Certainly! Let's solve the system of equations step by step.
The system of equations is:
[tex]\[ \left\{
\begin{array}{l}
\frac{x - 3y}{2} = 5 \quad \text{(Equation 1)} \\
4 - y + \frac{y}{2} = 1 \quad \text{(Equation 2)}
\end{array}
\right. \][/tex]
### Solving Equation 2:
First, simplify Equation 2:
[tex]\[ 4 - y + \frac{y}{2} = 1 \][/tex]
Combine like terms:
[tex]\[ 4 - y + \frac{y}{2} = 1 \][/tex]
[tex]\[ 4 - y + \frac{y}{2} = 1 \][/tex]
Rewrite [tex]\(-y\)[/tex] as [tex]\(-\frac{2y}{2}\)[/tex] to have a common denominator:
[tex]\[ 4 - \frac{2y}{2} + \frac{y}{2} = 1 \][/tex]
Combine the terms involving [tex]\(y\)[/tex]:
[tex]\[ 4 - \frac{2y - y}{2} = 1 \][/tex]
[tex]\[ 4 - \frac{y}{2} = 1 \][/tex]
Isolate the fraction involving [tex]\(y\)[/tex]:
[tex]\[ 4 - 1 = \frac{y}{2} \][/tex]
[tex]\[ 3 = \frac{y}{2} \][/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ y = 6 \][/tex]
So we have determined that:
[tex]\[ y = 6 \][/tex]
### Solving Equation 1:
With [tex]\(y\)[/tex] known, substitute [tex]\(y = 6\)[/tex] into Equation 1:
[tex]\[ \frac{x - 3(6)}{2} = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ \frac{x - 18}{2} = 5 \][/tex]
Multiply both sides by 2 to clear the fraction:
[tex]\[ x - 18 = 10 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = 28 \][/tex]
So, the solution to the system of equations is:
[tex]\[ x = 28 \][/tex]
[tex]\[ y = 6 \][/tex]
Thus, the ordered pair [tex]\((x, y)\)[/tex] that satisfies both equations is:
[tex]\[ (28, 6) \][/tex]
