Solve the following pair of linear equations: ax + by = c

Q) Solve the following pair of linear equations:

ax + by = c

bx + ay = 1 + c

Ans:

Given the equations:

a x + b y = c ….. (i)
b x + a y = 1 + c ….. (ii)

Let’s multiply equation (i) with a and equation (ii) with b and we get:

a2 x + ab y = a c ….. (iii)
b2 x + ab y = b + b c ….. (iv)

Let’s subtract equation (iv) from equation (iii), we get:

a² x – b² x = ac – b – b c

∴ x (a² – b² ) = c (a – b) – b

$\therefore x = \frac {c(a - b) - b}{a^2 - b^ 2}$

Let’s put $x = \frac {c(a - b) - b}{a^2 - b^ 2}$ in equation (i), we get:

$a (\frac {c(a - b) - b}{a^2 - b^ 2}) + b~y = c$

$\therefore b~y = c - a(\frac {c(a - b) - b}{a^2 - b^ 2})$

$\therefore b~y = \frac {c (a^2 - b^ 2) - a(c(a - b) - b)}{(a^2 - b^ 2)}$

$\therefore b~y = \frac {c (a^2 - b^ 2) - a(ac - bc - b)}{(a^2 - b^ 2)}$

$\therefore b~y = \frac {(a^2c - b^ 2c - a^2c + abc + ab)}{(a^2 - b^ 2)}$

$\therefore b~y = \frac {(\cancel{a^2c} - b^ 2c - \cancel{a^2c} + abc + ab)}{(a^2 - b^ 2)}$

$\therefore b~y = \frac {(- b^ 2c + abc + ab)}{(a^2 - b^ 2)}$

$\therefore b~y = \frac {b(- bc + ac + a)}{(a^2 - b^ 2)}$

$\therefore y = \frac {c(a - b) + a}{(a^2 - b^ 2)}$

Therefore, $x = \frac {c(a - b) - b}{a^2 - b^ 2}~and~y = \frac {c(a - b) + a}{(a^2 - b^ 2)}$

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