Let's analyze and solve the system of equations given:
[tex]\[
\begin{aligned}
& \text{Equation 1: } y = 3x + 9 \\
& \text{Equation 2: } 6x + 2y = 6
\end{aligned}
\][/tex]
### Step 1: Graph each equation
First, let's rewrite Equation 2 in slope-intercept form [tex]\(y = mx + b\)[/tex].
#### Equation 2:
[tex]\[ 6x + 2y = 6 \][/tex]
Subtract [tex]\(6x\)[/tex] from both sides:
[tex]\[ 2y = -6x + 6 \][/tex]
Divide by 2:
[tex]\[ y = -3x + 3 \][/tex]
Now we have both equations in the form [tex]\(y = mx + b\)[/tex]:
1. [tex]\(y = 3x + 9\)[/tex]
2. [tex]\(y = -3x + 3\)[/tex]
### Step 2: Graph the lines
Line 1: [tex]\(y = 3x + 9\)[/tex]
- The slope is 3, and the y-intercept is 9.
- This line rises steeply and crosses the y-axis at (0, 9).
Line 2: [tex]\(y = -3x + 3\)[/tex]
- The slope is -3, and the y-intercept is 3.
- This line falls steeply and crosses the y-axis at (0, 3).
### Step 3: Determine the intersection point
To find the intersection point, if any, we set the two equations equal to each other:
[tex]\[
3x + 9 = -3x + 3
\][/tex]
Solve for [tex]\(x\)[/tex]:
Add [tex]\(3x\)[/tex] to both sides:
[tex]\[ 6x + 9 = 3 \][/tex]
Subtract 9 from both sides:
[tex]\[ 6x = -6 \][/tex]
Divide by 6:
[tex]\[ x = -1 \][/tex]
Now, substitute [tex]\(x = -1\)[/tex] back into either original equation to find [tex]\(y\)[/tex]. Using [tex]\(y = 3x + 9\)[/tex]:
[tex]\[ y = 3(-1) + 9 \][/tex]
[tex]\[ y = -3 + 9 \][/tex]
[tex]\[ y = 6 \][/tex]
Thus, the unique solution to this system is [tex]\((-1, 6)\)[/tex].
### Step 4: Verify the type of solution
Since both lines are straight and have different slopes, they intersect at exactly one point. There are no other intersection points.
### Conclusion
Given the graphs and calculations:
- There is one unique solution [tex]\((-1, 6)\)[/tex].
So, the solution to the system of equations is [tex]\(\boxed{(-1, 6)}\)[/tex].
