Question

Use elimination to solve the system of equations. Enter your answers in the boxes.

3x – 2y = 24
x + 2y = 48

x =
; y =

Answer 1

To solve the system of equations using elimination, we have the two equations:

1. \( 3x - 2y = 24 \) (Equation 1)
2. \( x + 2y = 48 \) (Equation 2)

We can add the equations together after we manipulate them.

First, let's rearrange Equation 2 to eliminate \( y \). To do this, we can multiply Equation 2 by 1 (to keep it the same) and Equation 1 by 1 (to also keep it the same). However, to eliminate \( y \), we can transform Equation 2.

From Equation 2:

\[
x + 2y = 48
\]

Multiply Equation 2 by 1 (no change):

\[
1(x + 2y = 48) \implies x + 2y = 48
\]

Now, we can add both equations:

**Step 1:** Solve Equation 2 for \( y \):
\[
2y = 48 - x \implies y = \frac{48 - x}{2}
\]
Substituting \( y \) back into Equation 1:
\[
3x - 2\left(\frac{48 - x}{2}\right) = 24
\]
This simplifies:
\[
3x - (48 - x) = 24
\]
\[
3x - 48 + x = 24
\]
Combine like terms:
\[
4x - 48 = 24
\]

**Step 2:** Add 48 to both sides:
\[
4x = 72
\]

**Step 3:** Divide by 4:
\[
x = 18
\]

Now that we have \( x \), substitute it back into Equation 2 to find \( y \):
\[
18 + 2y = 48
\]
Subtract 18 from both sides:
\[
2y = 30
\]

**Step 4:** Divide by 2:
\[
y = 15
\]

Thus, the solution to the system of equations is:
\[
x = 18; \quad y = 15
\]

Final answers:
- \( x = 18 \)
- \( y = 15 \)