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Question
Solve the following pair of linear equations
ax + by = c
bx + ay = 1 + c
Solution
ax + by = c … (1)
bx + ay = 1 + c … (2)
Multiplying equation (1) by a and equation (2) by b, we obtain
a2x + aby = ac … (3)
b2x + aby = b + bc … (4)
Subtracting equation (4) from equation (3),
(a2 − b2) x = ac − bc − b
`x = (c(a-b)-b)/(a^2 - b^2)`
From equation (1), we obtain
ax + by = c
`a{((c(a-b)-b))/(a^2-b^2)}+by = c`
`(ac(a-b)-ab)/(a^2-b^2)+by=c`
`by = c - (ac(a-b)-ab)/(a^2-b^2)`
`by = (a^2c-b^2c-a^2c+abc+ab)/(a^2-b^2)`
`by= (abc-b^2c+ab)/(a^2-b^2)`
`by = (bc(a-b)+ab)/(a^2-b^2)`
`y= (c(a-b)+a)/(a^2-b^2)`
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