Beginning Algebra With Applications, Chapter 4, 4.1, Section 4.1, Problem 24

The sum of two numbers is two. The difference between eight and twice the smaller number is two less than four times the larger. Find the two numbers.

If we let $x$ and $y$ be the smaller and larger numbers, respectively, then we have

$x+y = 2$

$x = 2-y \qquad$ Equation 1

And


$
\begin{equation}
\begin{aligned}

8-2x =& 4y-2
\\
8+2 =& 2x+4y
\\
10 =& 2x+4y \qquad \text{Equation 2}

\end{aligned}
\end{equation}
$



By substituting equation 2 with 1, we get


$
\begin{equation}
\begin{aligned}

10 =& 2(2-y) + 4y
\\
10 =& 2(2) - 2(y) + 4y
\\
10 =& 4 - 2y + 4y
\\
10-4 =& 2y
\\
6 =& 2y
\\
\frac{6}{2} =& \frac{2y}{2}
\\
3 =& y

\end{aligned}
\end{equation}
$


Then, by applying back substitution, we get


$
\begin{equation}
\begin{aligned}

x = 2-y = 2-3 = -1

\end{aligned}
\end{equation}
$


Therefore, the two numbers are -1 and 3.