Turning word problems into equations

Four steps

Almost every algebra word problem yields to the same routine.

1. Name the unknown. Write "let $n$ = the number of notebooks bought" rather than just "$n$". Include units and, where it matters, a time: "Sana's age now".
2. Translate. Turn each sentence of the story into an equation or an expression, phrase by phrase.
3. Solve the equation or system.
4. Answer and check. Reread the question: is it asking for $n$, or for something built from $n$? Then put the answer back into the story, not into your equation.

Do not skip the fourth step. A problem that asks for the larger of two numbers will usually offer the smaller one as an option, because it is what you find first.

Translating phrase by phrase

Words	Algebra
is, was, will be, equals	$=$
of (with a fraction or percent)	$\times$: "a third of $x$" is $\frac{x}{3}$
5 more than $x$	$x + 5$
5 less than $x$	$x - 5$ (the order reverses)
$x$ less than 5	$5 - x$
twice as many as $x$	$2x$
$a$ exceeds $b$ by 4	$a - b = 4$
the sum of $x$ and $y$, tripled	$3(x + y)$

"Less than" is the phrase that catches people. "Nine less than five times a number" starts from five times the number and takes nine away: $5x - 9$. Read it as "five times the number, minus nine".

"Times as many" is the other one. "Ana has three times as many stamps as Ben" gives $A = 3B$: the multiplier goes with the smaller amount. A quick sense check with numbers (if Ben has 10, Ana has 30) confirms the direction.

Two kinds of quantity: how many and how much

Problems about tickets, coins or items at different prices involve two different totals: the number of things and their value. Each gives its own equation.

• Number: $a + c = 50$ (adult and child tickets).
• Value: $12a + 7c = 485$ (price times number, added).

Mixing them, for example adding prices to counts, is a common setup error. Write the units next to each equation (tickets, dollars) and check they match on both sides.

Ages

Age problems compare people at different times. Use one variable for each person's age now, then express other times from it: in 8 years, someone aged $n$ is $n + 8$; six years ago, $n - 6$. Shift every person by the same number of years. A table with a row for each person and a column for each time keeps this straight.

Consecutive integers

Call the smallest $n$. Consecutive integers are $n, n + 1, n + 2$; consecutive even or consecutive odd integers are $n, n + 2, n + 4$. Odd integers also go up in steps of 2, not 1. For an odd number of consecutive terms, the sum is the middle term times the number of terms, which gives a quick check: three consecutive odd integers summing to 87 have middle term $\frac{87}{3} = 29$.

Choose the variable to save work

Let the variable stand for the quantity the question asks for when you can, so the last step of solving is the answer. When the story has two linked unknowns, you can use two variables and a system, or one variable and an expression for the other ($c = 50 - a$). Both work; one variable usually means fewer lines.

Check against the story

Put the answer back into the words: do the tickets add up to 50, and do the prices add up to $485? Is an age positive, and a number of people a whole number? An answer that solves your equation but fails the story means the translation, not the algebra, went wrong.

Units

Convert everything to one unit before you write an equation. A problem that mixes minutes and hours, or cents and dollars, needs one conversion at the start, not a correction at the end.