Certainly! To solve the system of equations using the elimination method, let's follow these steps:
Given the system of equations:
[tex]\[
\begin{array}{l}
1.\quad 5x + 2y = 3 \\
2.\quad 4x - 8y = 12
\end{array}
\][/tex]
### Step 1: Make the Coefficients of [tex]\( y \)[/tex] Equal
First, we want to make the coefficients of [tex]\( y \)[/tex] in both equations equal so that we can eliminate [tex]\( y \)[/tex]. Notice that the second equation already has [tex]\(-8y\)[/tex], which is a multiple of the coefficient of [tex]\( y \)[/tex] in the first equation ([tex]\(+2y \times -4\)[/tex]).
### Step 2: Eliminate [tex]\( y \)[/tex]
We will multiply the entire first equation by 4 to make the coefficients of [tex]\( y \)[/tex] the same (in magnitude) but opposite in sign:
[tex]\[
4 \cdot (5x + 2y) = 4 \cdot 3
\][/tex]
[tex]\[
20x + 8y = 12 \quad \quad \text{New equation (3)}
\][/tex]
Now, our system of equations is:
[tex]\[
\begin{array}{l}
3.\quad 20x + 8y = 12 \\
2.\quad 4x - 8y = 12
\end{array}
\][/tex]
We can eliminate [tex]\( y \)[/tex] by adding equations (3) and (2):
[tex]\[
(20x + 8y) + (4x - 8y) = 12 + 12
\][/tex]
[tex]\[
20x + 4x + 8y - 8y = 24
\][/tex]
[tex]\[
24x = 24
\][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
[tex]\[
x = \frac{24}{24}
\][/tex]
[tex]\[
x = 1
\][/tex]
### Step 4: Substitute [tex]\( x \)[/tex] back to Find [tex]\( y \)[/tex]
We substitute [tex]\( x = 1 \)[/tex] back into the first equation to solve for [tex]\( y \)[/tex]:
[tex]\[
5(1) + 2y = 3
\][/tex]
[tex]\[
5 + 2y = 3
\][/tex]
[tex]\[
2y = 3 - 5
\][/tex]
[tex]\[
2y = -2
\][/tex]
[tex]\[
y = \frac{-2}{2}
\][/tex]
[tex]\[
y = -1
\][/tex]
### Conclusion
Thus, the solution to the system of equations is
[tex]\[
(x, y) = (1, -1)
\][/tex]
Among the given options, the correct answer is:
[tex]\[
(1, -1)
\][/tex]
