Question

Solve for $xandy$

$ax+by=a-b;bx-ay=a+b$

Answer 1

Step 1:Simplification to make the $\text{'}y\text{'}$ co-efficient as same

Given,

$$\begin{gathered} ax+by=a-b..........................................\left(1\right) \\ bx-ay=a+b...........................................\left(2\right) \end{gathered}$$$$\begin{gathered} equation\left(1\right)\times a\Rightarrow a^{2}x+aby=a^2-ab...................................\left(3\right) \\ equation\left(2\right)\times b\Rightarrow b^{2}x-aby=ab+b^2...................................\left(4\right) \end{gathered}$$

Step 2: Applying Elimination Method to find the value of x

$$\begin{gathered} Equation\left(3\right)+\left(4\right) \\ \Rightarrow a^{2}x+b^{2}x=a^2+b^2 \\ \Rightarrow a^{2}x+b^{2}x=a^2+b^2 \\ \Rightarrow x(a^2+b^2)=a^2+b^2 \\ \Rightarrow x=\frac{a^2+b^2}{a^2+b^2} \\ \Rightarrow x=1 \end{gathered}$$

Step 3:Find the value of $y$ by substitution method

$$\begin{gathered} Substitutex=1inequation\left(1\right) \\ \Rightarrow a+by=a-b \\ \Rightarrow by=a-b-a \\ \Rightarrow by=-b \\ \Rightarrow y=\frac{-b}{b} \\ \Rightarrow y=-1 \end{gathered}$$

Therefore the values of $x=1andy=-1$