2x − 1 = x + 4 has the unknown on both sides, and the right-hand one is written as a bare x. That bare x is where the trouble is: it has a coefficient of 1, and because the 1 is not written down it is easy to treat as though it were not there. Subtract x from both sides. On the left, 2x − x is 1x, ...
Type the equation as it is written, brackets and all. Three answers are possible and all three are real: one value, no value, or every value. When the unknown cancels out that is not a failure — it is the answer, and which of the two it is depends on the constant left behind. Anything that is not linear is refused rather than fudged: x² and an unknown in a denominator both get a reason, not a number. Decimals are read as exact fractions, so 0.1 + 0.2 is 3/10 here rather than 0.30000000000000004.
Brackets, fractions, decimals and the unknown on both sides all work. Any single letter can be the unknown.
ONE SOLUTION
x = 5
Both sides come to 9 at this value, checked against the equation as you typed it.
| As typed | 2x - 1 = x + 4 |
| Tidied | 2x − 1 = x + 4 |
| All on one side | x − 5 = 0 |
THE ANSWER PUT BACK INTO WHAT YOU TYPED
left side → 9
right side → 9 ✓
COEFFICIENT OF X
1
it survives, so there is one solution
CONSTANT LEFT OVER
−5
what is left after moving across
AS A DECIMAL
5
exact — the answer is a whole number
BOTH SIDES THERE
9
the two sides agree at the answer
WHAT IS HAPPENING
2x − 1 = x + 4
Brackets multiplied out, numbers combined, and everything written as a multiple of the unknown plus a constant. Nothing has been moved across the equals sign yet — this is only tidying.
x − 5 = 0
Subtracting the right-hand side from the left. What matters now is the coefficient of the unknown: if it survives there is one solution, and if it cancels the answer depends entirely on the constant left behind.
x = 5 ÷ 1 = 5
The coefficient of the unknown is not zero, so there is exactly one value that works. Kept as a fraction, because most solutions are not whole numbers and a rounded one does not satisfy the equation.
left 9 · right 9
Both sides are evaluated at the answer from the expressions as they were typed, not from the tidied version. If those two numbers differ, something went wrong earlier and this is where it shows.
Why there are exactly three answers. Each side of a linear equation describes a straight line, and asking where they are equal is asking where the lines meet. Two straight lines can cross once, run parallel and never meet, or be the same line and meet everywhere. There is no fourth possibility, which is why an equation like this can never have exactly two solutions — and why the case that looks like a dead end, the one where the unknown vanishes, is really the graph telling you the lines were parallel or identical all along.
Exact fractions throughout · the answer is substituted into the equation as you typed it
Read the bare x on the right as 1x. Writing the invisible coefficient in makes the next step obvious.
Subtract x from both sides: 2x − x leaves x on the left, and the right becomes 4.
Add 1 to both sides, giving x = 5. No division is needed because the coefficient is already 1.
Check both sides of the original: 2(5) − 1 = 9 and 5 + 4 = 9. They agree.
2x − 1 = x + 4 Write the right-hand x as 1x while you work, and the first move stops being ambiguous: 2x − 1 = 1x + 4 Subtract x from both sides: 2x − x − 1 = 1x − x + 4 x − 1 = 4 On the left, 2x − x = 1x = x. On the right the x terms cancel completely, which is the point of the move. Add 1 to both sides: x = 5 Check each side of the original separately: left: 2(5) − 1 = 10 − 1 = 9 right: 5 + 4 = 9 ✓ The mistake to avoid is treating 2x − x as 2x, as though a bare x had nothing to subtract. It has a coefficient of 1 like any other term. That slip gives 2x − 1 = 4, then 2x = 5 and x = 5/2 — an answer that looks reasonable and fails the check: 2(5/2) − 1 = 4, while 5/2 + 4 = 13/2. The two sides do not agree, and that disagreement is the check doing its job.
| Rule | What it says | Why |
|---|---|---|
| What counts as linear | the unknown to the first power | No x², no x under a division, no x inside a root. |
| One solution | the x terms do not cancel | The usual case. Divide by what is left of the coefficient. |
| No solution | 2x + 1 = 2x + 3 | The x terms cancel and the constants disagree. Parallel lines. |
| Every number works | 2(x+1) = 2x + 2 | Both sides are the same expression. An identity, not an equation. |
| Cancelling is not failure | it is the answer | Which of the two answers depends on what is left behind. |
| Do the same to both sides | add, subtract, multiply, divide | Any of the four, as long as you never divide by zero. |
| Clearing fractions | multiply by the common denominator | Legitimate because it is done to both sides equally. |
| Expanding brackets | multiply every term inside | The commonest slip is missing the last term of a long bracket. |
| A negative outside a bracket | −(x − 3) = −x + 3 | Both signs flip. The second one is the one people miss. |
| Most answers are fractions | 71.3% of small cases | A whole-number answer is the exception, not the norm. |
| Decimals are not decimal | 0.1 + 0.2 ≠ 0.3 in binary | Typed decimals are read as exact fractions here. |
| Two letters is a system | x + y = 3 | One equation with two unknowns needs a second equation. |
| Check by substituting | put the answer back in | Into the original, not the tidied version — that is the real check. |
| The graph reading | each side is a line | The solution is where they cross; parallel means none, identical means all. |
x = 5. Subtracting x from both sides gives x − 1 = 4, and adding 1 gives 5. Both sides of the original agree at that value: 2(5) − 1 = 9 and 5 + 4 = 9. No division is needed at the end, because the coefficient of x is already 1.
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Last updated: August 31, 2026 · Exact fractions throughout, never floating point · When the unknown cancels, that is the answer rather than a failure.