M7MATH-6.2

Two-Step Equations

Learn to solve two-step equations like px + q = r by undoing addition first and multiplication second, handle negative coefficients, and check every solution.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Two-Step Equations, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

What this lesson covers

You already know how to undo a single operation: if x+7=12x + 7 = 12, you subtract 7. But what happens when a variable is trapped under two operations at once, like 4x−9=154x - 9 = 15? The variable has been multiplied by 4 and then had 9 subtracted from it. To free it, you have to peel those operations off in the right order — and the right order is the reverse of the order they were applied.

In this lesson you will learn why you undo addition and subtraction first, then multiplication and division, how to keep track of negative coefficients without losing a sign, and how to check your answer by substitution so you always know whether you are right. These skills carry directly into writing equations from word problems and into solving inequalities later in this unit.

What a Two-Step Equation Is

A two-step equation has the form px+q=rpx + q = r, where pp, qq, and rr are numbers and pp is not zero. The variable has been operated on twice: multiplied by the coefficient pp, then added to (or subtracted from) by the constant qq.

Examples include 3x+5=203x + 5 = 20, −2n−7=9-2n - 7 = 9, and x4−6=1\frac{x}{4} - 6 = 1. That last one still counts as a two-step equation, because dividing by 4 is the same as multiplying by 14\frac{1}{4}.

Think of the equation as a machine that took a secret number xx and did two things to it. In 3x+5=203x + 5 = 20, the machine multiplied by 3, then added 5, and out came 20. To find the secret number, you run the machine backwards: start at 20, subtract 5, then divide by 3. That gives x=5x = 5.

This backwards-running idea is exactly why the order matters. When you get dressed you put on socks, then shoes; to undo it you take off shoes first, then socks. Multiplication happened first when the equation was built, so it gets undone last.

The rule that lets you do this legally is the properties of equality: whatever you do to one side of an equation you must do to the other side, and the equation stays true. An equation is a balance scale, and equal changes to both sides keep it level.

The Two Steps, In Order

Here is the procedure applied to 4x−9=154x - 9 = 15.
StepWhat you doResult
Start—4x−9=154x - 9 = 15
1. Undo the constantAdd 9 to both sides4x=244x = 24
2. Undo the coefficientDivide both sides by 4x=6x = 6
3. CheckSubstitute 64(6)−9=24−9=154(6) - 9 = 24 - 9 = 15 ✓
After step 1 you have a one-step equation, which you already know how to finish. That is the whole strategy: turn a two-step equation into a one-step equation.

A very common place students go wrong is dividing first. Someone looking at 4x−9=154x - 9 = 15 might divide everything by 4 and write x−9=3.75x - 9 = 3.75. That is not wrong because dividing is illegal — it is wrong because the 9 also has to be divided, giving x−94=154x - \frac{9}{4} = \frac{15}{4}, which is messier but still correct. Dividing only part of a side breaks the balance. Undoing the constant first avoids fractions entirely.

Another frequent slip is choosing the wrong inverse operation. The equation says −9-9, so you add 9. Students sometimes see the minus sign and subtract 9 again, getting 4x=64x = 6. Always ask: what operation is attached to the constant, and what undoes it?

Negative Coefficients and Negative Constants

Signs are where most errors live. Keep the sign glued to the number in front of it. In −5x+3=−12-5x + 3 = -12, the coefficient is −5-5, not 5, and the constant is +3+3.

Solve it: subtract 3 from both sides to get −5x=−15-5x = -15. Then divide both sides by −5-5: x=3x = 3. Notice that a negative divided by a negative is positive, so the answer is positive even though the equation looked heavily negative.

Now try 8−2x=208 - 2x = 20. The terms are out of the usual order, but the same reasoning works. Read it as −2x+8=20-2x + 8 = 20. Subtract 8: −2x=12-2x = 12. Divide by −2-2: x=−6x = -6. The mistake to avoid is dividing by 2 instead of −2-2 and reporting x=6x = 6; substitution catches this instantly, because 8−2(6)=−48 - 2(6) = -4, not 20.

