Solving an equation can end with the variable gone. If what is left is true, every number is a solution; if it is false, none is. Neither outcome is a mistake, and neither of them says x = 0.
Core ideas
An equation is a claim, and a claim about every value of x can be true always, false always, or true for some values only. There is no fourth possibility for a linear equation.
When the x terms cancel, the equation has finished telling you what it knows. What remains is a statement about numbers with no x in it, and that statement is either true or false — which is the answer.
2(x + 3) and 2x + 6 are the same expression; the distributive law says so. An equation between them is true for every value because there was only ever one expression in the room.
Reduce it to px = q. If p ≠ 0 there is exactly one solution. If p = 0 the statement is 0 = q, which does not mention x and so holds for every number or for no number at all.
The material
The names, the theorem behind them, and the two lines of working that get misread.
An equation is conditional when some values of the variable make it true and some make it false. It is an identity when every value makes it true. It is a contradiction when no value does.
The solution sets are, in order: some particular numbers; every real number; and nothing at all.
Every linear equation in x can be rewritten as px = q with p and q numbers. If p ≠ 0 it has exactly one solution, q/p. If p = 0 and q = 0 every number is a solution. If p = 0 and q ≠ 0 there are none.
Simplify both sides and subtract the right side from both. Every term is either a multiple of x or a constant, so what is left is px − q = 0 for some numbers p and q, that is px = q.
If p ≠ 0, divide both sides by p. Division by a nonzero number is reversible, so x = q/p is equivalent to the original and is the only solution.
If p = 0 the statement reads 0 = q. It contains no x, so its truth does not depend on x: it is true for every number when q = 0, and false for every number otherwise.
□
So the working itself tells you which case you are in. Simplify, subtract, and look at what survives.
Read the first line as: whatever x is, the two sides are the same number, so the claim holds. Read the second as: whatever x is, the left side is exactly 1 more than the right, so the claim never holds. The third constrains x to a single value.
0 = 0 is not x = 0. It is a true statement with no x in it, and the correct answer is "every real number". Reporting x = 0 names one solution out of infinitely many and rejects all the rest.
0 = 5 is not x = 5, and it is not x = 0 either. It is a false statement with no x in it, and the correct answer is "no solution". Both misreadings come from the same habit — expecting the last line of the working to have the shape x = something.
Draw each side as a graph, as the lab does. A conditional equation is two lines crossing at one point; an identity is one line drawn twice; a contradiction is two parallel lines that never meet. The graphs are a preview — lines are Lines, Functions and Graphs — and the argument above does not depend on them.
Two cautions about identities. First, an identity is what you are proving every time you expand a bracket, so they are not exotic. Second, x/2 + x/3 = (5/6)x is an identity because 1/2 + 1/3 = 3/6 + 2/6 = 5/6, and the only way to be sure of that is to do the addition.
3x + 5 = 3x + 5 + x. The two sides begin identically, which invites the answer "every number". Subtracting 3x + 5 from both sides leaves 0 = x, so it is conditional with the single solution x = 0. Check: at x = 0 both sides are 5; at x = 1 the left is 8 and the right is 9. The extra term on the right is the whole story, and only simplifying finds it.
Interactive
The three things an equation can be
Method
Do not judge by appearance. Simplify, subtract, then read.
Brackets out, like terms combined, fractions cleared if there are any. Two sides that look alike may differ by a term you have not multiplied out yet.
Everything on one side, 0 on the other. This is the step the theorem describes, and it is what produces px = q in a readable form.
Still a term in x: conditional, so finish solving it. 0 = 0 or any true numerical statement: identity. A false numerical statement such as 6 = 5: contradiction.
"Every real number", "no solution", or the number itself. Never x = 0 for an identity and never x = 5 for 0 = 5. Then test a couple of values in the original to confirm the verdict.
Between A and B only the right-hand constant changes, and that alone turns an identity into a contradiction. C changes the coefficient of x on the right, which is what makes a conditional equation possible at all.
In B the tests do not prove there is no solution — two failed values leave infinitely many untested. The proof is the line 6 = 5, which contains no x and is false, so no value of x can rescue it. The tests are there to catch an arithmetic slip in the simplifying.
D is the one to watch for. Distributing the minus correctly gives 3x − x + 4; distributing it over the first term only gives 3x − x − 4 = 2x − 4, and the equation would then look like a contradiction. A sign error can manufacture any of the three outcomes, which is why the simplifying comes first and gets checked.
For a faded classification, expand only the left side of 4(x − 2) + 3 = 4x − 5 to get 4x − 5. Stop there, predict the outcome and solution set, then subtract one whole side and test two different values before comparing with the lab.
Classify and solve these without the lab: 5x − 2 = 2x + 7, 3(x + 4) = 3x + 12, and 2(x − 1) = 2x + 1. For each, show the simplified remainder and write either the value, every real number, or no solution. Then use px = q to explain why no linear equation has exactly two solutions.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.