Lesson · Linear Equations and Inequalities · Equations

Variables on Both Sides

When x appears on both sides, no amount of arithmetic on one side alone will finish the job. The new move is to subtract a whole variable term from both sides, which is the addition property with an expression in place of a number, and the side you collect on is your choice.

Open the interactive lesson Practice

5x + 2 = 3x − 8 subtract 3x 2x + 2 = −8 x collected left 5x + 2 = 3x − 8 subtract 5x 2 = −2x − 8 x collected right both routes end at x = −5; the side you pick only decides a sign

One equation, two legal routes

Core ideas

What this lesson establishes

The move that makes this work is not new. What is new is applying it to something that contains the variable.

A variable term can be subtracted from both sides like any other quantity

Once x has a value, 3x is a number. The addition property of equality was stated for numbers and it never asked where the number came from, so "subtract 3x from both sides" is the same licence as "subtract 4 from both sides".

Which side you collect on is free

Both choices give the same solution. Collecting on the side that already has the larger coefficient leaves you dividing by a positive number at the end, which is worth a few seconds of thought before you start.

Simplify each side completely before moving anything across

Brackets multiplied out, like terms combined, on both sides. Moving a term out of an unsimplified side is how a term gets counted twice or dropped.

The material

Getting the unknown onto one side

Why the move is legal, why either direction works, and the sign that goes wrong.

5x + 2 = 3x − 8 cannot be finished by “One- and Two-Step Equations”'s routine. Undoing the + 2 leaves 5x = 3x − 10, which still has x in two places. Something has to remove one of them.

Theorem · Subtracting an expression from both sides

If A = B then A − C = B − C, whatever expression C is.

For any particular value of x, A, B and C are numbers, and the statement is then the ordinary addition property of equality. The move is reversible — add C back — so the solution set is unchanged.

Take C = 3x and the equation loses its right-hand variable term:

5x + 2 = 3x − 8 − 3x both sides 2x + 2 = −8 − 2 both sides 2x = −10 ÷ 2 both sides x = −5 check left 5·(−5) + 2 = −25 + 2 = −23 right 3·(−5) − 8 = −15 − 8 = −23

Now the same equation collecting on the right instead. Subtract 5x from both sides:

5x + 2 = 3x − 8 − 5x both sides 2 = −2x − 8 + 8 both sides 10 = −2x ÷ (−2) both sides −5 = x

Same answer, one extra sign to keep track of. −5 = x and x = −5 say the same thing; an equation may be read in either direction.

Choosing the side

Subtract the smaller of the two coefficients and the one that survives is positive. In 3x − 7 = 8x + 3 the coefficients are 3 and 8, so subtracting 3x leaves −7 = 5x + 3 and a division by +5. Subtracting 8x leaves −5x − 7 = 3 and a division by −5. Both are correct; one has fewer places to lose a minus sign.

Simplify first

Brackets and like terms come before any collecting, and the minus sign in front of a bracket reaches every term inside it:

2(x − 3) + 4 = 3(x + 1) − 5 left 2x − 6 + 4 = 2x − 2 right 3x + 3 − 5 = 3x − 2 2x − 2 = 3x − 2 − 2x both sides −2 = x − 2 + 2 both sides 0 = x check left 2(0 − 3) + 4 = −6 + 4 = −2 right 3(0 + 1) − 5 = 3 − 5 = −2

x = 0 is a solution like any other. It is a specific number that makes the statement true, and it is not the same as "no solution" — a confusion “Identities and Contradictions” exists to clear up.

The subtraction trap is 3x − (x − 4). The minus applies to the whole bracket: 3x − x + 4 = 2x + 4. Writing 3x − x − 4 changes the equation, and it is the error most likely to survive unnoticed, because the answer it produces looks perfectly ordinary.

Example · When every x disappears

Collecting on 6x + 1 = 6x − 4 gives 1 = −4, with no x left anywhere. Nothing has gone wrong: the equation is telling you it has no solution, since 1 is not −4 whatever x is. Compare 6x + 1 = 6x + 1, which collapses to 0 = 0 and is true for every number. “Identities and Contradictions” is about reading those two lines correctly.

Interactive

The variable on both sides

Collect it left or right — both work

Method

Solving with the variable on both sides

One extra step at the front; the rest is “One- and Two-Step Equations” unchanged.

Simplify both sides separately

Multiply out every bracket, distributing any minus sign in front of one over all of its terms, then combine like terms. Do this on both sides before anything moves.

Pick a side for the variable and subtract that variable term from both sides

Subtracting the smaller coefficient keeps the surviving one positive. Write the operation down: "subtract 3x from both sides" is a step you can check.

Move the constants the other way

Add or subtract so that every plain number is on the side without the variable. You are now looking at px = q, which is “One- and Two-Step Equations”'s last line.

Divide by the coefficient and check in the original

Sign included. Then substitute into the equation as it was first written — both sides had a variable in them, so both sides have to be evaluated.

Collecting the unknown

Score 0 / 0

Choose the answer you can justify.

One equation twice, and one with brackets

The first is solved both ways to show the answer does not depend on the choice. There is one answer, so it is checked once.

A. 5x + 2 = 3x − 8 collecting on the LEFT collecting on the RIGHT − 3x 2x + 2 = −8 − 5x 2 = −2x − 8 − 2 2x = −10 + 8 10 = −2x ÷ 2 x = −5 ÷ (−2) −5 = x check left 5·(−5) + 2 = −23 right 3·(−5) − 8 = −23 agrees B. 2(x − 3) + 4 = 3(x + 1) − 5 expand 2x − 2 = 3x − 2 − 2x −2 = x − 2 + 2 0 = x check left 2(0 − 3) + 4 = −2 right 3(0 + 1) − 5 = −2 agrees

In A the two routes differ only in the final division: by 2 on the left route, by −2 on the right. That is the whole cost of the choice, and it is why collecting where the coefficient is larger is worth doing.

In B the answer is x = 0. The check evaluates both sides at 0 and gets −2 twice, which is what makes it a solution. A reader who expected "no solution" from seeing a zero has misread the result, not the algebra.

For a faded pass, take 4x − 7 = 9x + 8. The smaller coefficient is 4, so begin by subtracting 4x from both sides. Finish the constant move and division, solve the same equation by subtracting 9x instead, and check that both routes give the same signed value.

Common mistakes

Adding a variable term to one side and subtracting it from the otherThe property says the same operation on both sides. Going from 5x + 2 = 3x − 8 to 8x + 2 = −8 has added 3x on the left and subtracted it on the right, and the resulting equation has a different solution.

Distributing a minus sign over only the first term of a bracket3x − (x − 4) is 3x − x + 4, that is 2x + 4. Writing 3x − x − 4 is a different expression, and everything downstream of it is answering a different question.

Reading x = 0 as "no solution"They are different results. x = 0 says exactly one number works and names it; "no solution" says none does. “Identities and Contradictions” gives the line of working that genuinely means the second thing, and it does not look like x = 0.

Check your understanding

✓

Finish when the choice of side is deliberate and every sign survives the collecting.

Solve 3(x − 2) + 4 = 5x − 8 without the lab. Expand first, choose the collecting side deliberately, carry the equation to x = a number, and check in the bracketed original. Then collect the other way and verify that the answer is unchanged.

Note: The lab runs both routes at once precisely so that the answer can be seen not to depend on the choice. If the two columns ever disagreed, one of them would contain an illegal step — and that is a useful way to think about what "legal" means here: a legal move is one that cannot change the answer.