Question

Solve the following pair of equations:

$\frac{x}{a}-\frac{y}{b}=0$, $ax+by=a^2+b^2$

• A. $x=b-a$ and $y=b$
• B. $x=a$ and $y=b$
• C. $x=0$ and $y=ba$
• D. $x=ba$ and $y=ba$

Answer 1

Answer: (B)

The equation $\frac{x}{a}-\frac{y}{b}=0$ can be rewritten as:

$bx-ay=\mathrm{0...........}\left(1\right)$

The other equation given is:

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

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $bx-ay=0$ and write it as

$x=\frac{ay}{b}$

Substitute the value of $x$ in equation $ax+by=a^2+b^2$. We get

$a\left(\frac{ay}{b}\right)+by=a^2+b^2\Rightarrow \frac{a^{2}y}{b}+by=a^2+b^2\Rightarrow \frac{a^{2}y+b^{2}y}{b}=a^2+b^2\Rightarrow (a^2+b^2)y=b(a^2+b^2)\Rightarrow y=b$

Therefore, $y=b$

Substituting this value of $y$ in the equation $x=\frac{ay}{b}$, we get

$x=\frac{ab}{b}=a$

Hence, the solution is $x=a,y=b$.