To find the [tex]\( x \)[/tex]-coordinate of the solution for the given system of equations
[tex]\[
\left\{\begin{array}{l}
3x + 3y = 3 \\
y = -\frac{1}{2} x + 2
\end{array}\right.
\][/tex]
we will solve this system step-by-step.
### Step 1: Simplify the first equation
First, let's simplify the first equation by dividing every term by 3:
[tex]\[
3x + 3y = 3 \quad \Rightarrow \quad x + y = 1
\][/tex]
This simplification makes the system:
[tex]\[
\left\{\begin{array}{l}
x + y = 1 \\
y = -\frac{1}{2}x + 2
\end{array}\right.
\][/tex]
### Step 2: Use substitution
We will use substitution to solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. Since we already have [tex]\( y \)[/tex] expressed in terms of [tex]\( x \)[/tex] from the second equation:
[tex]\[
y = -\frac{1}{2}x + 2
\][/tex]
we can substitute this into the first equation:
[tex]\[
x + \left( -\frac{1}{2}x + 2 \right) = 1
\][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Now, let's solve for [tex]\( x \)[/tex]:
[tex]\[
x - \frac{1}{2}x + 2 = 1
\][/tex]
[tex]\[
\frac{1}{2}x + 2 = 1
\][/tex]
Subtract 2 from both sides:
[tex]\[
\frac{1}{2}x = -1
\][/tex]
Multiply both sides by 2 to solve for [tex]\( x \)[/tex]:
[tex]\[
x = -2
\][/tex]
### Step 4: Solve for [tex]\( y \)[/tex]
Now that we have [tex]\( x = -2 \)[/tex], substitute this back into the second equation to find [tex]\( y \)[/tex]:
[tex]\[
y = -\frac{1}{2}(-2) + 2
\][/tex]
[tex]\[
y = 1 + 2
\][/tex]
[tex]\[
y = 3
\][/tex]
### Step 5: Verify the solution
Let's verify that [tex]\((x, y) = (-2, 3)\)[/tex] satisfies both original equations:
1. For the first equation [tex]\( 3x + 3y = 3 \)[/tex]:
[tex]\[
3(-2) + 3(3) = -6 + 9 = 3
\][/tex]
This is correct.
2. For the second equation [tex]\( y = -\frac{1}{2}x + 2 \)[/tex]:
[tex]\[
3 = -\frac{1}{2}(-2) + 2 = 1 + 2 = 3
\][/tex]
This is also correct.
Thus, the solution to the system of equations is [tex]\((x, y) = (-2, 3)\)[/tex].
### Answer
The [tex]\( x \)[/tex]-coordinate of the solution is [tex]\( \boxed{-2} \)[/tex].
