Question

Solve: $ax+by=c,bx+ay=1+c$

• A. $x=\frac{a-bc-b}{a^2-b^2},y=\frac{ac-bc+a}{a^2-a{b}^2}$
• B. $x=\frac{ac-bc-b}{a^2-b^2},y=\frac{ac-c+ba}{a^2-b^2}$
• C. $x=\frac{a-bc-b}{a^2-b^2},y=\frac{bc-c+ba}{a^2-c^2}$
• D. None of these

Answer 1

The correct option is D None of these

$ax+by=c$ ....(1)

$bx+ay=1+c$ ....(2)

Adding (1) and (2), we get

$(a+b)x+(a+b)y=2c+1$

$\Rightarrow x+y=\frac{2c+1}{a+b}$ ....(3)

Subtracting (2) and (1), we get

$(a-b)x-(a-b)y=-1$

$\Rightarrow x-y=\frac{-1}{a-b}$ .....(4)

Adding (3) and (4), we get

$2x=\frac{2c+1}{a+b}+\frac{-1}{a-b}=\frac{2ac-2bc+a-b-a-b}{a^2-b^2}$

$\Rightarrow 2x=\frac{2ac-2bc-2b}{a^2-b^2}$

$\Rightarrow x=\frac{ac-bc-b}{a^2-b^2}$

Subtracting (4) from (3) we get

$2y=\frac{2c+1}{a+b}+\frac{+1}{a-b}=\frac{2ac-2bc+a-b+a+b}{a^2-b^2}$

$\Rightarrow 2y=\frac{2ac-2bc+2a}{a^2-b^2}$

$\Rightarrow y=\frac{ac+bc+a}{a^2-b^2}$

Hence, $x=\frac{ac-bc-b}{a^2-b^2},y=\frac{ac-bc+a}{a^2-b^2}$