Answer :

To solve the system of linear equations:

[tex]\[ \left\{ \begin{array}{l} 6x - 5y = -3 \\ 3x + 2y = 12 \end{array} \right. \][/tex]

we will use the elimination method. Here is the step-by-step solution:

1. Multiply the equations to align coefficients:

Multiply the second equation by 2 to make the coefficient of \( x \) in the second equation equal to the coefficient of \( x \) in the first equation.

[tex]\[ 2 \times (3x + 2y = 12) \implies 6x + 4y = 24 \][/tex]

Now our system of equations looks like this:

[tex]\[ \begin{cases} 6x - 5y = -3 \\ 6x + 4y = 24 \end{cases} \][/tex]

2. Subtract the second equation from the first:

[tex]\[ (6x - 5y) - (6x + 4y) = -3 - 24 \][/tex]

Simplifying this, we get:

[tex]\[ 6x - 5y - 6x - 4y = -27 \][/tex]

[tex]\[ -9y = -27 \][/tex]

3. Solve for \( y \):

Divide both sides by -9:

[tex]\[ y = \frac{-27}{-9} = 3 \][/tex]

So, \( y = 3 \).

4. Substitute \( y = 3 \) into one of the original equations to solve for \( x \):

Let's use the second original equation \( 3x + 2y = 12 \):

[tex]\[ 3x + 2(3) = 12 \][/tex]

Simplify:

[tex]\[ 3x + 6 = 12 \][/tex]

[tex]\[ 3x = 12 - 6 \][/tex]

[tex]\[ 3x = 6 \][/tex]

Divide both sides by 3:

[tex]\[ x = \frac{6}{3} = 2 \][/tex]

So, \( x = 2 \).

Therefore, the solution to the system of equations is:

[tex]\[ x = 2 \quad \text{and} \quad y = 3 \][/tex]