⚔️ Algebra Kingdom · Algebra

Two-Step Equations

Solve equations of the form ax $+ b = c$ by undoing operations in reverse order.

In short

Undo operations in the reverse of the order they were applied: the constant comes off before the coefficient.
For ax $+ b = c$ , subtract b from both sides first, then divide both sides by a.
If you divide first you must divide every term on both sides, not only the right-hand side.
Dividing by a negative coefficient changes the sign of the answer, so carry the sign with the coefficient throughout.

Two things have happened to the unknown

In $2x + 7 = 19$ the unknown has been through two machines: first it was multiplied by 2, then 7 was added. The equation records the result, 19.

To get back to x you have to run the machines backwards — and, like undressing, in the reverse order. The last thing that happened is the first thing you undo.

Because the 7 was added last, the 7 comes off first. Only then do you deal with the 2 that is multiplying x.

Undo addition and subtraction first

The reliable recipe for ax $+ b = c$ is:

Step 1. Subtract b from both sides. This leaves ax $= c$ - b.
Step 2. Divide both sides by a. This leaves x on its own.

For $2x + 7 = 19$ : subtracting 7 gives $2x = 12$ , and dividing by 2 gives $x = 6$ .

If the equation is ax - $b = c$ , step 1 becomes "add b to both sides" — same idea, opposite operation. If it is $x/a + b = c$ , step 2 becomes "multiply both sides by a".

Why the order matters

Suppose you divide first in $2x + 7 = 19$ and write $x + 7 = 9.5$ . That is wrong, because dividing the left side by 2 has to divide every term on it, giving $x + 3.5$ , not $x + 7$ .

You $*may*$ divide first, as long as you divide everything: $2x + 7 = 19$ becomes $x + 3.5 = 9.5$ , and then $x = 6$ again. The answer agrees, which is reassuring, but the arithmetic is uglier.

The classic wrong answer is $19 / 2$ - $7 = 2.5$ . It comes from dividing only the right-hand side while the $+7$ is still sitting on the left. Clearing the constant first avoids the trap entirely.

Negative coefficients

An equation such as 12 - $5x = 47$ works the same way, but the term holding x is -5x.

Subtract 12 from both sides: $-5x = 35$ . Now divide both sides by -5, and remember that dividing by a negative flips the sign: $x = -7$ .

Two habits keep this safe. First, always write the coefficient with its sign as you go, so -5x stays -5x. Second, check the answer: 12 - $5 x (-7) = 12 + 35 = 47$ . Correct.

Fee plus rate: the everyday two-step

Most real two-step equations look like this: one fixed amount that never changes, plus a rate that depends on how many.

"The guild charges 40 gold to join and 12 gold a day" gives the expression $12n + 40$ . If the bill was 148 gold, then $12n + 40 = 148$ .

Take the fixed fee off first, because it is charged once, not per day: $12n = 108$ , so $n = 9$ days. A table makes the structure visible — 1 day costs $52, 2$ days cost $64, 3$ days cost 76 — going up by the same 12 each time, starting from 40.

Worked examples

Example 1

Solve for $x: 5x$ - $8 = 37$

Two operations have been applied to x: multiply by 5, then subtract 8.
Undo the subtraction first by adding 8 to both sides: $5x = 37 + 8 = 45$ .
Now undo the multiplication by dividing both sides by $5: x = 45 / 5$ .
$x = 9$ .
Check: $5 x 9$ - $8 = 45$ - $8 = 37$ . Correct.

Example 2

A ferry costs 15 gold to board plus 6 gold for every crate carried. Mira paid 63 gold. How many crates did she carry?

Let n be the number of crates. The 15 gold is paid once; the 6 gold is paid per crate.
Total cost: $6n + 15$ , and the story says that total was 63, so $6n + 15 = 63$ .
Subtract the fixed 15 from both sides: $6n = 48$ .
Divide both sides by $6: n = 8$ crates.
Check: $6 x 8 + 15 = 48 + 15 = 63$ gold. Correct.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Solve for $x: 4x + 3 = 35$

Answer: 8

$4x + 3 = 35$
$4x = 35$ - $3 = 32$ (subtract 3 from both sides)
$x = 32 / 4$ (divide both sides by 4)
$x = 8$
Check: $4 x 8 + 3 = 32 + 3 = 35$ . Correct.

Problem 2

Difficulty 3 of 5

Solve for $w: 2w$ - $48 = -30$

Answer: 9

2w - $48 = -30$
$2w = -30 + 48 = 18$ (add 48 to both sides)
$w = 18 / 2$ (divide both sides by 2)
$w = 9$
Check: $2 x 9$ - $48 = 18$ - $48 = -30$ . Correct.

Problem 3

Difficulty 4 of 5

Solve for $w: w/12 + 68 = 66$

Answer: -24

$w/12 + 68 = 66$
$w/12 = 66$ - $68 = -2$ (subtract 68 from both sides)
$w = -2 x 12$ (multiply both sides by 12)
$w = -24$
Check: $-24/12 + 68 = -2 + 68 = 66$ . Correct.

Common mistakes

Dividing before subtracting: answering $19/2$ - 7 for $2x + 7 = 19$ instead of clearing the $+7$ first.
Stopping one step early and giving the value of 2x rather than the value of x.
Dividing the whole total by the rate in a fee-plus-rate problem, forgetting that part of the total was the one-off fee.
Losing the minus sign on a negative coefficient and getting the right size with the wrong sign.

What you should be able to do

Undo addition or subtraction before undoing multiplication or division.
Solve two-step equations with negative and fractional coefficients.
Check solutions by substitution.
Model two-step word problems with an equation.

Where this fits in the curriculum

Common Core

7.EE.B.4.A
Grade 7 — Solve equations of the form px + q = r and p(x + q) = r fluently, and compare an algebraic solution with an arithmetic one.
7.EE.B.3
Grade 7 — Solve multi-step problems with positive and negative rational numbers in any form.

Ontario

G6.C2.3
Grade 6 — Solve equations that involve multiple terms and whole numbers, and verify the solution.
G7.C2.3
Grade 7 — Solve equations that involve multiple terms, whole numbers and decimal numbers, and verify the solution.