Certainly! Let's solve the system of linear equations using the elimination method step by step:
Given equations are:
[tex]\[
\begin{array}{r}
-3x + 7y = 12 \quad \text{(1)} \\
x - 2y = -3 \quad \text{(2)}
\end{array}
\][/tex]
Step 1: Make the coefficients of [tex]\(x\)[/tex] in both equations the same.
To do this, we need to multiply equation (2) by 3 so that the coefficient of [tex]\(x\)[/tex] will be -3, matching the coefficient of [tex]\(x\)[/tex] in equation (1).
Multiply equation (2) by 3:
[tex]\[ 3(x - 2y) = 3(-3) \][/tex]
[tex]\[ 3x - 6y = -9 \quad \text{(3)} \][/tex]
Now our system of equations looks like this:
[tex]\[
\begin{array}{r}
-3x + 7y = 12 \quad \text{(1)} \\
3x - 6y = -9 \quad \text{(3)}
\end{array}
\][/tex]
Step 2: Add equation (1) and equation (3) to eliminate [tex]\(x\)[/tex].
[tex]\[
(-3x + 7y) + (3x - 6y) = 12 + (-9)
\][/tex]
[tex]\[
-3x + 3x + 7y - 6y = 3
\][/tex]
[tex]\[
y = 3
\][/tex]
So, we have [tex]\( y = 3 \)[/tex].
Step 3: Substitute [tex]\( y = 3 \)[/tex] back into one of the original equations to solve for [tex]\( x \)[/tex].
Let's use equation (2):
[tex]\[
x - 2y = -3
\][/tex]
Substitute [tex]\( y = 3 \)[/tex]:
[tex]\[
x - 2(3) = -3
\][/tex]
[tex]\[
x - 6 = -3
\][/tex]
[tex]\[
x = 3
\][/tex]
So, the solution to the system of equations is:
[tex]\[
x = 3, \quad y = 3
\][/tex]
Therefore, the solution is:
[tex]\[
(x, y) = (3, 3)
\][/tex]
