i)
2 (x – 2) – 1.6y = 116 —- (ii)
From equation (i) =>
0.2x + 0.1y = 25
0.2/10 x + 0.1/10 y = 25
=> 2x + y/10 = 25
=> 2x + y = 25 × 10
=> 2x + y = 250
=> 2x = 250 – y
=> x = 250-y/2 —- (iii)
Substitute, x = 250-y/2 in equation (ii), we get
2 (x – 2) – 1.6y = 116
2 (250-y/2 – 2) – 1.6y = 116
=> 2 (250 – y – 4/2) – 1.6y = 116
=> 500-2y-8/2 – 1.6y = 116
=> 500-2y-8-3.2y/2 = 116
=> 500 – 8 – 2y – 3.2y = 116×2
=> 492 – 5.2y = 232
-5.2y = 232 – 492
=> -5.2y = -260
=> y = -260/-5.2
=> y = 50
Substitute, y = 50 in equation (iii), we get
x = 250-y/2
x = 250-50/2
x = 200/2
x = 100
∴ The value of x = 100 and y = 50
(5) 6x = 7y + 7
7y – x = 8
Solution:
Given equations are,
6x = 7y + 7 —- (i)
7y – x = 8 —– (ii)
From equation (ii) =>
7y – x = 8
7y = 8 + x
y = 8+x/7
Substitute y = 8+x/7 in equation (i), we get
6x = 7 × 8+x/7 + 7
= 8 + x + 7
6x = 15 + x
6x – x = 15
5x = 15
x = 15/5
x = 3
Substitute x = 3 in equation (i),
We get, 6x = 7y + 7
6 × 3 = 7y + 7
18 = 7y + 7
18 – 7 = 7y
11 = 7y
11/7 = y
∴ y = 11/7
∴ The value of x = 3 and y = 11/7
(6) y = 4x – 7
16x – 5y = 25
Solution:
Given equations are –
y = 4x – 7 —– (i)
16x – 5y = 25 —- (ii)
From equations (i) => y = 4x – 7
y + 7 = 4x
y + 7/4 = x
∴ x = y + 7/4
Substitute x = y + 7/4 in equation (ii),
We get,
16x – 5y = 25
16 × (y+7/4) – 5y = 25
16y+112/4 – 5y = 25
16y + 112 – 20y/4 = 25
– 4y + 112 = 25 × 4
– 4y + 112 = 100
-4y = 100 – 112
-4y = -12
y = -12/-4
y = 3
Substitute y = 3 in equation (i), we get,
y = 4x – 7
3 = 4x – 7
3 + 7 = 4x
10 = 4x
10/4 = x
5/2 = x
∴ x = 5/2
x = 2.5
∴ The value of x = 2.5 and y = 3.
(7) 2x + 7y = 39
3x + 5y = 31
Solution:
Given, equations are,
2x + 7y = 39 —– (i)
3x + 5y = 31 —– (ii)
From equation (i) =>
2x + 7y = 39
2x = 39 – 7y
x = 39 – 7y/2
Substitute x = 39 – 7y/2 in equation (ii), we get,
3x + 5y = 31
3 × (39 – 7y/2) + 5 = 31
117-21y/2 + 5y = 31
117 – 21y + 10y = 31
117 – 11y = 31×2
-11y = 62 – 117
-11y = -55
y = -55/-11
y = 55/11
y = 5
Substitute y = 5 in equation (i),
2x + 7 × 5 = 39
2x + 35 = 39
2x = 39 – 35
2x = 4
x = 4/2
x = 2
∴ The value of x = 2 and y = 5.
(8) 1.5x + 0.1y = 6.2
3x – 0.4y = 11.2
Solution:
1.5x + 0.1y = 6.2 —– (i)
3x – 0.4y = 11.2 —- (ii)
From equation (ii)=>
3x – 0.4y = 11.2
3x – 0.4/10 y = 11.2/10
3x – 4y/10 = 112/10
30x – 4y/10 = 112/10
30x – 4y = 112
30x = 112 + 4y
x = 112+4y/30
Substitute x = 112+4y/30 in equation (i), we get,
1.5x + 0.1y = 6.2
1.5× (112+4y/30) + 0.1y = 6.2
168 + 6y/30 + 0.1y = 6.2
168 + 6y + 3y/30 = 6.2
168 + 6y + 3y = 6.2 × 30
168 + 9y = 186
9y = 186 – 168
9y = 18
y = 18/9
y = 2
Substitute y = 2 in equation (i), we get
1.5x + 0.1 × 2 = 6.2
1.5x + 0.2 = 6.2
1.5x = 6.2 – 0.2
1.5x = 6
x = 6/1.5
x = 4
∴ The value of x = 4 and y = 2
(9) 2 (x – 3) + 3 (y – 5) = 0
5 (x – 1) + 4 (y – 4) = 0
Solution:
2 (x – 3) + 3 (y – 5) = 0 —– (i)
5 (x – 1) + 4 (y – 4) = 0 —– (ii)
From equation (i) =>
2 (x – 3) + 3 (y – 5) = 0
2x – 6 + 3y – 15 = 0
2x + 3y – 21 = 0
2x = 21 – 3y/2 —- (iii)
X = 21 – 3y/2
Substitute x = 21-3y/2 in equation (ii),
we get,
5 (x – 1) + 4 (y – 4) = 0
5x – 5 + 4y – 16 = 0
5x + 4y – 21 = 0
5 × (21-3y/2) + 4y – 21 = 0
105 – 15y/2 + 4y – 21 = 0
105 – 15y + 8y/2 = 21
105 – 7y = 21 × 2
105 – 7y = 42
-7y = 42 – 105
-7y = -63
y = 63/7 = 9
Substitute = 63/7 = 9
Substitute y = 9 in equation (iii) we get,
