Lesson · Linear Equations and Inequalities · Equations

One- and Two-Step Equations

An expression like 3x − 4 is built from x by a sequence of operations, and isolating x means undoing them in the opposite order. Every step is one operation applied to both sides, and the answer is an exact fraction, checked in the original equation.

Open the interactive lesson Practice

3x − 4 = 11 x was multiplied by 3, then 4 was subtracted so undo in reverse: add 4, then divide by 3 3x − 4 = 11 → 3x = 15 → x = 5 check: 3·5 − 4 = 11

Last operation applied, first operation undone

Core ideas

What this lesson establishes

There is one idea here and it is borrowed from the order of operations: what was done last has to be undone first.

Each step is one named operation applied to both sides

"Add 4 to both sides", "divide both sides by 3". If you cannot name the operation you are performing, you cannot check that it is one of the legal ones, and the step is a guess.

Undo in the reverse of the order the operations were applied

3x − 4 multiplies by 3 and then subtracts 4. Reversing that means adding 4 and then dividing by 3. This is the order of operations run backwards, and it is the only thing you have to remember.

The coefficient carries its sign into the division

From −2x = 12 the last move is division by −2, not by 2. Dropping the sign here produces 6 instead of −6 and the check catches it — but only if you do the check.

The material

Isolating the variable

The one-step shapes, the two-step routine, and what to do when the answer is not a whole number.

Read 3x − 4 as a recipe applied to x: multiply by 3, then subtract 4. Solving reverses the recipe. Because the subtraction happened last, it is undone first. The four one-step equations are the recipe with a single instruction in it, and each is undone by the one operation that reverses it, applied to both sides:

x + 7 = 2 subtract 7 from both sides x = −5 x − 7 = 2 add 7 to both sides x = 9 7x = 2 divide both sides by 7 x = 2/7 x/7 = 2 multiply both sides by 7 x = 14

The third line is the one people rewrite. x = 2/7 is a finished answer. Turning it into 0.2857 makes it a different number, and “What It Means to Solve an Equation” showed what happens to a different number in the check.

Two steps

With both a coefficient and a constant, deal with the constant first:

3x − 4 = 11 add 4 to both sides 3x = 15 divide both sides by 3 x = 5 check 3·5 − 4 = 15 − 4 = 11 and the right side is 11

Doing it the other way round is legal and gives the same answer with worse arithmetic: dividing first turns 3x − 4 = 11 into x − 4/3 = 11/3, and then adding 4/3 gives x = 15/3 = 5. The constant-first order is a convention chosen because it postpones fractions, not because the other route is wrong.

Brackets

2(x + 3) = 14 can be started in either of two ways, and both are legal moves on the whole side:

2(x + 3) = 14 divide both sides by 2 x + 3 = 7 x = 4 2(x + 3) = 14 distribute first 2x + 6 = 14 x = 4

What does not work is subtracting 3 from both sides in the hope of stripping the bracket. The move itself is legal — the addition property allows any number — but it does not do what it looks like it does. The 3 is inside the bracket, so it is not a term of the left side; the left side is a single product. Subtracting 3 from 2(x + 3) gives 2(x + 3) − 3, that is 2x + 3, not 2x.

Negative coefficients

The coefficient is the number multiplying x, and its sign belongs to it:

7 − 2x = 19 subtract 7 from both sides −2x = 12 divide both sides by −2 x = −6 check 7 − 2·(−6) = 7 + 12 = 19

Theorem · What the last step needs

An equation of the form px = q with p ≠ 0 has exactly one solution, x = q/p.

Division by p is legal because p ≠ 0, and it is reversible, so the solution set does not change. What happens when p = 0 is the subject of “Identities and Contradictions”.

Example · A fraction that is the answer

(3/4)x + 1/2 = 5/4. Subtract 1/2 from both sides: (3/4)x = 3/4, since 5/4 − 1/2 = 5/4 − 2/4 = 3/4. Divide both sides by 3/4: x = 1. “Equations with Fractions” shows a route through this kind of equation that avoids fractional arithmetic entirely.

Interactive

Solving, one operation at a time

Every step named, and the answer put back

Method

Solving a one- or two-step equation

Simplify before you undo anything; then undo in reverse order.

Simplify each side on its own

Multiply out brackets and combine like terms, left side and right side separately. You cannot reliably see which operations were applied to x until each side is a tidy expression.

Undo the addition or subtraction first

Whatever constant is attached to the variable term, remove it by doing the opposite to both sides. It was applied last, so it comes off first.

Divide both sides by the coefficient, sign included

−2x = 12 becomes x = −6. Write the answer as an exact fraction if it is not a whole number: x = 7/3, not 2.33.

Substitute the answer into the original equation

Both sides, evaluated separately, must come out to the same number. This is the step that turns a candidate into an answer, and it takes about ten seconds.

Undoing operations

Score 0 / 0

Choose the answer you can justify.

Three equations, each with its check

Every row names the operation that produced it.

A. 3x − 4 = 11 + 4 both sides 3x = 15 ÷ 3 both sides x = 5 check 3·5 − 4 = 11 right side 11 agrees B. 7 − 2x = 19 − 7 both sides −2x = 12 ÷ (−2) both sides x = −6 check 7 − 2·(−6) = 7 + 12 = 19 agrees C. x/4 + 3 = 1 − 3 both sides x/4 = −2 × 4 both sides x = −8 check (−8)/4 + 3 = −2 + 3 = 1 agrees

In B the division is by −2, and it is the only place the sign can be lost. Dividing by 2 instead gives x = 6, and the check then reads 7 − 12 = −5, which is not 19. The error is invisible in the working and obvious in the check.

In C the bracket in (−8)/4 matters when you write the check out. Substituting a negative value without brackets is how a correct answer gets rejected by its own verification.

For a faded pass, start 5 − 3x = −7 with the supplied first move: subtract 5 from both sides to get −3x = −12. Name and complete the remaining move, then check the value in 5 − 3x = −7 before opening the same equation in the lab.

Common mistakes

Applying the operation to one side onlyGoing from 3x − 4 = 11 to 3x = 11 adds 4 on the left and nothing on the right. The property of equality says both sides, and the two sides stop being equal the moment you treat them differently.

Dividing by the coefficient without its sign−2x = 12 gives x = −6. The minus is part of the number you are dividing by, not a decoration on the term, and this is the single most common arithmetic slip in the lesson.

Rounding an exact answer to a decimalx = 7/3 is the answer; 2.33 is a nearby number that fails the check. Fractions are exact, they are not harder to substitute, and the rest of this path expects them.

Check your understanding

✓

Finish when you can name every operation you apply and check the result without being asked.

Solve 9 − 4x = −6 without the lab: name each operation, keep the answer as an exact fraction, and substitute it into the original. Then solve 3(x + 2) = 21 once by division first and once by distribution; the two routes must agree. Everything in “Variables on Both Sides”, “Equations with Fractions” and “Literal Equations and Formulas” adds new opening moves and finishes exactly like this.

Note: The lab shows one operation per row on purpose. When an equation goes wrong it is almost never the strategy that failed — it is one row, and a trace where every row is named tells you which one. “Variables on Both Sides” adds the only genuinely new move on the course: subtracting a variable term.