Multi-Step Equations — Print

Distribute, collect, then undo. Every linear equation, however long, yields to that order.

What the practice questions cover

These questions use 4 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Three sealed bags and seven weights balance two sealed bags and twelve weights. Every bag holds the same amount.

You cannot open a bag. But the scale is level, and that is information.

take the same thing off both pans and it still balances x x x take two bags off each side x one bag left, and five weights — so one bag weighs five every legal step in algebra is that one move, written down

Lift two bags off the left pan. The scale tips — unless you lift two off the right pan as well.

Do both, and one bag plus seven weights now balances twelve weights. Take seven weights off each side.

One bag balances five weights. Without ever opening it, you know each bag holds five.

Every legal step in algebra is that one move: do the same thing to both sides. Distributing, collecting and dividing are just ways of tidying up between moves.

Solving three x plus five equals twenty, one step per line 3x + 5 = 20 the equation 3x = 15 subtract 5 from both sides x = 5 divide both sides by 3 3(5) + 5 = 20 ✓ check in the original

The fixed order

1. Distribute any brackets. 2. Collect like terms on each side. 3. Get all the variables onto one side. 4. Undo the constant, then undo the coefficient.

Doing the steps out of order is not wrong, but it is slower and it makes mistakes far more likely.

Whatever you do to one side, do to the other. That is the only rule underneath all of this.

One line, one move

Write a new line for every single step and say what you did. Four short lines beat one crossed-out line every time.

Variables on both sides

7x - 4 = 3x + 12. Subtract 3x from both sides to clear the smaller variable term: 4x - 4 = 12.

Then +4 to both sides: 4x = 16. Then ÷ 4: x = 4.

Moving the smaller coefficient keeps the remaining one positive, which is easier to work with.

Always check

Put x = 4 back in: left is 7(4) - 4 = 24, right is 3(4) + 12 = 24. Equal, so it is right. Checking takes ten seconds and catches nearly everything.

Brackets and fractions

Distribute first: 2(3x - 5) = 6x - 10. The multiplier hits every term inside, including the one being subtracted.

With a negative outside, every sign inside flips: -4(x - 2) = -4x + 8.

With fractions, multiply every term on both sides by the common denominator to clear them all at once.

The forgotten second term

3(x + 4) = 3x + 4 is wrong. It is 3x + 12. Draw two little arrows from the 3 to remind yourself there are two targets.

No solution, or every solution

Sometimes the variables all cancel. If what is left is false, like 5 = 9, there is no solution.

If what is left is true, like 7 = 7, then every number works — the two sides were the same expression all along.

Neither case is a mistake. They are real answers, and they show up on tests deliberately.

Turning words into an equation

Name the unknown with a letter and write down what it means: "let x be the number of hours".

Translate one phrase at a time. "Three more than twice a number" is 2x + 3, not 3 + 2 and not 2(x + 3).

Solve, then answer the question that was actually asked — sometimes the question wants x + 5, not x.

Worked examples

Variables on both sides

Solve 5x + 8 = 2x - 7.

  1. Subtract 2x from both sides: 3x + 8 = -7 Clear the smaller variable term.
  2. Subtract 8 from both sides: 3x = -15 Undo the constant next.
  3. Divide both sides by 3: x = -5 Undo the coefficient last.
  4. Check: 5(-5) + 8 = -17 and 2(-5) - 7 = -17 ✓ Both sides agree, so it is correct.

x = -5

Brackets on both sides

Solve 4(x - 3) = 2(x + 5).

  1. Distribute: 4x - 12 = 2x + 10 Both multipliers hit both terms.
  2. Subtract 2x: 2x - 12 = 10 Variables to one side.
  3. Add 12: 2x = 22 Then undo the constant.
  4. Divide by 2: x = 11 Check: 4(8) = 32 and 2(16) = 32 ✓

x = 11

A word problem

A taxi charges a $3 flag fee plus $2 per mile. A trip cost $27. How many miles was it?

