Let's solve the given system of equations using the elimination method.
The system of equations is:
[tex]\[
\begin{array}{l}
3y = 20 - 4x \quad \text{(1)} \\
2y = 12 - 3x \quad \text{(2)}
\end{array}
\][/tex]
First, we rewrite both equations in standard form [tex]\( Ax + By = C \)[/tex].
For equation (1):
[tex]\[
3y = 20 - 4x \\
4x + 3y = 20 \quad \text{(3)}
\][/tex]
For equation (2):
[tex]\[
2y = 12 - 3x \\
3x + 2y = 12 \quad \text{(4)}
\][/tex]
Next, we aim to eliminate one of the variables. Here, we will eliminate [tex]\(y\)[/tex] by making the coefficients of [tex]\(y\)[/tex] in both equations equal.
To make the coefficients of [tex]\(y\)[/tex] the same, we need to find a common multiple of the coefficients of [tex]\(y\)[/tex] (which are 3 and 2). The least common multiple of 3 and 2 is 6.
Multiply equation (3) by 2:
[tex]\[
2(4x + 3y) = 2(20) \\
8x + 6y = 40 \quad \text{(5)}
\][/tex]
Multiply equation (4) by 3:
[tex]\[
3(3x + 2y) = 3(12) \\
9x + 6y = 36 \quad \text{(6)}
\][/tex]
Now, subtract equation (5) from equation (6) to eliminate [tex]\(y\)[/tex]:
[tex]\[
(9x + 6y) - (8x + 6y) = 36 - 40 \\
9x + 6y - 8x - 6y = -4 \\
x = -4
\][/tex]
With [tex]\(x\)[/tex] determined, substitute [tex]\(x = -4\)[/tex] back into equation (3) to find [tex]\(y\)[/tex]:
[tex]\[
4(-4) + 3y = 20 \\
-16 + 3y = 20 \\
3y = 20 + 16 \\
3y = 36 \\
y = 12
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
(x, y) = (-4, 12)
\][/tex]
