Question

Question 9 (vi)
Solve the following pair of equations
$\frac{x}{a}+\frac{y}{b}=a+b,\frac{x}{a^2}+\frac{y}{b^2}=2,\text{where}a,b\ne 0$

Answer 1

Given pair of linear equations is:
$\frac{x}{a}+\frac{y}{b}=a+b...\left(i\right),\frac{x}{a^2}+\frac{y}{b^2}=\mathrm{2...}\left(ii\right),\text{where}a,b\ne 0$
$\Rightarrow$ Dividing the first equation by 'a' on both sides and subtracting it from the second equation, we get
$Firstequation:\frac{x}{a^2}+\frac{y}{ab}=1+\frac{b}{a},Subtracting,weget,y(\frac{1}{b^2}-\frac{1}{ab})=2-1-\frac{b}{a}y\left(\frac{a-b}{a{b}^2}\right)=2-1-\frac{b}{a}y\left(\frac{a-b}{a{b}^2}\right)=1-\frac{b}{a}y\left(\frac{a-b}{a{b}^2}\right)=\frac{a-b}{a}y=\frac{a{b}^2}{a}\Rightarrow y=b^2$
Now, put the value of $y$ in Eq. (ii), we get
$\frac{x}{a^2}+\frac{b^2}{b^2}=2\Rightarrow \frac{x}{a^2}=2-1=1\Rightarrow \frac{x}{a^2}=1\Rightarrow x=a^2$
Hence, the required values of $x$ and $y$ are $a^{2}and{b}^2$, respectively.