Question

Solve the following pair of equations:

$ax+by=c$
$bx+ay=1+c$

• A. $x=\frac{ac-b-bc}{b^2-a^2}$ and $y=\frac{bc-a-ac}{b^2-a^2}$
• B. $x=\frac{ab+b-ac}{a^2-b^2}$ and $y=\frac{bc-a-ac}{b^2-a^2}$
• C. $x=\frac{ac-b-bc}{a^2-b^2}$ and $y=\frac{bc-a-ac}{b^2-a^2}$
• D. $x=\frac{c-bc-a}{a^2-b^2}$ and $y=\frac{bc-a-ac}{b^2-a^2}$

Answer 1

The correct option is C $x=\frac{ac-b-bc}{a^2-b^2}$ and $y=\frac{bc-a-ac}{b^2-a^2}$

Given equations are,

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

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

Multiply equation (1) by $a$ and equation (2) by $b$, we get

$a^{2}x+aby=ac$ ....(3)

and $b^{2}x+aby=b+bc$ ....(4)

Subtract equations (3) and (4),

$x(a^2-b^2)=ac-b-bc$

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

Put this value in equation (1), we get

$a\left(\frac{ac-b-bc}{a^2-b^2}\right)+by=c$

$\Rightarrow a^{2}c-ab-abc-a^{2}c+b^{2}c=-by(a^2-b^2)$

$\Rightarrow b(bc-ab-a)=-by(a^2-b^2)$

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