Question

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

Answer 1

The given two equations

$$\begin{gathered} ax+by=c\dots \dots \dots \dots \dots \dots \dots ..\left(1\right) \\ bx+ay=1+c\dots \dots \dots \dots \dots \dots ..\left(2\right) \end{gathered}$$

Multiply equation $\left(1\right)$ by$a$ then we get

$a^{2}x+aby=ac\dots \dots \dots \dots \dots \dots \left(3\right)$

Multiply equation $\left(2\right)$ by $b$ then we get

$b^{2}x+aby=b+bc\dots \dots \dots \dots .\left(4\right)$

Subtracting the equation $\left(3\right)$ from equation $\left(4\right)$, we get

$$\begin{gathered} (a^{2}x+aby)–(b^{2}x+aby)=ac–(b+bc) \\ {a}^{2}x+aby–b^{2}x-aby=ac–b–bc \\ {a}^{2}x–b^{2}x=ac–b–bc \\ x(a^2–b^2)=ac–b–bc \end{gathered}$$

$x=\frac{ac–b–bc}{(a^2–b^2)}$

Multiply equation $\left(1\right)$ by $b$ then we get

$abx+b^{2}y=bc\dots \dots \dots \dots \dots ..\left(5\right)$

Multiply equation $\left(2\right)$ by $a$ then we get

$abx+a^{2}y=a+ac\dots \dots \dots ..\left(6\right)$

Subtracting equation $\left(5\right)$ from equation$\left(6\right)$, we get

$$\begin{gathered} abx+b^{2}y–abx–a^{2}y=bc–a–ac \\ {b}^{2}y–a^{2}y=bc–a–ac \\ y(b^2–a^2)=bc–a–ac \end{gathered}$$

$y=\frac{bc–a–ac}{(b^2–a^2)}$

Hence, $x=\frac{ac–b–bc}{(a^2–b^2)}$ and $y=\frac{bc–a–ac}{(b^2–a^2)}$ are the correct answer.