Solve simultaneous linear equations: (i) ¾ x – 2/3 y = 1

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Solve simultaneous linear equations:

(i) ¾ x – 2/3 y = 1
3/8 x – 1/6 y = 1
(ii) 2x – 3y – 3 = 0
2x/3 + 4y + ½ = 0.

Solution:

(i) ¾ x – 2/3 y = 1
3/8 x – 1/6 y = 1
We can write it as
¾ x – 2/3 y = 1
(9x – 8y)/ 12 = 1
By cross multiplication
9x – 8y = 12 ….. (1)
3/8 x – 1/6 y = 1
(9x – 4y)/ 24 = 1
By cross multiplication
9x – 4y = 24 ….. (2)
Subtracting equations (1) and (2)
– 4y = – 12
By division
y = – 12/ – 4 = 3
Substitute the value of y in (1)
9x – 8 × 3 = 12
By further calculation
9x – 24 = 12
9x = 12 + 24 = 36
By division
x = 36/ 9 = 4
Therefore, x = 4 and y = 3.
(ii) 2x – 3y – 3 = 0
2x/3 + 4y + ½ = 0
We can write it as
2x – 3y – 3 = 0
2x – 3y = 3 ….. (1)
2x/3 + 4y + ½ = 0
2x/3 + 4y = – ½
Taking LCM
(2x + 12y)/ 3 = – ½
By cross multiplication
2 (2x + 12y) = – 1 × 3
So we get
4x + 24y = – 3 …. (2)
Multiply equation (1) by 2
4x – 6y = 6
4x + 24y = – 3
By subtracting both the equations
– 30y = 9
So we get
y = -9/30 = – 3/10
Substitute the value of y in equation (1)
2x – 3 (-3/10) = 3
By further calculation
2x + 9/10 = 3
We can write it as
2x = 3 – 9/10
By taking LCM
2x = (30 – 9)/ 10
So we get
2x = 21/10
x = 21/20
Therefore, x = 21/20 and y = – 3/10.

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