Let's solve the given system of equations and write each equation in slope-intercept form.
The system of equations is:
[tex]\[
\begin{aligned}
1) & \quad 2x + y = -3, \\
2) & \quad -2y = 6 + 4x.
\end{aligned}
\][/tex]
### Step 1: Convert the first equation to slope-intercept form (y = mx + b)
The first equation is [tex]\(2x + y = -3\)[/tex].
To write it in slope-intercept form, solve for [tex]\( y \)[/tex]:
[tex]\[
2x + y = -3 \\
y = -2x - 3.
\][/tex]
So, the first equation in slope-intercept form is:
[tex]\[
y = -2x - 3.
\][/tex]
### Step 2: Convert the second equation to slope-intercept form
The second equation is [tex]\(-2y = 6 + 4x\)[/tex].
To write it in slope-intercept form, solve for [tex]\( y \)[/tex]:
[tex]\[
-2y = 6 + 4x \\
y = \frac{6 + 4x}{-2} \\
y = -2x - 3.
\][/tex]
So, the second equation in slope-intercept form is:
[tex]\[
y = -2x - 3.
\][/tex]
Thus, writing both equations in the slope-intercept form (y = mx + b), we get:
[tex]\[
\begin{array}{l}
y = -2x - 3 \\
y = -2x \quad + \square = -3.
\end{array}
\][/tex]
So in the second equation, the value of [tex]\(\square\)[/tex] is [tex]\(-3\)[/tex].
These two equations are essentially the same, meaning the lines overlap completely. The system of equations has infinitely many solutions since every point on the line [tex]\( y = -2x - 3 \)[/tex] is a solution to the system.
