Solving Systems of Equations: Any Method | Interactive Worksheet ... - Free Printable

The task involves solving systems of equations using any method. Here, I will solve a few of the systems as examples and explain the solution process step by step. You can apply similar methods to solve the rest.

Problem 1:

$$
\begin{aligned}
3x + y &= -8 \\
-6x + y &= 2
\end{aligned}
$$

#### Solution:
We can use the elimination method here.

1. Write the equations:
$$
\begin{aligned}
3x + y &= -8 \quad \text{(Equation 1)} \\
-6x + y &= 2 \quad \text{(Equation 2)}
\end{aligned}
$$

2. Subtract Equation 2 from Equation 1 to eliminate \( y \):
$$
(3x + y) - (-6x + y) = -8 - 2
$$
Simplify:
$$
3x + y + 6x - y = -10 \implies 9x = -10 \implies x = -\frac{10}{9}
$$

3. Substitute \( x = -\frac{10}{9} \) into Equation 1 to solve for \( y \):
$$
3\left(-\frac{10}{9}\right) + y = -8
$$
Simplify:
$$
-\frac{30}{9} + y = -8 \implies -\frac{10}{3} + y = -8
$$
Convert \(-8\) to a fraction with a denominator of 3:
$$
-\frac{10}{3} + y = -\frac{24}{3} \implies y = -\frac{24}{3} + \frac{10}{3} \implies y = -\frac{14}{3}
$$

4. The solution is:
$$
\boxed{\left( -\frac{10}{9}, -\frac{14}{3} \right)}
$$

---

Problem 4:

$$
\begin{aligned}
x + 4y &= -2 \\
3x + 2y &= -6
\end{aligned}
$$

#### Solution:
We can use the substitution method here.

1. Solve Equation 1 for \( x \):
$$
x + 4y = -2 \implies x = -2 - 4y
$$

2. Substitute \( x = -2 - 4y \) into Equation 2:
$$
3(-2 - 4y) + 2y = -6
$$
Simplify:
$$
-6 - 12y + 2y = -6 \implies -6 - 10y = -6
$$
Add 6 to both sides:
$$
-10y = 0 \implies y = 0
$$

3. Substitute \( y = 0 \) back into \( x = -2 - 4y \):
$$
x = -2 - 4(0) \implies x = -2
$$

4. The solution is:
$$
\boxed{(-2, 0)}
$$

---

Problem 7:

$$
\begin{aligned}
5x - 2y &= 14 \\
3x - 4y &= -24
\end{aligned}
$$

#### Solution:
We can use the elimination method here.

1. Write the equations:
$$
\begin{aligned}
5x - 2y &= 14 \quad \text{(Equation 1)} \\
3x - 4y &= -24 \quad \text{(Equation 2)}
\end{aligned}
$$

2. To eliminate \( y \), multiply Equation 1 by 2:
$$
2(5x - 2y) = 2(14) \implies 10x - 4y = 28 \quad \text{(New Equation 1)}
$$

3. Subtract Equation 2 from the new Equation 1:
$$
(10x - 4y) - (3x - 4y) = 28 - (-24)
$$
Simplify:
$$
10x - 4y - 3x + 4y = 28 + 24 \implies 7x = 52 \implies x = \frac{52}{7}
$$

4. Substitute \( x = \frac{52}{7} \) into Equation 1 to solve for \( y \):
$$
5\left(\frac{52}{7}\right) - 2y = 14
$$
Simplify:
$$
\frac{260}{7} - 2y = 14
$$
Convert 14 to a fraction with a denominator of 7:
$$
\frac{260}{7} - 2y = \frac{98}{7} \implies -2y = \frac{98}{7} - \frac{260}{7} \implies -2y = -\frac{162}{7}
$$
Divide by \(-2\):
$$
y = \frac{162}{7} \cdot \frac{1}{2} \implies y = \frac{81}{7}
$$

5. The solution is:
$$
\boxed{\left( \frac{52}{7}, \frac{81}{7} \right)}
$$

---

Problem 10:

$$
\begin{aligned}
y &= x + 12 \\
y &= 4x + 2
\end{aligned}
$$

#### Solution:
We can use the substitution method here.

1. Since both equations are solved for \( y \), set them equal to each other:
$$
x + 12 = 4x + 2
$$

2. Solve for \( x \):
$$
x + 12 = 4x + 2 \implies 12 - 2 = 4x - x \implies 10 = 3x \implies x = \frac{10}{3}
$$

3. Substitute \( x = \frac{10}{3} \) into \( y = x + 12 \):
$$
y = \frac{10}{3} + 12
$$
Convert 12 to a fraction with a denominator of 3:
$$
y = \frac{10}{3} + \frac{36}{3} \implies y = \frac{46}{3}
$$

4. The solution is:
$$
\boxed{\left( \frac{10}{3}, \frac{46}{3} \right)}
$$

---

Final Answers:

1. \(\boxed{\left( -\frac{10}{9}, -\frac{14}{3} \right)}\)
4. \(\boxed{(-2, 0)}\)
7. \(\boxed{\left( \frac{52}{7}, \frac{81}{7} \right)}\)
10. \(\boxed{\left( \frac{10}{3}, \frac{46}{3} \right)}\)

You can apply similar methods to solve the remaining problems. If you need further assistance, feel free to ask!