Math Core

A taxi charges $3 to start the ride plus $2 per mile, and your fare was $19. How far did you go? The equation is $2m + 3 = 19$ , and the variable has two operations attached to it. To free it, you undo both, in the right order.

Unwrapping the variable

Think about how the expression $2m + 3$ gets built if you know $m$ . The order of operations says:

Multiply $m$ by $2$ .
Then add $3$ .

Solving undoes those steps in reverse order, like unwrapping a present: the last thing wrapped is the first thing removed. So first subtract $3$ , then divide by $2$ .

\begin{aligned} 2m + 3 &= 19 \\ 2m + 3 - 3 &= 19 - 3 && \text{undo } +3 \\ 2m &= 16 \\ \frac{2m}{2} &= \frac{16}{2} && \text{undo } \times 2 \\ m &= 8 \end{aligned}

The ride was $8$ miles. Check: $2(8) + 3 = 16 + 3 = 19$ . ✓

Solving a two-step equation

Undo addition or subtraction first, so the variable term is alone on one side.
Then undo multiplication or division to get the variable alone.
Check the answer in the original equation.

In short: undo the operations in the reverse of the order of operations.

You could divide first, but then you would have to divide every term: $2m + 3 = 19$ becomes $m + 1.5 = 9.5$ . It still works, but it usually creates fractions. Clearing the constant first keeps the numbers clean.

Worked example: Subtract, then divide

Solve $5x - 8 = 27$ .

\begin{aligned} 5x - 8 &= 27 \\ 5x &= 35 && \text{add } 8 \text{ to both sides} \\ x &= 7 && \text{divide both sides by } 5 \end{aligned}

Check: $5(7) - 8 = 35 - 8 = 27$ . ✓

Negative coefficients and the variable on the right

A variable term with a negative coefficient is still just a product. Undo the constant first, then divide by the negative number. It also doesn't matter which side the variable ends up on.

Worked example: A negative coefficient

Solve $-9 = 11 - 4n$ .

The variable term is $-4n$ (the minus sign belongs to it), and $11$ is added to it.

\begin{aligned} -9 &= 11 - 4n \\ -20 &= -4n && \text{subtract } 11 \text{ from both sides} \\ 5 &= n && \text{divide both sides by } -4 \end{aligned}

Check: $11 - 4(5) = 11 - 20 = -9$ . ✓

Common mistake

In $11 - 4n$ , the term with the variable is $-4n$ , not $4n$ . A common mistake is to subtract $11$ and then divide by $4$ , getting $n = -5$ . The sign in front of a term belongs to that term, so divide by $-4$ .

Division and fraction bars

When the variable is divided by a number, undo the addition or subtraction first, then multiply.

Worked example: Divided by a number

Solve $\dfrac{y}{6} + 4 = 1$ .

\begin{aligned} \frac{y}{6} + 4 &= 1 \\ \frac{y}{6} &= -3 && \text{subtract } 4 \\ y &= -18 && \text{multiply by } 6 \end{aligned}

Check: $\dfrac{-18}{6} + 4 = -3 + 4 = 1$ . ✓

Watch where the constant sits, though. A fraction bar is a grouping symbol, just like parentheses. In $\dfrac{y + 4}{6}$ , the $4$ is added before dividing, so it gets undone after the division.

Worked example: The fraction bar groups

Solve $\dfrac{y + 4}{6} = 1$ .

Now the last operation done to $y$ is the division, so undo it first.

\begin{aligned} \frac{y + 4}{6} &= 1 \\ y + 4 &= 6 && \text{multiply by } 6 \\ y &= 2 && \text{subtract } 4 \end{aligned}

Check: $\dfrac{2 + 4}{6} = \dfrac{6}{6} = 1$ . ✓ Compare this with the previous example: same numbers, different structure, different answer.

Two-step word problems

Many situations have a starting amount plus a rate times a variable. That's the shape $ax + b = c$ .

Worked example: Saving up

Marcus has $45 saved and adds $15 each week. After how many weeks will he have $150?

Let $w$ be the number of weeks. Starting amount plus weekly savings equals the goal:

45 + 15w = 150.

Subtract $45$ : $15w = 105$ . Divide by $15$ : $w = 7$ . After $7$ weeks he will have $150. Check: $45 + 15(7) = 45 + 105 = 150$ . ✓

Tip

Before solving a word problem, decide what the variable stands for and write it down: "Let $w$ = number of weeks." At the end, answer the question in a sentence with units. It keeps you from reporting the wrong quantity.