To solve the system of equations [tex]\(2x + 3y = 40\)[/tex] and [tex]\(-2x + 2y = 20\)[/tex], we'll use the substitution or elimination method. Here's the step-by-step process:
1. Write down the given equations:
[tex]\[
2x + 3y = 40 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
-2x + 2y = 20 \quad \text{(Equation 2)}
\][/tex]
2. Add the two equations together to eliminate [tex]\(x\)[/tex]:
[tex]\[
(2x + 3y) + (-2x + 2y) = 40 + 20
\][/tex]
Simplifying this, we get:
[tex]\[
2x - 2x + 3y + 2y = 60
\][/tex]
[tex]\[
0x + 5y = 60
\][/tex]
[tex]\[
5y = 60
\][/tex]
3. Solve for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{60}{5}
\][/tex]
[tex]\[
y = 12
\][/tex]
4. Substitute [tex]\(y = 12\)[/tex] back into one of the original equations to solve for [tex]\(x\)[/tex]:
Using Equation 1:
[tex]\[
2x + 3(12) = 40
\][/tex]
[tex]\[
2x + 36 = 40
\][/tex]
[tex]\[
2x = 40 - 36
\][/tex]
[tex]\[
2x = 4
\][/tex]
[tex]\[
x = \frac{4}{2}
\][/tex]
[tex]\[
x = 2
\][/tex]
5. Verifying the solution with the second equation:
Substituting [tex]\(x = 2\)[/tex] and [tex]\(y = 12\)[/tex] into Equation 2:
[tex]\[
-2(2) + 2(12) = 20
\][/tex]
[tex]\[
-4 + 24 = 20
\][/tex]
[tex]\[
20 = 20 \quad \text{(True)}
\][/tex]
Therefore, the solution to the system of equations is [tex]\(x = 2\)[/tex] and [tex]\(y = 12\)[/tex].
The best answer is:
C. [tex]\(x=2, y=12\)[/tex]
