To solve the given system of equations:
[tex]\[
\begin{cases}
4x - 3y = 26 \\
3x + 2y = 11
\end{cases}
\][/tex]
we'll use the method of substitution or elimination. In this case, we'll use the elimination method.
First, let's label the equations for reference:
1) [tex]\( 4x - 3y = 26 \)[/tex]
2) [tex]\( 3x + 2y = 11 \)[/tex]
We want to eliminate one of the variables. Let's eliminate [tex]\(y\)[/tex]. To do this, we need to make the coefficients of [tex]\(y\)[/tex] in both equations equal in magnitude but opposite in sign. We can do this by multiplying each equation by the appropriate values to get the coefficients of [tex]\(y\)[/tex] to be 6 and -6.
Multiply equation (1) by 2:
[tex]\[ 2(4x - 3y) = 2(26) \][/tex]
[tex]\[ 8x - 6y = 52 \][/tex]
Multiply equation (2) by 3:
[tex]\[ 3(3x + 2y) = 3(11) \][/tex]
[tex]\[ 9x + 6y = 33 \][/tex]
Now we have the system:
[tex]\[ 8x - 6y = 52 \][/tex]
[tex]\[ 9x + 6y = 33 \][/tex]
Add these two equations together to eliminate [tex]\(y\)[/tex]:
[tex]\[ (8x - 6y) + (9x + 6y) = 52 + 33 \][/tex]
[tex]\[ 17x = 85 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{85}{17} \][/tex]
[tex]\[ x = 5 \][/tex]
Now that we have [tex]\(x = 5\)[/tex], we can substitute this value back into one of the original equations to solve for [tex]\(y\)[/tex]. Let's use equation (2):
[tex]\[ 3x + 2y = 11 \][/tex]
[tex]\[ 3(5) + 2y = 11 \][/tex]
[tex]\[ 15 + 2y = 11 \][/tex]
Subtract 15 from both sides:
[tex]\[ 2y = 11 - 15 \][/tex]
[tex]\[ 2y = -4 \][/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{-4}{2} \][/tex]
[tex]\[ y = -2 \][/tex]
Thus, the solution to the system of equations is [tex]\((x, y) = (5, -2)\)[/tex].
So, the correct answer is:
[tex]\((5, -2)\)[/tex]
