Question

Two numbers differ by $3.$ The sum of twice the smaller number and thrice the greater number is $19.$ Find the numbers.

Answer 1

Let the smaller number be ${}^{\prime }{x}^{\prime }$ and the larger number be ${}^{\prime }{y}^{\prime }.$
Given that the two numbers differ by $3$ so,
$y-x=\mathrm{3.....}\left(i\right)$
Also, the sum of twice the smaller number and thrice the greater number is $19.$
So, $2x+3y=19......\left(ii\right)$
The two equations obtained are
$y-x=3$
$2x+3y=19$
Multiplying $eq.\left(i\right)$ by $3$ we get
$3y-3x=\mathrm{9.....}\left(iii\right)$
On subtracting $eq.\left(iii\right)$ from $eq.\left(ii\right)$, we have

$2x+3y-(3y-3x)=19-9$

$\Rightarrow 2x+3y-3y+3x=10$

$\Rightarrow 5x=10$

$\Rightarrow x=\frac{10}{5}=2$

Putting the value of $x=2$ in $eq.\left(i\right)$, we get
$y-2=3$

$\Rightarrow y=3+2$

$\Rightarrow y=5$

Thus, the two numbers are $2$ and $5.$