Question

select all the equations that are equivalent to $y - 4 = \frac{2}{3}\left(x - 1\
ight)$.
a. $y = \frac{2}{3}x - \frac{10}{3}$
b. $y = \frac{2}{3}x + 3\frac{1}{3}$
c. $y = \frac{2}{3}x + 3$
d. $y = \frac{2}{3}x + \frac{10}{3}$
e. $y = \frac{2}{3}x - 3\frac{1}{3}$

Explanation:

Step1: Expand the given equation

We start with the equation \( y - 4=\frac{2}{3}(x - 1) \). Using the distributive property \( a(b - c)=ab - ac \), we get \( y-4=\frac{2}{3}x-\frac{2}{3} \).

Step2: Solve for y

To isolate \( y \), we add 4 to both sides of the equation. We know that \( 4=\frac{12}{3} \), so \( y=\frac{2}{3}x-\frac{2}{3}+\frac{12}{3} \).

Step3: Simplify the constant terms

Combining the constant terms, we have \( y=\frac{2}{3}x+\frac{10}{3} \). Now, we can convert \( \frac{10}{3} \) to a mixed number: \( \frac{10}{3}=3\frac{1}{3} \), and also \( -\frac{10}{3}=- 3\frac{1}{3} \) (but we have a plus sign here). Wait, let's re - check the expansion. Wait, no, when we add 4 (which is \( \frac{12}{3} \)) to \( -\frac{2}{3} \), we get \( \frac{- 2 + 12}{3}=\frac{10}{3}=3\frac{1}{3} \). Wait, but let's check the original expansion again. The original equation is \( y - 4=\frac{2}{3}(x - 1)=\frac{2}{3}x-\frac{2}{3} \). Then \( y=\frac{2}{3}x-\frac{2}{3}+4 \). \( 4=\frac{12}{3} \), so \( y=\frac{2}{3}x+\frac{-2 + 12}{3}=\frac{2}{3}x+\frac{10}{3} \). Now, \( \frac{10}{3}=3\frac{1}{3} \), and \( -\frac{10}{3}=-3\frac{1}{3} \). Wait, let's check option B: \( y = \frac{2}{3}x+3\frac{1}{3} \) (since \( 3\frac{1}{3}=\frac{10}{3} \)) and option D: \( y=\frac{2}{3}x+\frac{10}{3} \) (which is the same as \( y=\frac{2}{3}x + 3\frac{1}{3} \)). Wait, did I make a mistake earlier? Wait, no, let's re - do the expansion.

Wait, the original equation is \( y-4=\frac{2}{3}(x - 1) \). Distribute the \( \frac{2}{3} \): \( y-4=\frac{2}{3}x-\frac{2}{3} \). Then add 4 to both sides: \( y=\frac{2}{3}x-\frac{2}{3}+4 \). \( 4=\frac{12}{3} \), so \( y=\frac{2}{3}x+\frac{-2 + 12}{3}=\frac{2}{3}x+\frac{10}{3} \). \( \frac{10}{3}=3\frac{1}{3} \), so \( y=\frac{2}{3}x + 3\frac{1}{3} \) (option B) and \( y=\frac{2}{3}x+\frac{10}{3} \) (option D) are equivalent. Wait, but also, if we consider a sign error in the original problem? Wait, no, let's check the other way. Let's take the equation \( y - 4=\frac{2}{3}(x - 1) \). If we expand it correctly, \( y=\frac{2}{3}x-\frac{2}{3}+4=\frac{2}{3}x+\frac{10}{3} \). So \( \frac{10}{3}=3\frac{1}{3} \), so option B: \( y=\frac{2}{3}x + 3\frac{1}{3} \) and option D: \( y=\frac{2}{3}x+\frac{10}{3} \) are equivalent to the original equation. Wait, but let's check option A: \( y=\frac{2}{3}x-\frac{10}{3} \), option E: \( y=\frac{2}{3}x-3\frac{1}{3} \) (since \( 3\frac{1}{3}=\frac{10}{3} \)) are incorrect. Option C: \( y=\frac{2}{3}x + 3 \) is incorrect because \( 3=\frac{9}{3} eq\frac{10}{3} \).

Wait, I think I made a mistake in the sign when I first thought about the constant term. Let's re - derive:

Starting with \( y-4=\frac{2}{3}(x - 1) \)

1. Distribute \( \frac{2}{3} \) over \( (x - 1) \):
\( y-4=\frac{2}{3}x-\frac{2}{3} \)

2. Add 4 to both sides:
\( y=\frac{2}{3}x-\frac{2}{3}+4 \)

3. Convert 4 to thirds: \( 4=\frac{12}{3} \)
\( y=\frac{2}{3}x-\frac{2}{3}+\frac{12}{3} \)

4. Combine the constant terms:
\( y=\frac{2}{3}x+\frac{-2 + 12}{3}=\frac{2}{3}x+\frac{10}{3} \)

Now, \( \frac{10}{3}=3\frac{1}{3} \), so:

- Option B: \( y=\frac{2}{3}x + 3\frac{1}{3} \) (since \( 3\frac{1}{3}=\frac{10}{3} \)) is equivalent.
- Option D: \( y=\frac{2}{3}x+\frac{10}{3} \) is equivalent (same as option B, just in improper fraction form).

Wait, but let's check the other options:

Option A: \( y=\frac{2}{3}x-\frac{10}{3} \). The constant term here is \( -\frac{10}{3} \), which is not equal to \( \frac{10}{3} \), so it's not equivalent.

Option C: \( y=\frac{2}{3}x + 3 \). \( 3=\frac{9}{3} eq\frac{10}{3} \), so not equivalent.

Option E: \( y=\frac{2}{3}x-3\frac{1}{3} \). \( - 3\frac{1}{3}=-\frac{10}{3} eq\frac{10}{3} \), so not equivalent.

Answer:

B. \( y=\frac{2}{3}x + 3\frac{1}{3} \), D. \( y=\frac{2}{3}x+\frac{10}{3} \)