To solve the problem, we need to match each system of linear equations with its correct ordered pair solution. We will solve each system step by step.
System 1:
\[
\begin{cases}
2x + y = 7 \\
3x - y = 3
\end{cases}
\]
Step 1: Add the two equations to eliminate \( y \):
\[
(2x + y) + (3x - y) = 7 + 3
\]
\[
5x = 10
\]
\[
x = 2
\]
Step 2: Substitute \( x = 2 \) into the first equation to solve for \( y \):
\[
2(2) + y = 7
\]
\[
4 + y = 7
\]
\[
y = 3
\]
Solution: \((2, 3)\)
System 2:
\[
\begin{cases}
2x + 3y = 8 \\
3x + y = 5
\end{cases}
\]
Step 1: Multiply the second equation by 3 to align the coefficients of \( y \):
\[
3(3x + y) = 3(5)
\]
\[
9x + 3y = 15
\]
Step 2: Subtract the first equation from this new equation to eliminate \( y \):
\[
(9x + 3y) - (2x + 3y) = 15 - 8
\]
\[
7x = 7
\]
\[
x = 1
\]
Step 3: Substitute \( x = 1 \) into the second equation to solve for \( y \):
\[
3(1) + y = 5
\]
\[
3 + y = 5
\]
\[
y = 2
\]
Solution: \((1, 2)\)
System 3:
\[
\begin{cases}
5x + y = 15 \\
3x + 2y = 9
\end{cases}
\]
Step 1: Multiply the first equation by 2 to align the coefficients of \( y \):
\[
2(5x + y) = 2(15)
\]
\[
10x + 2y = 30
\]
Step 2: Subtract the second equation from this new equation to eliminate \( y \):
\[
(10x + 2y) - (3x + 2y) = 30 - 9
\]
\[
7x = 21
\]
\[
x = 3
\]
Step 3: Substitute \( x = 3 \) into the first equation to solve for \( y \):
\[
5(3) + y = 15
\]
\[
15 + y = 15
\]
\[
y = 0
\]
Solution: \((3, 0)\)
System 4:
\[
\begin{cases}
2x - 3y = 0 \\
-2x + 2y = -2
\end{cases}
\]
Step 1: Add the two equations to eliminate \( x \):
\[
(2x - 3y) + (-2x + 2y) = 0 + (-2)
\]
\[
-y = -2
\]
\[
y = 2
\]
Step 2: Substitute \( y = 2 \) into the first equation to solve for \( x \):
\[
2x - 3(2) = 0
\]
\[
2x - 6 = 0
\]
\[
2x = 6
\]
\[
x = 3
\]
Solution: \((3, 2)\)
Final Matching:
- \(\begin{cases} 2x + y = 7 \\ 3x - y = 3 \end{cases}\) → \((2, 3)\)
- \(\begin{cases} 2x + 3y = 8 \\ 3x + y = 5 \end{cases}\) → \((1, 2)\)
- \(\begin{cases} 5x + y = 15 \\ 3x + 2y = 9 \end{cases}\) → \((3, 0)\)
- \(\begin{cases} 2x - 3y = 0 \\ -2x + 2y = -2 \end{cases}\) → \((3, 2)\)
Thus, the final answer is:
\[
\boxed{(2, 3), (1, 2), (3, 0), (3, 2)}
\]
Parent Tip: Review the logic above to help your child master the concept of solving systems using elimination worksheet.
