Problem Description:
The task is to solve each of the given simple linear equations for the variable. The equations are listed in a grid format, and we need to isolate the variable on one side of the equation.
Solution Approach:
To solve each equation, we will use basic algebraic principles:
1.
Isolate the variable: Move all terms involving the variable to one side of the equation and constants to the other.
2.
Simplify: Divide or multiply both sides of the equation by the coefficient of the variable to solve for it.
Let's solve each equation step by step.
---
Equation 1: \( 6z = -54 \)
- Divide both sides by 6:
\[
z = \frac{-54}{6} = -9
\]
-
Solution: \( z = -9 \)
---
Equation 2: \( 6a = 30 \)
- Divide both sides by 6:
\[
a = \frac{30}{6} = 5
\]
-
Solution: \( a = 5 \)
---
Equation 3: \( 9x = -90 \)
- Divide both sides by 9:
\[
x = \frac{-90}{9} = -10
\]
-
Solution: \( x = -10 \)
---
Equation 4: \( 2a = -14 \)
- Divide both sides by 2:
\[
a = \frac{-14}{2} = -7
\]
-
Solution: \( a = -7 \)
---
Equation 5: \( 7y = -7 \)
- Divide both sides by 7:
\[
y = \frac{-7}{7} = -1
\]
-
Solution: \( y = -1 \)
---
Equation 6: \( 5c = -50 \)
- Divide both sides by 5:
\[
c = \frac{-50}{5} = -10
\]
-
Solution: \( c = -10 \)
---
Equation 7: \( 8b = 8 \)
- Divide both sides by 8:
\[
b = \frac{8}{8} = 1
\]
-
Solution: \( b = 1 \)
---
Equation 8: \( -4a = -20 \)
- Divide both sides by -4:
\[
a = \frac{-20}{-4} = 5
\]
-
Solution: \( a = 5 \)
---
Equation 9: \( -5c = 40 \)
- Divide both sides by -5:
\[
c = \frac{40}{-5} = -8
\]
-
Solution: \( c = -8 \)
---
Equation 10: \( -8b = 72 \)
- Divide both sides by -8:
\[
b = \frac{72}{-8} = -9
\]
-
Solution: \( b = -9 \)
---
Equation 11: \( -2x = -2 \)
- Divide both sides by -2:
\[
x = \frac{-2}{-2} = 1
\]
-
Solution: \( x = 1 \)
---
Equation 12: \( 5y = -40 \)
- Divide both sides by 5:
\[
y = \frac{-40}{5} = -8
\]
-
Solution: \( y = -8 \)
---
Equation 13: \( 9b = -27 \)
- Divide both sides by 9:
\[
b = \frac{-27}{9} = -3
\]
-
Solution: \( b = -3 \)
---
Equation 14: \( 7v = 35 \)
- Divide both sides by 7:
\[
v = \frac{35}{7} = 5
\]
-
Solution: \( v = 5 \)
---
Equation 15: \( -9z = 45 \)
- Divide both sides by -9:
\[
z = \frac{45}{-9} = -5
\]
-
Solution: \( z = -5 \)
---
Final Answers:
\[
\boxed{
\begin{aligned}
1. & \quad z = -9 \\
2. & \quad a = 5 \\
3. & \quad x = -10 \\
4. & \quad a = -7 \\
5. & \quad y = -1 \\
6. & \quad c = -10 \\
7. & \quad b = 1 \\
8. & \quad a = 5 \\
9. & \quad c = -8 \\
10. & \quad b = -9 \\
11. & \quad x = 1 \\
12. & \quad y = -8 \\
13. & \quad b = -3 \\
14. & \quad v = 5 \\
15. & \quad z = -5 \\
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of algebra 1 solving equations worksheet answers.
