Solve the linear equations:
(i) a + 3b = 5
7a – 8b = 6
(ii) 5x + 4y – 4 = 0
x – 20 = 12y
Solution:
(i) a + 3b = 5 …. (1)
7a – 8b = 6 ….. (2)
Now multiply equation (1) by 7
7a + 21b = 35 …. (3)
7a – 8b = 6 ….. (4)
By subtracting both the equations
29b = 29
So we get
b = 29/29 = 1
Now substituting b = 1 in equation (1)
a + 3 (1) = 5
By further calculation
a + 3 = 5
So we get
a = 5 – 3 = 2
Therefore, a = 2 and b = 1.
(ii) 5x + 4y – 4 = 0
x – 20 = 12y
We can write it as
5x + 4y = 4 …. (1)
x – 12y = 20 ….. (2)
Now multiply equation (2) by 5
5x + 4y = 4 …. (3)
5x – 60y = 100
By subtracting both the equations
64y = – 96
So we get
y = -96/64 = – 3/2
Now substitute the value of y in equation (1)
5x + 4 (-3/2) = 4
By further calculation
5x + 2 (-3) = 4
So we get
5x – 6 = 4
5x = 4 + 6 = 10
By division
x = 10/5 = 2
Therefore, x = 2 and y = – 3/2.
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