Answer :

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]