To solve the system of equations:
[tex]\[
\begin{cases}
2a - 3c = -6 \\
a + 2c = 11
\end{cases}
\][/tex]
we can use the method of substitution or elimination. Here, we'll use the elimination method to find the values of [tex]\(a\)[/tex] and [tex]\(c\)[/tex].
1. First Equation: [tex]\(2a - 3c = -6\)[/tex]
2. Second Equation: [tex]\(a + 2c = 11\)[/tex]
### Step-by-Step Solution:
Step 1: Align the coefficients
We have:
[tex]\[
2a - 3c = -6 \tag{1}
\][/tex]
[tex]\[
a + 2c = 11 \tag{2}
\][/tex]
Step 2: Eliminate one of the variables
To eliminate [tex]\(a\)[/tex], we can manipulate the equations to have the same coefficient for [tex]\(a\)[/tex]. Let's multiply Equation (2) by 2:
[tex]\[
2(a + 2c) = 2 \cdot 11
\][/tex]
[tex]\[
2a + 4c = 22 \tag{3}
\][/tex]
Now our system of equations looks like:
[tex]\[
2a - 3c = -6 \tag{1}
\][/tex]
[tex]\[
2a + 4c = 22 \tag{3}
\][/tex]
Step 3: Subtract one equation from the other
Next, subtract Equation (1) from Equation (3):
[tex]\[
(2a + 4c) - (2a - 3c) = 22 - (-6)
\][/tex]
[tex]\[
2a + 4c - 2a + 3c = 22 + 6
\][/tex]
[tex]\[
7c = 28
\][/tex]
Step 4: Solve for [tex]\(c\)[/tex]
[tex]\[
c = \frac{28}{7} = 4
\][/tex]
Step 5: Substitute [tex]\(c\)[/tex] back into one of the original equations
Now that we have [tex]\(c = 4\)[/tex], we substitute it back into Equation (2):
[tex]\[
a + 2c = 11
\][/tex]
[tex]\[
a + 2 \cdot 4 = 11
\][/tex]
[tex]\[
a + 8 = 11
\][/tex]
[tex]\[
a = 11 - 8
\][/tex]
[tex]\[
a = 3
\][/tex]
Step 6: Form the solution pair [tex]\((a, c)\)[/tex]
Thus, the solution to the system of equations is:
[tex]\[
(a, c) = (3, 4)
\][/tex]
### Conclusion:
The correct choice from the options presented is:
[tex]\((3, 4)\)[/tex]
So the solution [tex]\((a, c)\)[/tex] to the system of linear equations:
[tex]\[
\begin{cases}
2a - 3c = -6 \\
a + 2c = 11
\end{cases}
\][/tex]
is:
[tex]\((a, c) = (3, 4)\)[/tex].
