Beginning Algebra With Applications, Chapter 6, 6.2, Section 6.2, Problem 2

Use this system of equations:
$
\begin{equation}
\begin{aligned}

(1) y =& x+4
\\
(2) 3x+y =& 12

\end{aligned}
\end{equation}
$


To solve the system by substitution, substitute $\underline{x+4}$ for $y$ in equation (2): $3x+x+4 = 12$.

Solving the equation $3x+x+4 = 12$ for $x$ gives $x =$ $\underline{2}$

To find $y$, substitute this value of $x$ into equation (1): $y =$ $\underline{2}$ $+4 =$ $\underline{6}$.

The solution of the system of equations is $(\underline{2},$ $\underline{6})$.