  1. Let m be the number of miles. Name the unknown before anything else.
  2. Cost = 3 + 2m, so 3 + 2m = 27 Fixed fee plus a rate times the miles.
  3. Subtract 3: 2m = 24 Undo the flag fee.
  4. Divide by 2: m = 12 miles Check: 3 + 2(12) = 27 ✓

12 miles

Words to know

Like terms
Terms with exactly the same variable part: 5x and -2x, but not 5x and 5x2.
Distribute
Multiply a bracket out: 3(x + 4) = 3x + 12.
Inverse operation
The move that undoes another: subtract undoes add, divide undoes multiply.
Solution
A value that makes the two sides genuinely equal.

Mistakes to watch for

Distributing to only the first term. 3(x + 4) is 3x + 12, not 3x + 4, because the bracket means "three lots of the whole thing" — and the whole thing includes the 4. Draw two little arrows from the 3 so you can see both targets.

Forgetting that a negative multiplier flips every sign inside. -4(x - 2) is -4x + 8: the second sign changes too, because -4 × -2 is positive. Missing it is the single most common source of a sign error.

Doing something to one side only. If you subtract 3x on the left you must subtract it on the right, or the two sides stop being equal and every line after that is about a different problem.

Answering the wrong question. Some word problems ask for the total, not for x. Solve, then read the question again — it takes five seconds and it is where a lot of correct algebra loses its marks.

The word algebra comes from the Arabic al-jabr, meaning "the reunion of broken parts" — the act of moving a term to the other side to restore balance.

It is from the title of a book written around 820 AD by al-Khwārizmī in Baghdad: al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wa-l-muqābala.

His own name, worn down by centuries of Latin translation, gave us the word algorithm.

Problem 1

Solve 4x + 3 = 2x + 11.

Worked answer

4

2x + 3 = 11

2x = 8

x = 4

Problem 2

Solve 7y - 5 = 3y + 15.

Worked answer

5

4y - 5 = 15

4y = 20

y = 5

Problem 3

Solve 3(x + 4) = 27.

Worked answer

5

3x + 12 = 27

3x = 15

x = 5

Problem 4

Solve 2(n - 7) = 10.

Worked answer

12

2n - 14 = 10

2n = 24

n = 12

Problem 5

Solve 5x + 2x - 6 = 22.

Worked answer

4

7x - 6 = 22

7x = 28

x = 4

Problem 6

Solve 5(2x - 3) = 3(x + 6).

Worked answer

33/7

10x - 15 = 3x + 18

7x = 33

x = 337

Problem 7

Solve 9 - 4x = 2x - 15.

Worked answer

4

9 = 6x - 15

24 = 6x

x = 4

Problem 8

Solve -3(x - 5) = 21.

Worked answer

-2

-3x + 15 = 21

-3x = 6

x = -2

Problem 9

Solve x4 + 3 = 10.

Worked answer

28

x4 = 7

x = 28

Problem 10

Solve 2x + 15 = 3.

Worked answer

7

2x + 1 = 15

2x = 14

x = 7

Problem 11

How many solutions does 4(x + 2) = 4x + 8 have?

  1. Infinitely many
  2. Exactly one
  3. None
  4. Exactly two
Worked answer

Infinitely many

4x + 8 = 4x + 8

The two sides are identical

Every value of x works

Problem 12

Solve 2(x + 3) = 2x + 1. If there is no solution, answer "none".

Worked answer

none

2x + 6 = 2x + 1

6 = 1, which is false

So there is no solution

Problem 13

A rental costs $25 plus $7 per hour. The bill was $67. How many hours?

Worked answer

6

7h = 42

h = 6

6 hours

Problem 14

Two numbers differ by 8 and add to 46. Find the larger one.

Worked answer

27

x + (x + 8) = 46

2x = 38, so x = 19

The larger is 27

Problem 15

Solve 6 - 2(3x - 4) = 4x + 9.

Worked answer

1/2

6 - 6x + 8 = 4x + 9

14 - 6x = 4x + 9

5 = 10x

x = 12

Problem 16

A phone plan charges $30 plus $0.05 per text. Another charges $20 plus $0.09 per text. At how many texts do they cost the same?

Worked answer

250

30 + 0.05t = 20 + 0.09t

10 = 0.04t

t = 250 texts

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