Sure, let's solve the system of equations using the elimination method:
1. The system of equations is:
[tex]\[
\begin{array}{l}
5x + 6y = 12 \quad \text{(Equation 1)}\\
2x - 4y = 8 \quad \text{(Equation 2)}
\end{array}
\][/tex]
2. First, we'll try to eliminate one of the variables by making the coefficients of [tex]\(x\)[/tex] or [tex]\(y\)[/tex] in both equations the same.
3. We’ll start by eliminating [tex]\(x\)[/tex]. The coefficients of [tex]\(x\)[/tex] in Equation 1 and Equation 2 are 5 and 2, respectively. To make them the same, we can multiply Equation 2 by 5 and Equation 1 by 2:
[tex]\[
2(5x + 6y) = 2(12) \implies 10x + 12y = 24 \quad \text{(Equation 3)}
\][/tex]
[tex]\[
5(2x - 4y) = 5(8) \implies 10x - 20y = 40 \quad \text{(Equation 4)}
\][/tex]
4. Now, subtract Equation 4 from Equation 3 to eliminate [tex]\(x\)[/tex]:
[tex]\[
(10x + 12y) - (10x - 20y) = 24 - 40
\][/tex]
Simplifying, we get:
[tex]\[
10x + 12y - 10x + 20y = 24 - 40
\][/tex]
[tex]\[
32y = -16
\][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{-16}{32} = -\frac{1}{2}
\][/tex]
5. Now that we have the value of [tex]\(y\)[/tex], we can substitute it back into either Equation 1 or Equation 2 to find the value of [tex]\(x\)[/tex]. We'll use Equation 2:
[tex]\[
2x - 4(-\frac{1}{2}) = 8
\][/tex]
Simplifying, we get:
[tex]\[
2x + 2 = 8
\][/tex]
[tex]\[
2x = 8 - 2
\][/tex]
[tex]\[
2x = 6
\][/tex]
[tex]\[
x = \frac{6}{2} = 3
\][/tex]
6. Therefore, the solution to the system of equations is [tex]\(x = 3\)[/tex] and [tex]\(y = -\frac{1}{2}\)[/tex], which matches the point [tex]\((3, -0.5)\)[/tex].
So, the correct answer is [tex]\((3, -0.5)\)[/tex].
