Linear Equations

Tap a term to take it off both sides and watch the equation change. The strip under it shows the solutions of the equation as it stands, and a legal move never moves them.

Worked example — Linear Equations

Solution	3
Checked against the original	13 ✓
Moves made	0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Linear Equations358 words

Every linear equation in one unknown, however it is written, can be reduced by legal moves to the single form ax = b, with a and b known numbers. Everything the equation can do is then decided by those two.

What each symbol means

a: the coefficient of the unknown after every x term has been collected onto one side. It is the number that decides which of the three outcomes below you are in.
b: the constant, everything left on the other side. It only matters when a is 0.
x: the unknown. It appears to the first power and nowhere inside a function; that is what "linear" means here.

Where the formula comes from

Collect: move every term containing x to one side and every term without it to the other. Both are additions, so both are always reversible.
You now have ax = b. If a ≠ 0 divide by it, and there is exactly one solution, x = b/a.
If a = 0 the unknown is gone and what is left is a statement about numbers alone: false when b ≠ 0, true when b = 0. So which of the three outcomes you are in is settled by the pair (a, b) before any solving happens at all.

Assumes: That a and b are exact. Where they are measured rather than given, a coefficient near zero is a warning rather than a value: x = b/a divides by it, so a small uncertainty in a becomes a large one in x. That is the same difficulty this path meets again at step 5 with two unknowns, where the determinant plays the part a plays here.
Breaks when: Nowhere, which is the useful part. One unknown admits no fourth case: one solution, none, or all of them, and which you get is settled by looking at a and b before solving anything. That completeness is the one-unknown instance of the rank condition that governs every linear system, however many unknowns it has, and steps 5 and 6 of this path are the same trichotomy with more letters.

Two of the four buttons are the same move 🖖

Subtracting 7 from both sides is adding −7 to both sides, and dividing by 2 is multiplying by a half. The panel offers four buttons because that is how people think, but underneath there are only two moves: add something, or multiply by something. The two behave differently. Adding is always safe, whatever you add. Multiplying is safe for every number except one, and the exception is not a technicality — press × with a value of 0 and watch what happens to an equation that had a perfectly good answer a moment ago.

Multiplying by zero is the move that is not reversible 🖖

Take x = 3 and multiply both sides by 0. You get 0 = 0, which is true. It is also true for x = 5, x = −1 and every other number, so the equation has stopped saying anything about x at all. Every other move can be walked back — add 7, subtract 7, and you are where you started — and that reversibility is the actual guarantee behind "do the same thing to both sides". The rule is not that both sides must match. It is that the move must be undoable, and multiplying by zero is the one arithmetic operation that is not.

Not every equation has an answer, and that is information 🖖

Press No x works. Subtract 2x from both sides of 2x + 5 = 2x + 9 and the x disappears, leaving 5 = 9. Nothing went wrong: the equation is simply false, and no value of x will rescue it. Press Every x works and the same thing happens with a different ending, 5 = 5, which is true regardless. Both are what happens when the x terms cancel, and which one you land on is decided entirely by the constants. So "solve for x" has three possible answers, not one, and a solver that only ever reports a number is hiding two thirds of the outcomes.

Problems solved in full

Taxis charging 4 plus 3 a mile and 10 plus 1 7 steps
A taxi charges 4 to start plus 3 a mile. Another charges 10 to start plus 1 a mile. At what distance do they cost the same, and what does the algebra do if you get the question slightly wrong?
1. Write both fares as expressions in the same unknown. The unknown has to be the same thing in both, which is where most word problems go wrong before any algebra starts.
  
  $4 + 3 x 10 + 1 x$
2. Setting them equal is the whole of the modelling step. Everything after it is mechanical.
  
  $4 + 3 x = 10 + x$
3. Collect the x terms. Subtracting x from both sides is legal and reversible, so the solutions are untouched.
  
  $4 + 2 x = 10$
4. Collect the constants, then divide by what is left in front of x. Three miles, and both fares come to 13.
  
  $2 x = 6 \Rightarrow x = 3 \Rightarrow 4 + 9 = 13 = 10 + 3$
5. Now get the question wrong on purpose. Suppose the second firm also charged 3 a mile. The x terms cancel and you are left with 4 = 10, which is false, so there is no distance at which they agree — one is simply always dearer by 6.
  
  $4 + 3 x = 10 + 3 x \Rightarrow 4 = 10 \Rightarrow \emptyset$
6. And if the second firm charged 3 a mile with the same 4 standing charge, the same cancellation leaves 4 = 4, true for every distance. The two fares are the same fare written twice.
  
  $4 + 3 x = 4 + 3 x \Rightarrow 4 = 4 \Rightarrow x \in R$
7. Which is where the algebra stops and judgement starts. All three outcomes are correct answers to a well-posed question, and the tool will report them faithfully. What it cannot tell you is that a negative solution means the crossover happened before the journey started, or that a fractional mile may not be a thing you can buy. The equation does not know what x is.
  
