Sure, let's solve the system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
y = -\frac{1}{2}x + 3 \\
y = 6 + x
\end{array}
\right.
\][/tex]
### Step-by-step Solution:
1. Set the Equations Equal to Each Other:
Since both expressions are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[
-\frac{1}{2}x + 3 = 6 + x
\][/tex]
2. Solve for [tex]\( x \)[/tex]:
First, let's eliminate the fractions by multiplying the entire equation by 2 to make calculations easier:
[tex]\[
2 \left( -\frac{1}{2}x + 3 \right) = 2(6 + x)
\][/tex]
[tex]\[
-x + 6 = 12 + 2x
\][/tex]
Next, we will add [tex]\( x \)[/tex] to both sides of the equation to begin isolating [tex]\( x \)[/tex]:
[tex]\[
6 = 12 + 3x
\][/tex]
Now, subtract 12 from both sides of the equation:
[tex]\[
6 - 12 = 3x
\][/tex]
[tex]\[
-6 = 3x
\][/tex]
To isolate [tex]\( x \)[/tex], divide both sides by 3:
[tex]\[
x = -2
\][/tex]
3. Solve for [tex]\( y \)[/tex] by Substituting [tex]\( x \)[/tex]:
Substitute [tex]\( x = -2 \)[/tex] back into either of the original equations to find [tex]\( y \)[/tex]. Let's use the equation [tex]\( y = 6 + x \)[/tex]:
[tex]\[
y = 6 + (-2)
\][/tex]
[tex]\[
y = 4
\][/tex]
So, the solution to the system of equations is [tex]\((x, y) = (-2, 4)\)[/tex].
### Verification:
To ensure our solution is correct, let's substitute [tex]\( x = -2 \)[/tex] and [tex]\( y = 4 \)[/tex] back into the original equations.
For the first equation [tex]\( y = -\frac{1}{2} x + 3 \)[/tex]:
[tex]\[
4 = -\frac{1}{2}(-2) + 3
\][/tex]
[tex]\[
4 = 1 + 3
\][/tex]
[tex]\[
4 = 4 \quad \text{(True)}
\][/tex]
For the second equation [tex]\( y = 6 + x \)[/tex]:
[tex]\[
4 = 6 + (-2)
\][/tex]
[tex]\[
4 = 4 \quad \text{(True)}
\][/tex]
Both equations are satisfied. Therefore, the solution [tex]\((x, y) = (-2, 4)\)[/tex] is correct.