A third pattern is a bare negative variable, as in −x+4=10-x + 4 = 10. The coefficient of −x-x is −1-1. Subtract 4 to get −x=6-x = 6, then divide by −1-1 to get x=−6x = -6.
EquationCoefficientLast stepSolution
−5x+3=−12-5x + 3 = -12−5-5divide by −5-5x=3x = 3
8−2x=208 - 2x = 20−2-2divide by −2-2x=−6x = -6
−x+4=10-x + 4 = 10−1-1divide by −1-1x=−6x = -6
x−3−1=2\frac{x}{-3} - 1 = 2−13-\frac{1}{3}multiply by −3-3x=−9x = -9

Verifying by Substitution

Checking is not optional extra work — it is how you know your answer is right without asking anyone. To verify, put your value back into the original equation and evaluate each side separately.

Check x=−6x = -6 in 8−2x=208 - 2x = 20:8−2(−6)=8+12=208 - 2(-6) = 8 + 12 = 20The left side equals 20 and the right side is 20, so the solution checks.

Three habits make checking reliable. First, always substitute into the original equation, not into a line you rewrote — if you made an error while rewriting, checking against the bad line will confirm the error. Second, use parentheses around the substituted value: writing −2(−6)-2(-6) instead of −2−6-2-6 prevents sign disasters. Third, follow the order of operations, doing the multiplication before the addition or subtraction.

If the two sides do not match, you have a signal, not a failure. Go back and look at your two moves. Did you use the correct inverse operation? Did you apply it to both sides? Did you carry the negative sign on the coefficient? Most two-step errors are one of those three things.

Also get comfortable with solutions that are not whole numbers. If 6x+5=126x + 5 = 12, then 6x=76x = 7 and x=76x = \frac{7}{6}. A fraction is a perfectly good answer; there is no rule that says variables must equal integers.

Where Two-Step Equations Show Up

Two-step equations are the standard shape for a real situation with a fixed starting amount plus a repeated amount. A gym charges a 30 dollar sign-up fee plus 15 dollars per month, and you have 165 dollars to spend, so 15m+30=16515m + 30 = 165 gives 15m=13515m = 135 and m=9m = 9 months.

The fixed part becomes qq; the repeated per-unit part becomes the coefficient pp; the total becomes rr. Recognizing this shape lets you write the equation quickly, which is the focus of the next lesson.

Negative coefficients model quantities that shrink. If a 500 gallon tank drains 25 gallons per hour, the amount left after hh hours is 500−25h500 - 25h. Asking when 150 gallons remain gives 500−25h=150500 - 25h = 150, so −25h=−350-25h = -350 and h=14h = 14 hours.

One last caution about interpreting answers in context: the arithmetic can produce a value that does not make sense for the situation. If solving for a number of months gives m=−4m = -4, the algebra may be fine while the situation is impossible, which usually means the equation was set up backwards. Always reread the question after you solve, and attach units to your answer: 9 months, 14 hours, 6 tickets.

Key terms

Two-step equation.
An equation of the form px+q=rpx + q = r in which the variable has been multiplied (or divided) by a number and then had a number added (or subtracted), requiring two inverse operations to solve.
Coefficient.
The number multiplying the variable. In −7x+2=9-7x + 2 = 9 the coefficient is −7-7, including its sign.
Constant term.
A number in the equation with no variable attached, such as the +2+2 in −7x+2=9-7x + 2 = 9.
Inverse operation.
The operation that undoes another: addition and subtraction undo each other, and multiplication and division undo each other.
Properties of equality.
The rules stating that adding, subtracting, multiplying, or dividing both sides of an equation by the same nonzero value keeps the equation true.
Solution of an equation.
A value for the variable that makes both sides of the equation equal when substituted.
Substitution check.
Replacing the variable with your solution in the original equation and evaluating both sides to confirm they are equal.
Isolate the variable.
To use inverse operations until the variable stands alone on one side of the equation with coefficient 1.

Worked example

Solve −6x+11=−25-6x + 11 = -25 and verify the solution by substitution.
Identify the parts first. The coefficient is −6-6 and the constant term is +11+11. The variable is trapped by a multiplication by −6-6 and then an addition of 11.

Step 1: undo the addition. Subtract 11 from both sides.−6x+11−11=−25−11-6x + 11 - 11 = -25 - 11−6x=−36-6x = -36The left side is now a one-step equation. Note that −25−11-25 - 11 means going 11 further in the negative direction, giving −36-36, not −14-14. Getting −14-14 here is one of the most common slips.

Step 2: undo the multiplication. Divide both sides by −6-6.−6x−6=−36−6\frac{-6x}{-6} = \frac{-36}{-6}x=6x = 6A negative divided by a negative is positive, so the solution is x=6x = 6.

