SOLVING SYSTEMS OF EQUATIONS BY SUBSTITUTION EXAMPLES

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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