Question 1 :
Solve, using the method of substitution
(i) 2x − 3y = 7; 5x + y = 9
Solution :
2x − 3y = 7 --(1)
5x + y = 9 ---(2)
From (2),
y = 9 - 5x
By applying the value of y in the 1st equation, we get
2x - 3(9 - 5x) = 7
2x - 27 + 15x = 7
17x = 7 + 27
17x = 34
x = 34/17 = 2
When x = 2, then y = 9 - 5(2)
y = 9 - 10
y = -1
Hence the solution is (2, -1).
(ii) 1.5x + 0.1y = 6.2, 3x - 0.4y = 11.2
Solution :
1.5x + 0.1y = 6.2 ----(1)
3x - 0.4y = 11.2 -----(2)
By multiplying the 1st and 2nd equation by 10, we get
15x + 1y = 62
30x - 4y = 112
From (1) ==>
y = 62 - 15x
By applying the value of y in (2), we get
30x - 4(62 - 15x) = 112
30x - 248 + 60x = 112
90x = 112 + 248
90x = 360
x = 360/90
x = 4
When x = 40, then y = 62 - 15(4)
y = 62 - 60
y = 2
Hence the solution is (4, 2).
Question 2 :
10% of x + 20% of y = 24; 3x − y = 20
Solution :
10% of x + 20% of y = 24
3x − y = 20
0.1x + 0.2y = 24 ----(1)
3x - y = 20 ------(2)
From (2),
y = 3x - 20
By applying the value of y in (1), we get
0.1 x + 0.2(3x - 20) = 24
0.1x + 0.6x - 4 = 24
0.7x = 28
x = 28/0.7
x = 280/7
x = 40
When x = 40, y = 3(40) - 20
y = 120 - 20
y = 100
Hence the solution is (40, 100).
(iv) √2 x − √3 y = 1; √3x − √8 y = 0
Solution :
√2 x − √3 y = 1 ----(1)
√3x − √8 y = 0 ----(2)
√3 y = √2 x - 1
y = (√2 x - 1)/√3
By applying the value of y in (2), we get
√3x − √8((√2 x - 1))/√3 = 0
√3x − (√16 x - √8)/√3 = 0
3x - 4x + 2√2 = 0
-x = -2√2
x = 2√2 = √8
When x = √8, y = (√2(√8) - 1))/√3
y = (√16) - 1)/√3
y = (4 - 1)/√3
y = (3/√3)⋅(√3/√3)
y = 3√3/3
y = √3
Hence the solution is (√8, √3).
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