Sure, let's solve the system of linear equations:
[tex]\[
\begin{aligned}
1. & \quad x + 2y = 13 \\
2. & \quad 3x + 3y = 29
\end{aligned}
\][/tex]
We'll use the method of elimination to solve this system. Let's start by eliminating one of the variables.
First, we can simplify the second equation by dividing the entire equation by 3:
[tex]\[
3x + 3y = 29 \implies x + y = \frac{29}{3}
\][/tex]
Now our simplified system of equations looks like this:
[tex]\[
\begin{aligned}
1. & \quad x + 2y = 13 \\
2. & \quad x + y = \frac{29}{3}
\end{aligned}
\][/tex]
Next, we will subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[
(x + 2y) - (x + y) = 13 - \frac{29}{3}
\][/tex]
Simplify this step by step:
[tex]\[
x + 2y - x - y = 13 - \frac{29}{3}
\][/tex]
[tex]\[
y = 13 - \frac{29}{3}
\][/tex]
To proceed, we need a common denominator on the right-hand side. The common denominator of 13 and [tex]\(\frac{29}{3}\)[/tex] is 3, so we convert 13 to a fraction over 3:
[tex]\[
13 = \frac{39}{3}
\][/tex]
Substituting back:
[tex]\[
y = \frac{39}{3} - \frac{29}{3} = \frac{39 - 29}{3} = \frac{10}{3}
\][/tex]
So, we have found [tex]\( y = \frac{10}{3} \)[/tex].
Now, substitute [tex]\( y = \frac{10}{3} \)[/tex] back into the simplified second equation [tex]\( x + y = \frac{29}{3} \)[/tex]:
[tex]\[
x + \frac{10}{3} = \frac{29}{3}
\][/tex]
Subtract [tex]\(\frac{10}{3}\)[/tex] from both sides to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{29}{3} - \frac{10}{3} = \frac{29 - 10}{3} = \frac{19}{3}
\][/tex]
Thus, the solutions are:
[tex]\[
\begin{aligned}
x &= \frac{19}{3} \\
y &= \frac{10}{3}
\end{aligned}
\][/tex]
So, the solution to the system of equations is:
[tex]\[
\left( x, y \right) = \left( \frac{19}{3}, \frac{10}{3} \right)
\][/tex]
