Simple Algebra worksheet for solving basic equations to find the values of x and y.

To solve the given algebraic equations and find the values of \( x \) and \( y \), we will solve each equation step by step. Let's go through each equation one by one.

Equation 1: \( x + 4 = 6 \)

1. Subtract 4 from both sides:
\[
x + 4 - 4 = 6 - 4
\]
\[
x = 2
\]

Equation 2: \( y - 4 = 8 \)

1. Add 4 to both sides:
\[
y - 4 + 4 = 8 + 4
\]
\[
y = 12
\]

Equation 3: \( y + 5 = 12 \)

1. Subtract 5 from both sides:
\[
y + 5 - 5 = 12 - 5
\]
\[
y = 7
\]

Equation 4: \( x - 4 = 3 \)

1. Add 4 to both sides:
\[
x - 4 + 4 = 3 + 4
\]
\[
x = 7
\]

Equation 5: \( x + 0 = 11 \)

1. Simplify (adding 0 does not change the value):
\[
x = 11
\]

Equation 6: \( y - 3 = 9 \)

1. Add 3 to both sides:
\[
y - 3 + 3 = 9 + 3
\]
\[
y = 12
\]

Equation 7: \( y + 3 = 15 \)

1. Subtract 3 from both sides:
\[
y + 3 - 3 = 15 - 3
\]
\[
y = 12
\]

Equation 8: \( x - 7 = 10 \)

1. Add 7 to both sides:
\[
x - 7 + 7 = 10 + 7
\]
\[
x = 17
\]

Equation 9: \( x + 6 = 24 \)

1. Subtract 6 from both sides:
\[
x + 6 - 6 = 24 - 6
\]
\[
x = 18
\]

Equation 10: \( y - 10 = 15 \)

1. Add 10 to both sides:
\[
y - 10 + 10 = 15 + 10
\]
\[
y = 25
\]

Summary of Solutions:

- From Equation 1: \( x = 2 \)
- From Equation 2: \( y = 12 \)
- From Equation 3: \( y = 7 \)
- From Equation 4: \( x = 7 \)
- From Equation 5: \( x = 11 \)
- From Equation 6: \( y = 12 \)
- From Equation 7: \( y = 12 \)
- From Equation 8: \( x = 17 \)
- From Equation 9: \( x = 18 \)
- From Equation 10: \( y = 25 \)

Each equation provides a different value for \( x \) or \( y \). Since the problem asks to find the values of \( x \) and \( y \), and there are multiple equations, we can conclude that the values depend on the specific equation being solved. However, if we are looking for consistent solutions across all equations, we notice that some values repeat (e.g., \( y = 12 \) in multiple equations).

Final Answer:

\[
\boxed{x = 2, y = 12}
\]
(Note: This is based on the first occurrence of consistent values. Other equations provide different values, so the solution depends on the context or specific equation being considered.)