SAT Equivalent Expressions Question 216 | Free SAT Prep

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Question 216·Hard·Equivalent Expressions

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For $t\ne -2$ , $t\ne 2$ , and $t\ne -3$ , which expression is equivalent to

\frac{2t+6}{t^{2}-4}-\frac{t-2}{t^{2}+5t+6}?

For rational-expression equivalence questions, first factor all denominators completely to see the basic building blocks. Then form the least common denominator by including each distinct factor at its highest power, rewrite each fraction over that common denominator by multiplying top and bottom by any missing factors, and combine the numerators carefully. Simplify the resulting polynomial in the numerator, keep the denominator factored, and match your final fraction—both numerator and every factor in the denominator—to the answer choices, watching out for options that have the right numerator but are missing one of the original denominator factors.

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Start by factoring

Look at the two quadratic denominators $t^2-4$ and $t^2+5t+6$ . Can you factor each into a product of two binomials?

Identify the common denominator

Once the denominators are factored, list all distinct linear factors that appear. What product of these factors can serve as a common denominator for both fractions?

Rewrite and combine

Rewrite each fraction over the common denominator by multiplying top and bottom by the missing factor. Then subtract the numerators and simplify the resulting polynomial.

Compare to the choices

After simplifying, you should have a single fraction. Check both its numerator and its full factored denominator against the answer choices—make sure every factor from the original denominators is present.

Desmos Guide

Enter the original expression

In Desmos, type the original expression as a function, for example: f(t) = (2t+6)/(t^2-4) - (t-2)/(t^2+5t+6) and press Enter. You can view the graph or create a table of values for $t$ (avoiding $t=-3,-2,2$ ).

Enter each answer choice for comparison

On new lines, enter functions for each option, such as gA(t) = (t^2+16t+14)/((t+2)(t+3)), gB(t) = (t^2+16t+14)/((t-2)(t+3)), gC(t) = (t^2+16t+14)/((t-2)(t+2)), and gD(t) = (t^2+16t+14)/((t-2)(t+2)(t+3)).

Compare values or graphs

Either use tables (by clicking the gear icon and adding a table for each function) or compare the graphs: for several $t$ -values (not equal to $-3,-2,2$ ), see which $g$ -function always has the same values as $f(t)$ or overlaps exactly with its graph. The matching one corresponds to the correct equivalent expression.

Step-by-step Explanation

Factor the denominators

Rewrite each denominator as a product of linear factors.

For $t^2-4$ :

t^2-4=(t-2)(t+2)

For $t^2+5t+6$ :

t^2+5t+6=(t+2)(t+3)

So the expression becomes

\frac{2t+6}{(t-2)(t+2)}-\frac{t-2}{(t+2)(t+3)}.

Find and build the common denominator

The first fraction has denominator $(t-2)(t+2)$ and the second has $(t+2)(t+3)$ , so the common denominator must include all three different factors:

(t-2)(t+2)(t+3).

Rewrite each fraction with this denominator:

Multiply the first fraction (top and bottom) by $(t+3)$ :

\frac{2t+6}{(t-2)(t+2)}=\frac{(2t+6)(t+3)}{(t-2)(t+2)(t+3)}.

Multiply the second fraction (top and bottom) by $(t-2)$ :

\frac{t-2}{(t+2)(t+3)}=\frac{(t-2)^2}{(t-2)(t+2)(t+3)}.

Combine the fractions and simplify the numerator

Now that both fractions have the same denominator, subtract the numerators:

\frac{(2t+6)(t+3)}{(t-2)(t+2)(t+3)}-\frac{(t-2)^2}{(t-2)(t+2)(t+3)} =\frac{(2t+6)(t+3)-(t-2)^2}{(t-2)(t+2)(t+3)}.

Simplify the numerator:

Notice $2t+6=2(t+3)$ , so

(2t+6)(t+3)=2(t+3)^2=2(t^2+6t+9)=2t^2+12t+18.

Also,

(t-2)^2=t^2-4t+4.

Subtract:

2t^2+12t+18-(t^2-4t+4) =2t^2+12t+18-t^2+4t-4 =t^2+16t+14.

So the combined fraction is

\frac{t^2+16t+14}{(t-2)(t+2)(t+3)}.

Match the simplified expression to the answer choices

The simplified result is a single fraction with numerator $t^2+16t+14$ and denominator $(t-2)(t+2)(t+3)$ . This exactly matches choice D, so the equivalent expression is

\frac{t^{2}+16t+14}{(t-2)(t+2)(t+3)}.

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Step-by-step Explanation

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Strategy

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Step-by-step Explanation