Problem: Solve each equation
We will solve each equation step by step. The goal is to isolate the variable on one side of the equation.
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####
Equation 1: \( y + 2 = 5 \)
1. Subtract 2 from both sides:
\[
y + 2 - 2 = 5 - 2
\]
2. Simplify:
\[
y = 3
\]
Solution: \( y = 3 \)
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####
Equation 2: \( 3 + x = -5 \)
1. Subtract 3 from both sides:
\[
3 + x - 3 = -5 - 3
\]
2. Simplify:
\[
x = -8
\]
Solution: \( x = -8 \)
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####
Equation 3: \( -2 + y = -5 \)
1. Add 2 to both sides:
\[
-2 + y + 2 = -5 + 2
\]
2. Simplify:
\[
y = -3
\]
Solution: \( y = -3 \)
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####
Equation 4: \( y - 4 = 2 \)
1. Add 4 to both sides:
\[
y - 4 + 4 = 2 + 4
\]
2. Simplify:
\[
y = 6
\]
Solution: \( y = 6 \)
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####
Equation 5: \( 3 + y = 3 \)
1. Subtract 3 from both sides:
\[
3 + y - 3 = 3 - 3
\]
2. Simplify:
\[
y = 0
\]
Solution: \( y = 0 \)
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####
Equation 6: \( x - 5 = -3 \)
1. Add 5 to both sides:
\[
x - 5 + 5 = -3 + 5
\]
2. Simplify:
\[
x = 2
\]
Solution: \( x = 2 \)
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####
Equation 7: \( -5 + x = 5 \)
1. Add 5 to both sides:
\[
-5 + x + 5 = 5 + 5
\]
2. Simplify:
\[
x = 10
\]
Solution: \( x = 10 \)
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####
Equation 8: \( x + 3 = -4 \)
1. Subtract 3 from both sides:
\[
x + 3 - 3 = -4 - 3
\]
2. Simplify:
\[
x = -7
\]
Solution: \( x = -7 \)
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####
Equation 9: \( -3 + y = 5 \)
1. Add 3 to both sides:
\[
-3 + y + 3 = 5 + 3
\]
2. Simplify:
\[
y = 8
\]
Solution: \( y = 8 \)
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####
Equation 10: \( y - 5 = -2 \)
1. Add 5 to both sides:
\[
y - 5 + 5 = -2 + 5
\]
2. Simplify:
\[
y = 3
\]
Solution: \( y = 3 \)
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####
Equation 11: \( y - 2 = 4 \)
1. Add 2 to both sides:
\[
y - 2 + 2 = 4 + 2
\]
2. Simplify:
\[
y = 6
\]
Solution: \( y = 6 \)
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####
Equation 12: \( x + 5 = 3 \)
1. Subtract 5 from both sides:
\[
x + 5 - 5 = 3 - 5
\]
2. Simplify:
\[
x = -2
\]
Solution: \( x = -2 \)
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Final Answers:
\[
\boxed{
\begin{aligned}
1. & \ y = 3 \\
2. & \ x = -8 \\
3. & \ y = -3 \\
4. & \ y = 6 \\
5. & \ y = 0 \\
6. & \ x = 2 \\
7. & \ x = 10 \\
8. & \ x = -7 \\
9. & \ y = 8 \\
10. & \ y = 3 \\
11. & \ y = 6 \\
12. & \ x = -2
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of algebra problems.