Sure! Let's solve the given system of equations using the elimination method and find the correct solution.
We are given:
[tex]\[
\begin{array}{l}
6x + 4y = 42 \quad \text{... (1)} \\
-3x + 3y = -6 \quad \text{... (2)}
\end{array}
\][/tex]
Our goal is to eliminate one of the variables. In this case, let's eliminate [tex]\(x\)[/tex] by aligning the coefficients of [tex]\(x\)[/tex]. To do this, we can manipulate equation (2) so that the coefficient of [tex]\(x\)[/tex] matches the coefficient in equation (1).
Step 1: Multiply equation (2) by 2. This gives us:
[tex]\[
-3x \cdot 2 + 3y \cdot 2 = -6 \cdot 2
\][/tex]
which simplifies to:
[tex]\[
-6x + 6y = -12 \quad \text{... (3)}
\][/tex]
Step 2: Now add equation (1) and equation (3):
[tex]\[
(6x + 4y) + (-6x + 6y) = 42 + (-12)
\][/tex]
Simplifying this, we get:
[tex]\[
6x - 6x + 4y + 6y = 42 - 12
\][/tex]
[tex]\[
0x + 10y = 30
\][/tex]
[tex]\[
10y = 30
\][/tex]
Step 3: Solve for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{30}{10}
\][/tex]
[tex]\[
y = 3
\][/tex]
Step 4: Substitute the value of [tex]\(y\)[/tex] back into equation (1) to solve for [tex]\(x\)[/tex]:
[tex]\[
6x + 4y = 42
\][/tex]
Substitute [tex]\(y = 3\)[/tex]:
[tex]\[
6x + 4 \cdot 3 = 42
\][/tex]
[tex]\[
6x + 12 = 42
\][/tex]
Subtract 12 from both sides:
[tex]\[
6x = 30
\][/tex]
Divide both sides by 6:
[tex]\[
x = \frac{30}{6}
\][/tex]
[tex]\[
x = 5
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
(x, y) = (5, 3)
\][/tex]
Let's verify the result against the given multiple-choice answers:
- (27,30)
- (3.6,4.6)
- (4.6,3.6)
- (3,5)
- (5,3)
- (30,27)
The solution [tex]\((x, y) = (5, 3)\)[/tex] matches one of the options exactly:
[tex]\[
(5, 3)
\][/tex]
Hence, the correct solution to the system of equations is:
[tex]\((5, 3)\)[/tex].
