Question

Find two numbers such that the sum of twice the first and thrice the second is $92$, and four times the first exceeds seven times the second by $2$.

Answer 1

Given : two numbers are such that the sum of twice the first and thrice the second is $92$, and four times the first exceeds seven times the second by $2$.

To find : the two numbers

Let $x$ and $y$ be the two numbers required.

According to the question :

$\Rightarrow 2x+3y=92$ ........$\left(1\right)$

$\Rightarrow 4x-7y=2$ ........$\left(2\right)$

multiply the first equation by $2$ , and subtract eqn $\left(1\right)$ from eqn $\left(2\right)$

$4x+6y=184$

$-(4x-7y=2)$ , we get

$\Rightarrow 13y=182$

$\Rightarrow y=\frac{182}{13}=14$

Put $y=14$ in $\left(1\right)$

$2x+3y=92$

$\Rightarrow 2x+3\times 14=92$

$\Rightarrow 2x=92-42=50$

$\therefore x=\frac{50}{2}=25$

$\therefore x=25$ and $y=14$