Question

Solve the following system of linear equations for $x\text{and}y$

$\frac{x}{a}-\frac{y}{b}=a-b;ax+by=a^3+b^3$

Answer 1

Step 1:Simplifying the equation by taking the LCM

Given,

$$\begin{gathered} \frac{x}{a}-\frac{y}{b}=a-b \\ \Rightarrow \frac{bx-ay}{ab}=a-b\left[\text{taking ab as LCM}\right] \\ \Rightarrow bx-ay=ab[a-b] \\ \Rightarrow bx-ay=ab[a-b]..........\left(1\right) \end{gathered}$$

Also,

$ax+by=a^3+b^3..........\left(2\right)$

Step 2: Making ‘y’ coefficient same to find the value of $x$

We get,

$$\begin{gathered} \left(1\right)xb\Rightarrow b^{2}x-aby=a{b}^2[a-b].........\left(3\right) \\ \left(2\right)xa\Rightarrow a^{2}x+aby=a[a^3+b^3]..........\left(4\right) \\ \left(3\right)+\left(4\right)\Rightarrow b^{2}x+a^{2}x=a^2{b}^2-a{b}^3+a^4+a{b}^3 \\ \Rightarrow x[a^2+b^2]=a^2[a^2+b^2] \\ \Rightarrow x=\frac{a^2[a^2+b^2]}{[a^2+b^2]}=a^2 \end{gathered}$$

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

Substituting $x=a^2\text{in eq}\left(2\right)$

$$\begin{gathered} \left(2\right)\Rightarrow a^3+by=a^3+b^3 \\ \Rightarrow by=a^3+b^3-a^3 \\ \Rightarrow by=b^3 \\ \Rightarrow y=b^2 \end{gathered}$$

Hence the values of $x=a^2\text{and}y=b^2$