Sure, let's solve the given system of equations step-by-step.
The system of equations is:
[tex]\[
\begin{aligned}
2x + y &= -3 \\
-2y &= 6 + 4x
\end{aligned}
\][/tex]
First, let's transform each equation into slope-intercept form, which is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
### Transforming the First Equation
The first equation is already quite close to the slope-intercept form:
[tex]\[ 2x + y = -3 \][/tex]
To isolate [tex]\( y \)[/tex]:
[tex]\[ y = -2x - 3 \][/tex]
### Transforming the Second Equation
The second equation is:
[tex]\[ -2y = 6 + 4x \][/tex]
First, solve for [tex]\( y \)[/tex] by dividing everything by [tex]\(-2\)[/tex]:
[tex]\[
y = \frac{6 + 4x}{-2}
\][/tex]
Distribute the division on the right-hand side:
[tex]\[
y = -2x - 3
\][/tex]
### Final Slope-Intercept Form of Each Equation
So, the slope-intercept form for each equation is:
[tex]\[
\begin{array}{l}
y = -2x - 3 \\
y = -2x - 3
\end{array}
\][/tex]
### Conclusion
Both equations are exactly the same. This means that rather than describing two lines that intersect at a point, these equations describe the same line.
Thus, the system of equations is actually dependent, meaning that any point [tex]\((x, y)\)[/tex] on the line [tex]\(y = -2x - 3\)[/tex] is a solution to the system.
So, the solution to the system is the set of all points [tex]\((x, y)\)[/tex] that satisfy the equation:
[tex]\[ y = -2x - 3 \][/tex]
