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Solve the following system of equations using elimination:

[tex]\[
\begin{array}{l}
-2x + 3y = -8 \\
5x - 2y = -2
\end{array}
\][/tex]

A. [tex]$(2, -4)$[/tex]
B. [tex]$(2, 4)$[/tex]
C. [tex]$(-2, 4)$[/tex]
D. [tex]$(-2, -4)$[/tex]
E. No Solution
F. Infinite Solutions

Sagot :

GinnyAnswer
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]
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