Solving, and why the moves are legal
45 min
Two hosts talk the lesson through. The voices are synthetic; the script was written from this lesson and checked against it, and asserts nothing the lesson does not.
- Solve a linear equation in one unknown, and justify each step by what it does to both sides
- Handle brackets, an unknown on both sides, and a negative term without losing the sign
- Check every solution by substituting it back into the original equation
Lesson 2 left you with a balance and a rule: whatever you do to one pan, you do to the other, and the beam stays level. This lesson turns that into a method.
Solving an equation means finding the value that makes the claim true. 3x + 5 = 20 is true for exactly one number, and the job is to find it without guessing.
The goal, stated once
You want to end up with the letter alone on one side: x = something. That shape is an equation you can read the answer off.
Every move you make is chosen to get closer to that shape, and every move is legal for exactly one reason: it does the same thing to both pans, so the claim survives. There's no second reason. Textbooks state the same thing as properties of equality, and OpenStax's chapter 2 has them written out if you want the formal names.
Solving one, every step annotated
Solve 3x + 5 = 20.
| Line | What was done | Why it's allowed |
|---|---|---|
3x + 5 = 20 |
the equation we were given | |
3x = 15 |
took 5 off both sides | the pans were equal, so they still are |
x = 5 |
divided both sides by 3 | same |
Now the step most people skip, and the one that makes algebra safe: check it. Put 5 back into the original equation. 3(5) + 5 is 15 + 5, which is 20, and the right side is 20. The two sides are the same size, so the claim is true, so 5 is the solution.
Notice what the check does. It doesn't tell you that your method was good; it tells you that your answer is right, which is the thing you actually care about. You can get an answer by any route at all, including guessing, and the check will still confirm or destroy it.
Solve 2(x + 3) = 14, then check. There are two sensible routes; take either.
Show the answer
Four. Route one: divide both sides by 2, giving x + 3 = 7, then take 3 off both sides, giving x = 4. Route two: expand the bracket first, giving 2x + 6 = 14, then take 6 off both, then divide by 2. Both are legal because both do the same thing to both sides, and both arrive at the same place, which is a useful thing to notice: there is rarely one correct order. Check: 2(4 + 3) is 2 × 7, which is 14, and the right side is 14.
The harder shape: a bracket, both sides, and a negative
Most real equations are messier than the first one in three ways at once: a bracket, the unknown on both sides, and a negative term. All three at once:
Solve 4(x - 2) = 2x + 6.
| Line | What was done |
|---|---|
4(x - 2) = 2x + 6 |
given |
4x - 8 = 2x + 6 |
expanded the bracket: 4 times x and 4 times -2 |
2x - 8 = 6 |
took 2x off both sides |
2x = 14 |
added 8 to both sides |
x = 7 |
divided both sides by 2 |
Check: 4(7 - 2) is 4 × 5, which is 20. And 2(7) + 6 is 14 + 6, which is 20. The sides match, so 7 is right.
Two things in that working are worth slowing down for.
Expanding the bracket. 4(x - 2) means four lots of (x - 2), so it's 4x - 8, not 4x - 2. The multiplier reaches everything inside the bracket, including the second term and its sign. Test it with a number if you ever doubt it: at x = 3, 4(3 - 2) is 4, and 4x - 8 is 12 - 8, which is 4, while 4x - 2 would be 10.
Taking 2x off both sides. You can subtract an expression as well as a number. The balance does not care what the weight is made of, only that you removed the same weight from both pans.
Below is 5x - 3 = 2x + 9, worked to the second to last line. Take 2x off both sides to get 3x - 3 = 9. Add 3 to both sides to get 3x = 12. Finish it, and check your answer.
Show the answer
Divide both sides by 3, so x = 4. Check: 5(4) - 3 is 20 - 3, which is 17, and 2(4) + 9 is 8 + 9, which is 17. Both sides are 17, so 4 is the solution. If you finished it without needing the check, do the check anyway; it takes ten seconds and it's the step that cannot lie to you.
The minus sign belongs to the term
The most common source of arithmetic slips in solving is a sign, and it's worth meeting directly rather than being warned about.1
Look at 7 - 3x = 1. Solve it in your head before you read the working.
What is x?
Show the answer
Two. If you got 4, or something involving a 4, you saw a 7 and a 3 and subtracted, which is the trap this section is about.
It is tempting to see a 7 and a 3 and produce a 4. That's wrong, and the reason is that the minus sign is not an instruction sitting between two numbers; it belongs to the 3x. The left side is "seven, plus negative three lots of x".
