Intermediate Algebra, Chapter 4, 4.1, Section 4.1, Problem 24

Solve the system $\begin{equation}
\begin{aligned}

& -3x + y = -5 \\
& x + 2y = 0

\end{aligned}
\end{equation}
$ by substitution. If the system is inconsistent or has dependent equations.

We solve for $y$ in equation 1


$
\begin{equation}
\begin{aligned}

-3x + y =& -5
&& \text{Given equation}
\\
y =& 3x - 5
&& \text{Add each side by $3x$}

\end{aligned}
\end{equation}
$


Since equation 1 is solved for $y$, we substitute $3x - 5$ for $y$ in equation 2.


$
\begin{equation}
\begin{aligned}

x + 2(3x - 5) =& 0
&& \text{Substitute $y = 3x - 5$}
\\
x + 6x - 10 =& 0
&& \text{Distributive Property}
\\
7x - 10 =& 0
&& \text{Combine like terms}
\\
7x =& 10
&& \text{Add each side by $10$}
\\
x =& \frac{10}{7}
&& \text{Divide each side by $7$}

\end{aligned}
\end{equation}
$


We found $x$. Now we solve for $y$ in equation 1.


$
\begin{equation}
\begin{aligned}

y =& 3 \left( \frac{10}{7} \right) - 5
&& \text{Substitute } x = \frac{10}{7}
\\
\\
y =& \frac{30}{7} - 5
&& \text{Multiply}
\\
\\
y =& \frac{30-35}{7}
&& \text{Get the LCD}
\\
\\
y =& - \frac{5}{7}
&&

\end{aligned}
\end{equation}
$