Intermediate Algebra, Chapter 4, 4.1, Section 4.1, Problem 28

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

& 3x - y = 10 \\
& 2x + 5y = 1

\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 = 10
&& \text{Given equation}
\\
& -y = -3x + 10
&& \text{Subtract each side by $3x$}
\\
& y = 3x - 10
&& \text{Multiply each side by $-1$}

\end{aligned}
\end{equation}
$


Since equation 1 is solved for $y$, we substitute $3x - 10$ for $y$ in equation 2.


$
\begin{equation}
\begin{aligned}

2x + 5 (3x - 10) =& 1
&& \text{Substitute $y =3x - 10$}
\\
2x + 15x - 50 =& 1
&& \text{Distributive Property}
\\
17x - 50 =& 1
&& \text{Combine like terms}
\\
17x =& 51
&& \text{Add each side by $50$}
\\
x =& 3
&& \text{Divide each side by $17$}

\end{aligned}
\end{equation}
$


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


$
\begin{equation}
\begin{aligned}

y =& 3(3) - 10
&& \text{Substitute } x = 3
\\
y =& 9 - 10
&& \text{Multiply}
\\
y =& -1
&& \text{Subtract}

\end{aligned}
\end{equation}
$