So solve it properly:
| Line | What was done |
|---|---|
7 - 3x = 1 |
given |
-3x = -6 |
took 7 off both sides |
x = 2 |
divided both sides by -3 |
Check: 7 - 3(2) is 7 - 6, which is 1. Correct.
And notice the last division. Dividing both sides by a negative is legal like every other move: it's the same operation on both pans. Minus six divided by minus three is positive two, because a negative divided by a negative is positive.
If you ever lose track of a sign, the check catches it immediately, which is the argument for checking every single time rather than only when you are unsure.
Solve 9 - 2x = x - 3 before reading on, and check it.
Show the answer
Four. Add 2x to both sides: 9 = 3x - 3. Add 3 to both sides: 12 = 3x. Divide by 3: 4 = x, which is the same claim as x = 4, since the sign means the two sides are the same size and does not care which way round they are written. Check: 9 - 2(4) is 9 - 8, which is 1, and 4 - 3 is 1. Both sides are 1. If you got minus four, you probably combined the two x terms as -3x instead of -x, which is the commonest slip in this shape.
Collecting like terms, which is why simplifying exists
5x - 3 = 2x + 9 had x terms on both sides, and the first move gathered them. That gathering has a name, collecting like terms, and it exists in service of solving rather than as a topic of its own.
Like terms are terms with the same letter part. 3x and 5x are like terms and add to 8x. 3x and 5y are not, and can't be combined at all. 3x and 5x² are not either, because x and x² are different things: at x = 3, x is 3 and x² is 9.
The test, as always, is a number. Does 3x + 5x = 8x? At x = 2: the left is 6 + 10, which is 16, and the right is 16. Yes. Does 3x + 5y = 8xy? At x = 2 and y = 1: the left is 6 + 5, which is 11, and the right is 8 × 2 × 1, which is 16. No.
What people get wrong
Doing something to one side only. Not a move. The equation you end up with isn't the one you started with, so its answer is not the answer to your problem.
4(x - 2) becoming 4x - 2. The multiplier reaches every term inside the bracket.
Losing the sign on a moved term. 9 - 2x = x - 3 becomes 9 = 3x - 3, not 9 = -x - 3. If you are unsure, add the term to both sides explicitly rather than "moving it over", which is a description of the result rather than a move.
Reading 7 - 3x as a subtraction between 7 and 3. The minus belongs to the 3x.
Not checking. Every solution can be verified in ten seconds by substitution, and the check is the only step that can't be fooled by a confident wrong method.
Solve each, then check by substituting your answer into the original equation. Do not skip the checks; they are the point.
- A delivery costs a $7 fee taken off a $13 credit, giving
4x - 7 = 13. Solve for the number of itemsx 5x - 8 = 3x + 2x/4 + 2 = 63(2x - 1) = 15- A pizza is cut so that one share is twice another and the whole is six slices, giving
6 - x = 2x. Solve it 2(x + 4) = 3x + 1
Answers to the practice set, once you have done all six
Show the answer
x = 5, since4(5) - 7is 13.x = 5, since5(5) - 8is 17 and3(5) + 2is 17.x = 16, since16/4 + 2is 6.x = 3, since3(2 × 3 - 1)is3 × 5, which is 15.x = 2, since6 - 2is 4 and2 × 2is 4.x = 7, since2(7 + 4)is 22 and3(7) + 1is 22.
Connections
Lesson 2's balance is the whole justification for every move here. If you ever forget why a step is allowed, put the equation back on the scales and ask what you did to each pan.
Lesson 1's habit of substituting a number is what the check is, and it's also how you settle any question about whether two expressions are the same.
Lesson 4 draws relationships as lines, and solving turns up there as finding where a line reaches a particular height.
The lesson after that writes equations from sentences, which is the hard half of this subject: solving an equation you were handed is mechanical, and producing the right equation is where the thinking is.
Sources
[1] The lesson's shape comes from John Sweller and Graham Cooper, "The use of worked examples as a substitute for problem solving in learning algebra", Cognition and Instruction 2(1), 1985. Across five experiments with study time held equal, learners who studied worked examples went on to solve similar problems in about half the time and with roughly a fifth of the errors. That is why every idea here arrives as a worked example first, then as one with the last step removed for you to supply, and only then as a problem.
Go deeper
- OpenStax, Elementary Algebra 2e, chapter 2, free online, which works through the same moves with many more examples and states them as formal properties of equality.
- Khan Academy, solving equations, for practice with immediate feedback.
Check your understanding
This lesson has a 6-question quiz. Pass it and the questions come back on a schedule in Review, so what you learned stays learned. Your progress is saved in your browser; no account needed.