Alright, let's solve the given system of linear equations using the elimination method. We have:
[tex]\[
\begin{cases}
4x + 5y = 2 \quad \text{(Equation 1)}\\
2x + 3y = 0 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
Step 1: Eliminate one of the variables.
We will eliminate [tex]\( x \)[/tex] by making the coefficients of [tex]\( x \)[/tex] the same in both equations and subtracting one equation from the other.
To do this, we can multiply Equation 2 by 2 to align the [tex]\( x \)[/tex] terms:
[tex]\[
\begin{cases}
4x + 5y = 2 \quad \text{(Equation 1)}\\
4x + 6y = 0 \quad \text{(Equation 3: \(2 \times (2x + 3y = 0) \))}
\end{cases}
\][/tex]
Step 2: Subtract Equation 1 from Equation 3.
[tex]\[
(4x + 6y) - (4x + 5y) = 0 - 2
\][/tex]
Simplifying this, we eliminate [tex]\( 4x \)[/tex]:
[tex]\[
4x + 6y - 4x - 5y = -2
\][/tex]
[tex]\[
y = -2
\][/tex]
So, we have found that [tex]\( y = -2 \)[/tex].
Step 3: Substitute [tex]\( y \)[/tex] back into one of the original equations to solve for [tex]\( x \)[/tex].
We substitute [tex]\( y = -2 \)[/tex] into Equation 2:
[tex]\[
2x + 3(-2) = 0
\][/tex]
[tex]\[
2x - 6 = 0
\][/tex]
[tex]\[
2x = 6
\][/tex]
[tex]\[
x = 3
\][/tex]
So, we have found [tex]\( x = 3 \)[/tex].
Conclusion: The solution to the system of equations is
[tex]\[
(x, y) = (3, -2)
\][/tex]
