Question 168·Medium·Advanced Math

Define $f(x)$ as the original expression, then type $x_1=0$ and $f(x_1)$. Desmos shows $4.333\ldots$; its fraction button shows $\frac{13}{3}$. At $x=0$, every $x$-term vanishes, so $\frac{13}{3}$ is the original's constant term.

At $x=0$, each choice equals the number it shows without $x$. The choices with $-\frac{13}{3}$ and $\frac53$ don't match $\frac{13}{3}$, so eliminate both. Don't treat the constant from $-\frac23(x-4)$ as negative: its two minus signs make it positive.

Use $x_2=6$, which works neatly with the sixths. Type $f(x_2)$ and the two remaining expressions with $x_2$ in place of $x$. Desmos gives $\frac{46}{3}$ for the original, $\frac{16}{3}$ for $\frac16x+\frac{13}{3}$, and $\frac{46}{3}$ for $\frac{11}{6}x+\frac{13}{3}$ (use the fraction buttons). All three are linear, so matching at both $0$ and $6$ identifies the same line. The equivalent expression is $\frac{11}{6}x+\frac{13}{3}$. Choice D.

The distributive property says the factor outside parentheses multiplies both terms inside. Multiply each term by $\frac56$: $\frac56(3x+2)=\frac{15}{6}x+\frac{10}{6}$. Reduce the fractions:

The minus applies to both terms in $(x-4)$. Distribute $-\frac23$; subtracting a negative makes a positive:

Like terms have the same variable part. Group the $x$-terms together and the numbers together: $\left(\frac52x+\frac53\right)+\left(-\frac23x+\frac83\right)=\left(\frac52-\frac23\right)x+\left(\frac53+\frac83\right)$. Put the $x$-fractions over $6$: $\left(\frac{15-4}{6}\right)x+\frac{5+8}{3}$. Add the numerators: $\frac{11}{6}x+\frac{13}{3}$. That is the equivalent expression. Choice D.