x = 21 – 3 × (9)/2
= 21 – 27/2
= -6/2
x = -3
∴ The value of x = -3 and y = 9
(10) (2x+1/7) + (5y-3/3) = 12
(3x+2/2) – (4y+3/9) = 13
Solution:
(2x+1)/7 + (5y-3)/3 = 12 —– (i)
(3x+2)/2 – (4y+3)/9 = 13 —– (ii)
From equation (i) =>
(2x+1)/7 + (5y-3)/3 = 12
=> 3 (2x + 1) + 7 (5y – 3)/21 = 12
=> 6x + 3 + 35y – 21/21 = 12
=> 6x + 35y – 18 = 12×12
=> 6x + 35y – 18 = 252
6x + 35y = 252 + 18
6x + 35y = 270
6x = 270 – 35y
x = 270 – 35y/6 —– (iii)
Substitute, x = 270-35y/6 in equation (ii) we get,
(3x+2)/2 – (4y+3)/9 = 13
9(3x+12)-2(4y+3)/18 = 13
27x+18-8y-6/18 = 13
27x – 8y + 12 = 13×18
27x – 8y + 12 = 234
27x – 8y = 234 – 12
27x – 8y = 222
27× (270-35y/6) – 8y = 222
7290-945y/6 – 8y = 222
7290 – 945y – 48y = 222×6
7290 – 993y = 1332
– 993y = 1332 – 7290
-993y = -5958
y = -5958/993
y = 6
Put y = 6 in equation (iii), we get
x = 270-35(6)/6
= 270-210/6
= 60/6
x = 10
The value of x = 10 and y = 6
(11) 3x + 2y = 11
2x – 3y + 10 = 0
Solution:
3x + 2y = 11 —- (i)
2x – 3y = -10 —— (ii)
From equation (i) =>
3x + 2y = 11
3x = 11 – 2y
x = 11 – 2y/3 —– (iii)
Substitute x = 11 – 2y/3 in equation (ii), we get,
2 × (11-2y/3) – 3y = -10
22 – 4y/3 – 3y = -10
22 – 4y-9y/3 = -10
22-4y-9y = -10×3
22 – 4y – 9y = -30
22 – 13y = -30
-13y = -30 -22
-13y = -52
y = -52/-13
y = 4
Put y = 4 in equation (iii)
x = 11-2×4/3
= 11-8/3
= 3/3
x = 1
The value of x = 1 and y = 4
(12) 2x – 3y + 6 = 0
2x + 3y – 18 = 0
Solution:
2x – 3y + 6 = 0 —– (i)
2x + 3y – 18 = 0 —– (ii)
From equation (i) =>
2x – 3y = -6
2x = -6 + 3y
x = -6+3y/2 —– (iii)
Substitute x = -6+3y/2 in equation (ii),
2 × (-6+3y/2) + 3y – 18 = 0
-12+6y/2 + 3y = 18
-12+6y+6y/2 = 18
-12 + 12y = 18×2
-12 + 12y = 36
12y = 36 + 12
12y = 48
y = 48/12
y = 4
Put y = 4 in equation (iii), we get,
x = -6+3×4/2
x = -6+12/2
x = 6/2
x = 3
∴ The value of x = 3 and y = 4
(13) (3x/2) – (5y/3) + 2 = 0
x/3 + y/2 = 2 1/6
Solution:
(3x/2) – (5y/3) + 2 = 0 —– (i)
x/3 + y/2 = 12+1/6
x/3 + y/2 = 13/6 —– (ii)
From equation (i) =>
3x/2 – 5y/3 = -2
3×3x-2×5y/6 = -2
9x – 10y = -2×6
9x – 10y = -12
9x = -12 + 10y
x = -12 + 10y/9 —– (iii)
Substitute x = -12+10y/9 in equation (ii), we get,
x/3 + y/2 = 13/6
-12+10y/9×3 + y/2 = 13/6
-12+10y/27 + y/2 = 13/6
2 (-12+10y) + 27y/54 = 13/6
-24+20y+27y/54 = 13/6
6 (-24+47y)/54) = 13
-144 + 282y = 13×54
-144 + 282y = 702
282y = 702 + 144
282y = 846
y = 846/282
y = 3
Put y = 3 in equation (iii), we get,
x = -12+10y/9
= -12+10×3/9
= -12+30/9
x = 18/9
x = 2
∴ The value of x = 2 and y = 3
(14) x/6 + y/15 = 4
x/3 – y/12 = 4 3/4
Solution:
x/6 + y/15 = 4 —– (i)
x/3 – y/12 = 16+3/4
x/3 – y/12 = 19/4 —- (ii)
From equation (i),
x/6 + y/15 = 4
15x + 6y/90 = 4
15x + 6y = 4 × 90
15x + 6y = 360
15x = 360 – 6y
x = 360 – 6y/15 —– (iii)
Substitute x = 360-6y/15 in equation (ii), we get,
x/3 – y/12 = 19/4
360-6y/15×3 – y/12 = 19/4
12(360-6y)-45y/540 = 19/4
4320-72y-45y/540 = 19/4
4320-117y/540 = 19/4
4 (4320 – 117y) = 19×540
17,280 – 468y = 10,260
-468y = 10,260 – 17,280
-468y = -7020
y = -7020/468
y = 15
Put y = 15 in equation (iii),
x = 360 – 6×15/15
= 360-90/15
= 270/15
x = 18
∴ The value of x = 18 and y = 15
Here is your solution of Selina Concise Class 9 Maths Chapter 6 Simultaneous Equations Exercise 6A
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