  $x = - 2 \Rightarrow x \in / [0, \infty)$
Answer

Three miles, at a fare of 13 each. Change one coefficient and the same algebra returns "no distance" or "every distance" instead, and both are answers rather than failures.
x on both sides: 5x − 2 = 3x + 8 5 steps
Solve 5x − 2 = 3x + 8. There is an x on each side, so nothing can be read off until they are in one place.
1. Get the x terms together first. Taking 3x off the right also takes 3x off the left, and the equation still says the same thing about x.
  
  $5 x - 2 = 3 x + 8 - 3 x$
2. Now the left side is 2x − 2. Add 2 to both sides to clear the number away from the x.
  
  $2 x - 2 = 8 + 2$
3. Two lots of x make 10, so one lot makes 5.
  
  $2 x = 10 \div 2 ⟹ x = 5$
4. Check it in the equation you started with, not in the one you finished with. Both sides come to 23, so x = 5 satisfies the original.
  
  $5 (5) - 2 = 23, 3 (5) + 8 = 23$
5. You did not have to collect on the left. Take 5x off both sides instead and you get −2 = −2x + 8, which looks different and gives the same x = 5. Which side you collect on is free; the answer is not.
  
  $5 x - 2 = 3 x + 8 - 5 x ⟹ - 2 = - 2 x + 8 ⟹ 2 x = 10$
Answer

x = 5, and both sides come to 23. Collecting the x terms on the other side is a different-looking route to the same number.
When the answer is negative: 4x + 9 = x − 6 5 steps
Solve 4x + 9 = x − 6. Nothing here is unusual until the last line, where the answer comes out below zero.
1. Take x off both sides to collect the x terms on the left.
  
  $4 x + 9 = x - 6 - x$
2. Take 9 off both sides to clear the number.
  
  $3 x + 9 = - 6 - 9$
3. Three lots of x make −15, so one lot makes −5. A minus sign in the answer is not a sign that something went wrong.
  
  $3 x = - 15 \div 3 ⟹ x = - 5$
4. Check in the original: the left side is −20 + 9 = −11, and the right side is −5 − 6 = −11. They agree, so −5 really is the solution.
  
  $4 (- 5) + 9 = - 20 + 9 = - 11, (- 5) - 6 = - 11$
5. Whether −5 is a sensible answer depends on what x was. As a temperature it is an ordinary Tuesday; as a number of tickets it is nothing at all. The equation cannot tell the difference and will not warn you.
  
  $x = - 5 \Rightarrow 5 tickets short of none$
Answer

x = −5, with both sides at −11. A negative solution is an answer to the equation; whether it is an answer to the question is a separate judgement, and only you can make it.
When the answer is not a whole number: 6x + 1 = 2x + 4 5 steps
Solve 6x + 1 = 2x + 4. The same three moves as before, and this time they do not land on a whole number.
1. Take 2x off both sides.
  
  $6 x + 1 = 2 x + 4 - 2 x$
2. Take 1 off both sides.
  
  $4 x + 1 = 4 - 1$
3. Four lots of x make 3, so one lot is 3 divided by 4, which is 0.75 exactly.
  
  $4 x = 3 \div 4 ⟹ x = \frac{3}{4} = 0.75$
4. Check in the original: 6 × 0.75 + 1 = 5.5, and 2 × 0.75 + 4 = 5.5.
  
  $6 (0.75) + 1 = 5.5, 2 (0.75) + 4 = 5.5$
5. Three quarters happens to be exact as a decimal. Two thirds is not: solve 3x + 1 = 3 and the answer is 2⁄3, which no decimal ever finishes writing. The fraction is the answer, and 0.6667 is a rounding of it.
  
  $3 x + 1 = 3 ⟹ x = \frac{2}{3} = 0.6667 \dots$
Answer

x = 0.75, or three quarters, with both sides at 5.5. Fractions are ordinary answers. Where a fraction has no exact decimal, the fraction is the exact answer and the decimal is not.
Brackets, two ways: 3(x + 4) = 24 4 steps
Solve 3(x + 4) = 24. There are two legal routes, and they are not the same amount of work.
1. The bracket says three lots of (x + 4). Divide both sides by 3 first and the bracket simply goes: x + 4 = 8, so x = 4.
  
  $3 (x + 4) = 24 \div 3 ⟹ x + 4 = 8 ⟹ x = 4$
2. Or expand first. Three lots of (x + 4) is 3x + 12, then take 12 off both sides and divide by 3. Same x = 4.
  
