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]
