Both methods, Tonya's and Deon's, will result in equivalent equations, but let's break down each method step-by-step to see how they arrive at their solutions.
### Tonya's Method: Using the Distributive Property
Tonya's first step is to apply the Distributive Property to the left side of the equation [tex]\(\frac{1}{2}(2y + 4) = -6\)[/tex].
1. Distribute the [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[
\frac{1}{2} \cdot 2y + \frac{1}{2} \cdot 4
\][/tex]
2. Simplify each term:
[tex]\[
\frac{1}{2} \cdot 2y = y
\][/tex]
[tex]\[
\frac{1}{2} \cdot 4 = 2
\][/tex]
3. Combine the simplified terms:
[tex]\[
y + 2 = -6
\][/tex]
Thus, Tonya's method results in the equation:
[tex]\[
y + 2 = -6
\][/tex]
### Deon's Method: Multiplying Each Side by 2
Deon's first step is to multiply each side of the original equation by 2 to eliminate the fraction:
[tex]\[
\frac{1}{2}(2y + 4) = -6
\][/tex]
1. Multiply both sides by 2:
[tex]\[
2 \cdot \frac{1}{2}(2y + 4) = 2 \cdot -6
\][/tex]
2. Simplify both sides:
[tex]\[
2y + 4 = -12
\][/tex]
Thus, Deon's method results in the equation:
[tex]\[
2y + 4 = -12
\][/tex]
### Comparing the Two Methods
Both methods yield equations that are equivalent and can be solved to find the same value for [tex]\(y\)[/tex].
- Tonya's equation: [tex]\(y + 2 = -6\)[/tex]
- Deon's equation: [tex]\(2y + 4 = -12\)[/tex]
We can solve each equation step-by-step to check that they lead to the same solution.
#### Solving Tonya's Equation:
1. Subtract 2 from both sides:
[tex]\[
y + 2 - 2 = -6 - 2
\][/tex]
[tex]\[
y = -8
\][/tex]
#### Solving Deon's Equation:
1. Subtract 4 from both sides:
[tex]\[
2y + 4 - 4 = -12 - 4
\][/tex]
[tex]\[
2y = -16
\][/tex]
2. Divide by 2:
[tex]\[
\frac{2y}{2} = \frac{-16}{2}
\][/tex]
[tex]\[
y = -8
\][/tex]
### Conclusion
Both Tonya's method and Deon's method result in equivalent equations, and both methods ultimately lead to the same solution for [tex]\(y\)[/tex]. Therefore, either method can be used to solve the original equation effectively.
