Plugging in numbers and working backward from the choices

There are two ways to use the answer choices. Plugging in means you choose your own numbers for the letters. Working backward means you test the numbers the choices give you.

Read the last line first and write down what it asks for. Then look at the choices before you write any algebra.

A short equation is quicker to solve. Testing four choices in $4x+9=89$ takes longer than this:

1. Subtract 9 from both sides.

   $$4x=80$$

2. Divide both sides by 4.

   $$x=20$$

A grid-inA question where you type your own answer instead of picking one of four.Read the lesson has no choices, so you can't work backward. Plugging in still works there when the answer doesn't depend on the number you pick, as in a percent question with no starting amount. Otherwise, solve it and put your answer back into the question to check it. The grid-in lesson covers typing the answer in.

Here's a question with letters in the choices.

“Mia reads $p$ pages a night for $d$ nights. Then she reads 5 more pages a night than that for another $d$ nights. Which expression gives the total number of pages she reads?”

• A. $2pd+5$
• B. $2pd+5p$
• C. $pd+5d$
• D. $2pd+5d$

1. Pick a different number for each letter. Skip 0, 1 and the 5 from the question.

   $$p=4,\ d=3$$

2. First stretch, from the question's words: 4 pages a night for 3 nights.

   $$4\times3=12$$

3. Second stretch: 9 pages a night for 3 more nights.

   $$(4+5)\times3=27$$

4. The question asks for total pages. Write down 39 pages.

   $$12+27=39$$

Now put $p=4$ and $d=3$ into every choice.

Only D gives 39, so D is the answer. Check all four even after you find a match. If a second choice also gave 39, you'd need new numbers.

C gives 27, the pages from the second stretch alone. A choice can land on a number from partway through your work, so write down what your target counts and match only that.

Small numbers keep the arithmetic quick, but some of them make different choices give the same value. Here's what happens in Mia's question:

Zero wipes out every term it multiplies, and 1 makes $p$, $p^2$ and $p^3$ all equal. Using one number for both letters hides a choice like B, which puts $p$ where $d$ belongs.

Skip numbers the question already uses. For $(x+3)(x-3)$, $x=3$ makes the right choice, $x^2-9$, and the wrong one, $x^2-6x+9$, both 0. At $x=4$ they give 7 and 1.

Stay inside any condition the question gives. If it says $2\lt k\lt4$, use $k=3$. For $x\gt9$, use 10 or 12. Use a whole number if the question needs one, and never a number that makes a denominatorThe bottom number of a fraction, like the 4 in 3/4.Read the lesson zero.

When a formula ties the letters together, pick all but one and work out the last one. Say the question gives $V=\pi r^2h$ and asks for $h$ in terms of $V$ and $r$.

1. Pick numbers for $r$ and $h$.

   $$r=3,\ h=2$$

2. Work out $V$ from the formula.

   $$V=\pi(3^2)(2)=18\pi$$

3. Try a choice with $V=18\pi$ and $r=3$. It gives back $h=2$, so it stays.

   $$\frac{V}{\pi r^2}=\frac{18\pi}{9\pi}=2$$

4. This choice gives 3, so it's out.

   $$\frac{V}{2\pi r}=\frac{18\pi}{6\pi}=3$$

If two choices give your target, don't guess between them. Try new numbers on just those two. Say you'd picked $p=3$ and $d=3$ for Mia's question, and B and D both gave 33. Switch to $p=4$ and $d=3$, and the target is 39. B gives 44 and D gives 39, so D is the answer.

One number can knock a choice out, but a match at one number doesn't prove a choice is always right. That's why you test all four. When three fail, the fourth is the answer.

A question that asks what must be true for every value of a letter needs algebra instead, since one number only checks one case. The equivalent expressions lesson shows how to multiply out and compare.

With two letters, type each choice with your numbers in place of the letters. For choice D in Mia's question, type 2*4*3+5*3 and Desmos prints 39.

With one letter, type f(x)= and the original on one line, g(x)= and a choice on the next, then compare f(3) with g(3). The equivalent expressions lesson walks through the keystrokes.

For arithmetic you can do in your head, like a single multiplication, skip the calculator. Desmos basics covers typing expressions in.

“A store raises a price by 40%, then lowers the new price by 25%. The final price is $k$ times the original price. What is $k$?”

No price is given, so pick 100.

1. Start at 100.

   $$100$$

2. Up 40%.

   $$100\times1.40=140$$

3. Down 25%, taken of the new price, 140.

   $$140\times0.75=105$$

4. Compare the end with your starting 100, so $k=1.05$.

   $$105\div100=1.05$$

Read which form the question wants. A final price of 105 is 105% of the original, which is a 5% increase. As a multiplier, that's $k=1.05$. A choice of 105 for $k$ forgot to divide by the 100 you started with.

In Desmos, 100*1.4*0.75 prints 105. If the choices use a letter for the starting amount, like $n$, put $n=100$ through the steps and into every choice. The percentages lesson has more on changes in a row.

When the choices are numbers, take one and check it against everything the question says.

“A small theater's seats are set out in rows. Each row has 4 more seats than there are rows, and the theater has 140 seats. How many rows are there?”

• A. $7$
• B. $10$
• C. $14$
• D. $35$

The choices count rows. They run in order, so start with the second or third. If it misses, a total that's too big or too small tells you which way to go.

1. Try C. With 14 rows, each row has 18 seats.

   $$14+4=18$$

2. Find the total. That's more than 140, so you need fewer rows.

   $$14\times18=252$$

3. Try B. With 10 rows, each row has 14 seats.

   $$10+4=14$$

4. Find the total. It matches, so B is the answer.

   $$10\times14=140$$

Solving it the other way means writing $r^2+4r-140=0$ and factoring. Two tests were quicker. Once a choice fits every fact in the question, you're done.

In the theater, more rows always meant more seats, so “too many” told you to go smaller. Some questions don't move steadily like that. For $(x-5)^2$, $x=3$ and $x=7$ both give 4. On a quadraticAn expression whose biggest power is a square, like x squared plus 3x minus 4.Read the lesson, an absolute valueHow far a number is from zero, so it’s never negative, like 5 for both 5 and -5.Read the lesson, or a question asking which value “CANNOT be,” test the choices one at a time.

For “the greatest number” or “the least number,” test the choice and its neighbor. Say each bus seats 44 students, 300 students are going on a trip, and the question asks for the least number of buses.

1. Try 6 buses. That's fewer than 300 seats, so 6 doesn't work.

   $$6\times44=264$$

2. Try 7. That's enough, so 7 is the least number.

   $$7\times44=308$$

The least number has to work while the next number down fails. For the greatest number, the answer works and the next number up fails. That also settles whether to round up or down.

Working backward also fits a mean or medianThe middle value once the numbers are put in order.Read the lesson with one value missing. Put the choice in, then work out the mean again or put the numbers in order. With two prices and one total, let the choice be the count the question asks for, work out the other count, and check the total. For a trip split between two speeds or two fuel rates, split it the way the choice says and check both parts.

Work through the three examples, then try five more. Before each one, look at the choices and decide how you'll use them, or whether it's quicker to just solve it.