Question

Solve the pair of equations : $21x+47y=110$ and $47x+21y=162$

Answer 1

$21x+47y=110.....\left(1\right)$

$47x+21y=162.....\left(2\right)$

Multiplying equation $\left(1\right)$ by $21$, we get

$441x+987y=2310.....\left(3\right)$

Also multiplying equation $\left(2\right)$ by $47$, we get

$2209x+987y=7614.....\left(4\right)$

Now, subtracting equation $\left(3\right)$ from $\left(4\right)$, we have

$(2209x+987y)-(441x+987y)=7614-2310$

$1768x=5304$

$\Rightarrow x=\frac{5304}{1768}=3$

Substituting the value of $x$ in equation $\left(1\right)$, we have

$21\times 3+47y=110$

$\Rightarrow 47y=110-63$

$\Rightarrow y=\frac{47}{47}=1$

Therefore,

$x=3$ and $y=1$.