Class work worksheet with nine systems of equations to solve using substitution.

To solve the given systems of equations using the substitution method, we will follow these steps:

1. Identify one equation where a variable is already isolated or can be easily isolated.
2. Substitute the expression for that variable into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found back into one of the original equations to solve for the other variable.

Let's solve each system step by step.

---

Problem 1:

\[
\begin{cases}
x = -2 \\
x + y = 5
\end{cases}
\]

- From the first equation, \( x = -2 \).
- Substitute \( x = -2 \) into the second equation:
\[
-2 + y = 5
\]
- Solve for \( y \):
\[
y = 5 + 2 = 7
\]

Solution: \((x, y) = (-2, 7)\)

---

Problem 2:

\[
\begin{cases}
x = 5 \\
2x - 3y = 13
\end{cases}
\]

- From the first equation, \( x = 5 \).
- Substitute \( x = 5 \) into the second equation:
\[
2(5) - 3y = 13
\]
- Simplify:
\[
10 - 3y = 13
\]
- Solve for \( y \):
\[
-3y = 13 - 10 \implies -3y = 3 \implies y = -1
\]

Solution: \((x, y) = (5, -1)\)

---

Problem 3:

\[
\begin{cases}
y = 2x \\
-3x + 5y = -21
\end{cases}
\]

- From the first equation, \( y = 2x \).
- Substitute \( y = 2x \) into the second equation:
\[
-3x + 5(2x) = -21
\]
- Simplify:
\[
-3x + 10x = -21 \implies 7x = -21
\]
- Solve for \( x \):
\[
x = -3
\]
- Substitute \( x = -3 \) back into \( y = 2x \):
\[
y = 2(-3) = -6
\]

Solution: \((x, y) = (-3, -6)\)

---

Problem 4:

\[
\begin{cases}
y = 2x - 3 \\
-2x + 3y = 7
\end{cases}
\]

- From the first equation, \( y = 2x - 3 \).
- Substitute \( y = 2x - 3 \) into the second equation:
\[
-2x + 3(2x - 3) = 7
\]
- Simplify:
\[
-2x + 6x - 9 = 7 \implies 4x - 9 = 7
\]
- Solve for \( x \):
\[
4x = 7 + 9 \implies 4x = 16 \implies x = 4
\]
- Substitute \( x = 4 \) back into \( y = 2x - 3 \):
\[
y = 2(4) - 3 = 8 - 3 = 5
\]

Solution: \((x, y) = (4, 5)\)

---

Problem 5:

\[
\begin{cases}
x = -3y - 4 \\
2x + 5y = -6
\end{cases}
\]

- From the first equation, \( x = -3y - 4 \).
- Substitute \( x = -3y - 4 \) into the second equation:
\[
2(-3y - 4) + 5y = -6
\]
- Simplify:
\[
-6y - 8 + 5y = -6 \implies -y - 8 = -6
\]
- Solve for \( y \):
\[
-y = -6 + 8 \implies -y = 2 \implies y = -2
\]
- Substitute \( y = -2 \) back into \( x = -3y - 4 \):
\[
x = -3(-2) - 4 = 6 - 4 = 2
\]

Solution: \((x, y) = (2, -2)\)

---

Problem 6:

\[
\begin{cases}
x = -2y + 1 \\
-4x - 7y = -1
\end{cases}
\]

- From the first equation, \( x = -2y + 1 \).
- Substitute \( x = -2y + 1 \) into the second equation:
\[
-4(-2y + 1) - 7y = -1
\]
- Simplify:
\[
8y - 4 - 7y = -1 \implies y - 4 = -1
\]
- Solve for \( y \):
\[
y = -1 + 4 \implies y = 3
\]
- Substitute \( y = 3 \) back into \( x = -2y + 1 \):
\[
x = -2(3) + 1 = -6 + 1 = -5
\]

Solution: \((x, y) = (-5, 3)\)

---

Problem 7:

\[
\begin{cases}
x + 5y = 13 \\
2x - 4y = -16
\end{cases}
\]

- From the first equation, solve for \( x \):
\[
x = 13 - 5y
\]
- Substitute \( x = 13 - 5y \) into the second equation:
\[
2(13 - 5y) - 4y = -16
\]
- Simplify:
\[
26 - 10y - 4y = -16 \implies 26 - 14y = -16
\]
- Solve for \( y \):
\[
-14y = -16 - 26 \implies -14y = -42 \implies y = 3
\]
- Substitute \( y = 3 \) back into \( x = 13 - 5y \):
\[
x = 13 - 5(3) = 13 - 15 = -2
\]

Solution: \((x, y) = (-2, 3)\)

---

Problem 8:

\[
\begin{cases}
4x - 3y = 9 \\
-2x + y = -5
\end{cases}
\]

- From the second equation, solve for \( y \):
\[
y = -5 + 2x
\]
- Substitute \( y = -5 + 2x \) into the first equation:
\[
4x - 3(-5 + 2x) = 9
\]
- Simplify:
\[
4x + 15 - 6x = 9 \implies -2x + 15 = 9
\]
- Solve for \( x \):
\[
-2x = 9 - 15 \implies -2x = -6 \implies x = 3
\]
- Substitute \( x = 3 \) back into \( y = -5 + 2x \):
\[
y = -5 + 2(3) = -5 + 6 = 1
\]

Solution: \((x, y) = (3, 1)\)

---

Problem 9:

\[
\begin{cases}
-2x - 6y = 2 \\
x - 4y = -15
\end{cases}
\]

- From the second equation, solve for \( x \):
\[
x = -15 + 4y
\]
- Substitute \( x = -15 + 4y \) into the first equation:
\[
-2(-15 + 4y) - 6y = 2
\]
- Simplify:
\[
30 - 8y - 6y = 2 \implies 30 - 14y = 2
\]
- Solve for \( y \):
\[
-14y = 2 - 30 \implies -14y = -28 \implies y = 2
\]
- Substitute \( y = 2 \) back into \( x = -15 + 4y \):
\[
x = -15 + 4(2) = -15 + 8 = -7
\]

Solution: \((x, y) = (-7, 2)\)

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ (x, y) = (-2, 7) \\
2. & \ (x, y) = (5, -1) \\
3. & \ (x, y) = (-3, -6) \\
4. & \ (x, y) = (4, 5) \\
5. & \ (x, y) = (2, -2) \\
6. & \ (x, y) = (-5, 3) \\
7. & \ (x, y) = (-2, 3) \\
8. & \ (x, y) = (3, 1) \\
9. & \ (x, y) = (-7, 2)
\end{aligned}
}
\]