Step 3: verify by substituting 6 into the original equation, using parentheses.−6(6)+11=−36+11=−25-6(6) + 11 = -36 + 11 = -25The left side equals −25-25 and the right side is −25-25, so the two sides match and x=6x = 6 is confirmed.

Practice questions

What is the solution to 9−4x=379 - 4x = 37?
  1. x=−7x = -7
  2. x=7x = 7
  3. x=−11.5x = -11.5
  4. x=11.5x = 11.5

Answer: x=−7x = -7

Rewrite as −4x+9=37-4x + 9 = 37. Subtract 9 from both sides: −4x=28-4x = 28. Now divide by the coefficient, which is −4-4, not 4: x=−7x = -7. Choosing x=7x = 7 means dividing by positive 4 and dropping the sign; a check shows 9−4(7)=9−28=−199 - 4(7) = 9 - 28 = -19, which is not 37. Checking the correct answer: 9−4(−7)=9+28=379 - 4(-7) = 9 + 28 = 37. ✓
A student solves 5x+8=435x + 8 = 43 by first dividing every term by 5, writing x+8=8.6x + 8 = 8.6, then subtracting 8 to get x=0.6x = 0.6. Explain the error, then solve the equation correctly and verify your answer.

Answer: The student divided only the 5x5x term by 5 and left the 8 unchanged. Solving correctly: subtract 8 to get 5x=355x = 35, then divide by 5 to get x=7x = 7. Check: 5(7)+8=35+8=435(7) + 8 = 35 + 8 = 43. ✓

Dividing a side of an equation by 5 means dividing every term on that side, so the correct result of dividing would have been x+85=435x + \frac{8}{5} = \frac{43}{5}, not x+8=8.6x + 8 = 8.6. Dividing only part of a side destroys the balance. Undoing the constant first is easier anyway, because it keeps the numbers whole. Substitution catches the student's answer immediately: 5(0.6)+8=3+8=115(0.6) + 8 = 3 + 8 = 11, not 43.
Solve x−3+5=2\frac{x}{-3} + 5 = 2 and state which operation you undo first and why.

Answer: x=9x = 9. Undo the +5+5 first, because addition was the last operation applied to the variable expression, so it is the first one to reverse.

Subtract 5 from both sides: x−3=−3\frac{x}{-3} = -3. Then undo the division by multiplying both sides by −3-3: x=(−3)(−3)=9x = (-3)(-3) = 9. Check in the original equation: 9−3+5=−3+5=2\frac{9}{-3} + 5 = -3 + 5 = 2. ✓ Dividing by −3-3 is the same as multiplying by −13-\frac{1}{3}, so this is still a two-step equation of the form px+q=rpx + q = r.

FAQ

Why do I undo addition and subtraction before multiplication and division?
Because you are reversing the order in which the operations were applied to the variable. In 3x+5=203x + 5 = 20, the variable was multiplied by 3 first and had 5 added second. To unwrap it you remove the outer layer first, so you subtract 5 and then divide by 3. Dividing first is not illegal, but it forces you to divide the constant too, which usually creates fractions for no reason.
What do I do when the variable has a negative sign in front of it, like −x+4=10-x + 4 = 10?
Treat −x-x as −1⋅x-1 \cdot x, so the coefficient is −1-1. Subtract 4 from both sides to get −x=6-x = 6, then divide both sides by −1-1 to get x=−6x = -6. Never just erase the negative sign; dividing by −1-1 is a real step.
Is a fraction or decimal answer allowed?
Yes. If 4x+3=104x + 3 = 10, then 4x=74x = 7 and x=74x = \frac{7}{4}, or 1.75. Nothing requires solutions to be whole numbers. In a real-world context, though, reread the question: if the variable counts buses or tickets, a fractional answer may need to be rounded up or interpreted, and a negative answer usually means the equation was set up incorrectly.
How do I know my answer is right without an answer key?
Substitute your value into the original equation and evaluate each side separately using order of operations, keeping parentheses around negatives. If both sides give the same number, your solution is correct. If they do not match, check three things: did you use the correct inverse operation, did you apply it to both sides, and did you keep the sign on the coefficient?

Learn this with a teacher, not a page

The Crimsora tutor teaches Two-Step Equations live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.