  $3 (x + 4) = 3 x + 12 = 24 - 12 ⟹ 3 x = 12 ⟹ x = 4$
3. Check in the original: 3 × 4 + 12 = 24. The tool takes equations in the form ax + b = cx + d, so it wants the expanded line.
  
  $3 (4) + 12 = 24$
4. Divide-first was shorter only because 24 divides by 3 exactly. Change the 24 to a 25 and the same route hands you x + 4 = 25⁄3 — still correct, and now the messier of the two.
  
  $3 (x + 4) = 25 ⟹ x + 4 = \frac{25}{3} ⟹ x = \frac{13}{3}$
Answer

x = 4 by either route. Both are legal, so pick after looking at the numbers rather than by rule.
Two phone plans that never cost the same 4 steps
Plan A is £8 a month plus £2 a gigabyte. Plan B is £12 a month plus £2 a gigabyte. At how many gigabytes do they cost the same?
1. Write each cost with x for the gigabytes used, then set them equal — which is what "cost the same" means.
  
  $A : 2 x + 8 B : 2 x + 12$
2. Both sides carry 2x, so take 2x off both.
  
  $2 x + 8 = 2 x + 12 - 2 x$
3. The x has gone and what is left is 8 = 12, which is false. No value of x can make it true, so the two plans never cost the same.
  
  $8 = 12 — false, for every x$
4. The useful answer is the one the equation stumbled over: B is £4 more than A at every single x. The plans do not cross because they climb at the same rate, and £4 apart they stay.
  
  $(2 x + 12) - (2 x + 8) = 4 at every x$
Answer

No x works. Equal per-gigabyte rates and different monthly fees means a fixed gap for ever — here £4, whatever you use.
The same offer, written twice 4 steps
One shop charges 2(x + 3). Another charges 2x + 6. Which is cheaper, and at how many books?
1. Expand the first one. Two lots of (x + 3) is 2x + 6, which is what the other shop wrote down.
  
  $2 (x + 3) = 2 x + 6$
2. Setting them equal and taking 2x off both sides leaves 6 = 6 — true, and true without knowing anything about x.
  
  $2 x + 6 = 2 x + 6 - 2 x ⟹ 6 = 6$
3. So every number works. Try 0, 1, 7, −100: the two sides agree every time.
  
  $x = 0, 1, 7, - 100 all true$
4. That is worth naming: the difference between the two prices is not small, it is exactly zero at every x. The two shops wrote one price twice.
  
  $2 (x + 3) - (2 x + 6) = 0 identically$
Answer

Every x works, so neither is cheaper — ever. When both sides are the same expression in disguise, "solve for x" is the wrong question and the right one is "are these actually different?".
Ages: the algebra is easy, the setting up is not 5 steps
Ben is three times as old as Mia. In eight years he will be twice her age. How old are they now?
1. Call Mia's age now x. Then Ben is 3x. In eight years Mia is x + 8 and Ben is 3x + 8.
  
  $Mia = x, Ben = 3 x$
2. "Twice her age" means twice her age then, not twice her age now. That single reading is the whole difficulty: 3x + 8 = 2(x + 8), which expands to 3x + 8 = 2x + 16.
  
  $3 x + 8 = 2 (x + 8) = 2 x + 16$
3. Take 2x off both sides and then 8, and x = 8. Mia is 8 and Ben is 24.
  
  $3 x + 8 = 2 x + 16 - 2 x ⟹ x + 8 = 16 ⟹ x = 8$
4. Check the story, not just the algebra: in eight years Mia is 16 and Ben is 32, and 32 really is twice 16.
  
  $Mia 8 \to 16, Ben 24 \to 32, 32 = 2 \times 16$
5. Read the sentence the other way — as twice her age now — and you get 3x + 8 = 2x, so x = −8. The algebra is just as obedient and the answer is a girl who will be born in eight years.
  
  $3 x + 8 = 2 x ⟹ x = - 8$
Answer

Mia is 8 and Ben is 24; in eight years they are 16 and 32. Three moves of algebra, and one line of reading that decides whether any of them mean anything.

Learning path

Solving for x, from the minus sign upward

Leads to Inequalities the move list — the operations you may do to both sides without changing which values satisfy the equation.

By the 42Math team · Updated 2026-08-26

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

x on both sides - 3x + 4 = x + 10. Subtract x, subtract 4, divide by 2, and x = 3. Then check it against the original: both sides come to 13.
A negative answer - 2x + 9 = 5x. The answer is x = 3 here, but change d and the solution goes negative without anything breaking — a negative solution is an answer, not an error.
Every x works - 2x + 5 = 2x + 5. The two sides are the same expression, so every x satisfies it. There is nothing to solve, and "x = " is the wrong question.
No x works - 2x + 5 = 2x + 9. Subtract 2x and you are left with 5 = 9, which no x can rescue. The equation is not hard; it is false.