Answer:
[tex]\textsf{D.} \quad y=\dfrac{2}{3}x+\dfrac{10}{3}[/tex]
Step-by-step explanation:
Given equation:
[tex]y-4=\dfrac{2}{3}(x-1)[/tex]
Distribute the parentheses on the right side of the equation:
[tex]\implies y-4=\dfrac{2}{3}x-\dfrac{2}{3}[/tex]
Add 4 to both sides:
[tex]\implies y-4+4=\dfrac{2}{3}x-\dfrac{2}{3}+4[/tex]
[tex]\implies y=\dfrac{2}{3}x-\dfrac{2}{3}+4[/tex]
Rewrite 4 as 12/3:
[tex]\implies y=\dfrac{2}{3}x-\dfrac{2}{3}+\dfrac{12}{3}[/tex]
[tex]\implies y=\dfrac{2}{3}x+\dfrac{12}{3}-\dfrac{2}{3}[/tex]
[tex]\textsf{Apply the fraction rule} \quad \dfrac{a}{c}-\dfrac{b}{c}=\dfrac{a-b}{c}:[/tex]
[tex]\implies y=\dfrac{2}{3}x+\dfrac{12-2}{3}[/tex]
[tex]\implies y=\dfrac{2}{3}x+\dfrac{10}{3}[/tex]
Therefore, the equation that is equivalent to the given equation is:
[tex]\boxed{y=\dfrac{2}{3}x+\dfrac{10}{3}}[/tex]
