Question

The sum of the successors of two numbers is $42$ and the difference of predecessors is $12$. Find the numbers.

Answer 1

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

Then,

$\begin{array}{cc}(x+1)+(y+1)& =42\\ x+y& =40\dots \left(i\right)\end{array}$

Also,

$\begin{array}{cc}(x-1)-(y-1)& =12\\ x-y& =12\dots \left(ii\right)\end{array}$

Add equations $\left(i\right)$ and $\left(ii\right)$,

$\begin{array}{cc}x+y+x-y& =40+12\\ 2x& =52\\ x& =26\end{array}$

Substitute $26$ for $x$ in equation $\left(i\right)$,

$\begin{array}{cc}26+y& =40\\ y& =14\end{array}$

Hence, the two numbers are $26$ and $14$.