Let's solve the given system of equations step-by-step using the Elimination method:
The system of equations is:
[tex]\[
\begin{array}{l}
-2x + 3y = -8 \\
5x - 2y = -2 \\
\end{array}
\][/tex]
1. Multiply Each Equation:
To eliminate [tex]\( y \)[/tex], we need to make the coefficients of [tex]\( y \)[/tex] in both equations equal in magnitude but opposite in sign. Therefore, we'll multiply the first equation by 2 and the second equation by 3:
[tex]\[
\begin{aligned}
&2 \times (-2x + 3y) = 2 \times (-8) \quad \Rightarrow \quad -4x + 6y = -16 \quad \quad \text{[Equation 1 multiplied by 2]} \\
&3 \times (5x - 2y) = 3 \times (-2) \quad \Rightarrow \quad 15x - 6y = -6 \quad \quad \text{[Equation 2 multiplied by 3]}
\end{aligned}
\][/tex]
2. Add the Two Equations:
Now add the two new equations to eliminate [tex]\( y \)[/tex]:
[tex]\[
\begin{aligned}
(-4x + 6y) + (15x - 6y) &= -16 + (-6) \\
-4x + 15x + 6y - 6y &= -22 \\
11x &= -22
\end{aligned}
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{-22}{11} = -2
\][/tex]
3. Substitute [tex]\( x \)[/tex] Back into One of the Original Equations:
Use the first original equation to find [tex]\( y \)[/tex]:
[tex]\[
\begin{aligned}
-2(-2) + 3y &= -8 \\
4 + 3y &= -8 \\
3y &= -8 - 4 \\
3y &= -12 \\
y &= \frac{-12}{3} = -4
\end{aligned}
\][/tex]
So, the solution to the system of equations is [tex]\( x = -2 \)[/tex] and [tex]\( y = -4 \)[/tex].
Therefore, the correct solution for the system of equations [tex]\( \begin{array}{l} -2x + 3y = -8 \\ 5x - 2y = -2 \end{array} \)[/tex] is:
[tex]\[
\boxed{(-2, -4)}
\][/tex]
