To solve the given system of equations and write each equation in slope-intercept form, follow these steps:
### Starting Equations:
[tex]\[
\begin{aligned}
(1) \quad 2x + y &= -3 \\
(2) \quad -2y &= 6 + 4x
\end{aligned}
\][/tex]
### Step-by-Step Solution:
#### First Equation: [tex]\( 2x + y = -3 \)[/tex]
We need to rewrite this in slope-intercept form [tex]\( y = mx + b \)[/tex]:
1. Isolate [tex]\( y \)[/tex] on one side:
[tex]\[
y = -2x - 3
\][/tex]
So the first equation in slope-intercept form is:
[tex]\[
y = -2x - 3
\][/tex]
#### Second Equation: [tex]\( -2y = 6 + 4x \)[/tex]
Again, we need to rewrite this in slope-intercept form [tex]\( y = mx + b \)[/tex]:
1. Divide both sides by -2 to isolate [tex]\( y \)[/tex]:
[tex]\[
y = -\frac{4x + 6}{2}
\][/tex]
2. Simplify the right side:
[tex]\[
y = -2x - 3
\][/tex]
So the second equation in slope-intercept form is:
[tex]\[
y = -2x - 3
\][/tex]
### Summary:
The system of equations in slope-intercept form is:
[tex]\[
\begin{aligned}
y &= -2x - 3 \\
y &= -2x - 3
\end{aligned}
\][/tex]
Both equations describe the same line, so they are not two distinct lines but rather the same line repeated. This suggests that every point on the line [tex]\( y = -2x - 3 \)[/tex] is a solution to the system.
The solution to the system of equations can be written as:
[tex]\[
y = -2x - 3
\][/tex]
So in the format you requested:
[tex]\[
\begin{array}{l}
y = -2x - 3 \\
y = -2x - 3
\end{array}
\][/tex]
Every point [tex]\((x, y)\)[/tex] that lies on the line [tex]\( y = -2x - 3 \)[/tex] is a solution to the system.
