To solve the given system by elimination, let's go through the process step-by-step:
1. Write down the system of equations:
[tex]\[
\begin{cases}
2x - 3y = 19 \quad \text{(Equation 1)} \\
5x + 4y = -10 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
2. Eliminate one variable by adding or subtracting the equations. To do this, we need to make the coefficients of one variable (either [tex]\( x \)[/tex] or [tex]\( y \)[/tex]) equal in both equations. Let's eliminate [tex]\( y \)[/tex].
3. Find the common multiple of the coefficients of [tex]\( y \)[/tex]: The coefficients of [tex]\( y \)[/tex] are [tex]\(-3\)[/tex] in Equation 1 and [tex]\(4\)[/tex] in Equation 2. The least common multiple of 3 and 4 is 12. Thus, we will multiply the entire Equation 1 by [tex]\(4\)[/tex] and the entire Equation 2 by [tex]\(3\)[/tex] so the coefficients of [tex]\( y \)[/tex] become [tex]\( -12 \)[/tex] and [tex]\( 12 \)[/tex] respectively.
[tex]\[
4 \cdot (2x - 3y) = 4 \cdot 19
\][/tex]
[tex]\[
3 \cdot (5x + 4y) = 3 \cdot (-10)
\][/tex]
This yields:
[tex]\[
8x - 12y = 76 \quad \text{(Equation 3)}
\][/tex]
[tex]\[
15x + 12y = -30 \quad \text{(Equation 4)}
\][/tex]
4. Add Equation 3 and Equation 4 to eliminate [tex]\( y \)[/tex]:
[tex]\[
(8x - 12y) + (15x + 12y) = 76 + (-30)
\][/tex]
Simplifying this, we get:
[tex]\[
23x = 46
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{46}{23}
\][/tex]
[tex]\[
x = 2
\][/tex]
6. Substitute [tex]\( x = 2 \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]. We will use Equation 1:
[tex]\[
2(2) - 3y = 19
\][/tex]
Simplifying this, we get:
[tex]\[
4 - 3y = 19
\][/tex]
[tex]\[
-3y = 19 - 4
\][/tex]
[tex]\[
-3y = 15
\][/tex]
[tex]\[
y = \frac{15}{-3}
\][/tex]
[tex]\[
y = -5
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
\boxed{(2, -5)}
\][/tex]
So, the correct choice is:
A. The solution set is [tex]\(\{ (2, -5) \} \)[/tex].
