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Mathematics

Basic Algebra Practice Questions with Worked Answers

Twelve questions on collecting like terms, substitution and one- and two-step equations. Every solution shows the inverse operation and ends with a check.

The short answer

Twelve questions covering collecting like terms, substitution and one- and two-step equations, ending with two that have the unknown on both sides. Every solution names the inverse operation applied to both sides and finishes by substituting the answer back into the original equation.

The questions

  1. Question 1Foundation1 marks

    Simplify 5a + 3a - 2a

    Show the working

    6a

    1. All three terms are lots of a, so they can be combined.
    2. 5 + 3 - 2 = 6
    3. Answer: 6a
  2. Question 2Foundation2 marks

    Simplify 4x + 7y - x + 2y

    Show the working

    3x + 9y

    1. Group the x terms and the y terms separately. They are not like terms and cannot be combined with each other.
    2. 4x - x = 3x
    3. 7y + 2y = 9y
    4. Answer: 3x + 9y
  3. Question 3Foundation2 marks

    a = 4. Work out the value of 3a + 5.

    Show the working

    17

    1. 3a means 3 × a, so replace a with 4: 3 × 4 + 5.
    2. Multiplication comes before addition: 3 × 4 = 12.
    3. 12 + 5 = 17
  4. Question 4Foundation2 marks

    Solve x + 9 = 14

    Show the working

    x = 5

    1. 9 has been added to x, so subtract 9 from both sides.
    2. x + 9 - 9 = 14 - 9
    3. x = 5
    4. Check: 5 + 9 = 14
  5. Question 5Foundation2 marks

    Solve 6x = 42

    Show the working

    x = 7

    1. x has been multiplied by 6, so divide both sides by 6.
    2. 6x ÷ 6 = 42 ÷ 6
    3. x = 7
    4. Check: 6 × 7 = 42
  6. Question 6Core2 marks

    p = 3 and q = -2. Work out the value of 2p - 5q.

    Show the working

    16

    1. Substitute with brackets so the signs stay visible: 2(3) - 5(-2).
    2. 2 × 3 = 6
    3. -5 × -2 = +10, because a negative multiplied by a negative gives a positive.
    4. 6 + 10 = 16
  7. Question 7Core3 marks

    Solve 3x + 4 = 19

    Show the working

    x = 5

    1. x has been multiplied by 3 and then had 4 added. Undo in reverse order.
    2. Subtract 4 from both sides: 3x = 15.
    3. Divide both sides by 3: x = 5.
    4. Check: 3 × 5 + 4 = 15 + 4 = 19
  8. Question 8Core3 marks

    Solve x/4 - 3 = 2

    Show the working

    x = 20

    1. x has been divided by 4 and then had 3 subtracted. Undo the subtraction first.
    2. Add 3 to both sides: x/4 = 5.
    3. Multiply both sides by 4: x = 20.
    4. Check: 20 ÷ 4 - 3 = 5 - 3 = 2
  9. Question 9Core3 marks

    Expand and simplify 3(2x + 5) - 4x

    Show the working

    2x + 15

    1. Multiply both terms inside the bracket by 3: 3 × 2x = 6x and 3 × 5 = 15.
    2. The expression becomes 6x + 15 - 4x.
    3. Collect the x terms: 6x - 4x = 2x.
    4. Answer: 2x + 15
  10. Question 10Core3 marks

    Solve 5x - 7 = 2x + 8

    Show the working

    x = 5

    1. Move the smaller x term: subtract 2x from both sides, giving 3x - 7 = 8.
    2. Add 7 to both sides: 3x = 15.
    3. Divide both sides by 3: x = 5.
    4. Check each side separately: 5(5) - 7 = 18 and 2(5) + 8 = 18.
  11. Question 11Stretch3 marks

    Solve 4 - 2x = 10

    Show the working

    x = -3

    1. The term in x is -2x, with the minus sign belonging to it.
    2. Subtract 4 from both sides: -2x = 6.
    3. Divide both sides by -2: x = 6 ÷ (-2) = -3.
    4. Check: 4 - 2(-3) = 4 + 6 = 10
  12. Question 12Stretch4 marks

    Solve 2(x - 3) = 5x + 9

    Show the working

    x = -5

    1. Expand the bracket first: 2x - 6 = 5x + 9.
    2. Subtract 2x from both sides: -6 = 3x + 9.
    3. Subtract 9 from both sides: -15 = 3x.
    4. Divide both sides by 3: x = -5.
    5. Check: left side 2(-5 - 3) = 2 × -8 = -16; right side 5(-5) + 9 = -25 + 9 = -16.

Where these go wrong

  • Collecting 4x and 7y into 11xy. Terms with different letters are not like terms and never combine.
  • Substituting q = -2 into -5q and writing -10. Two negatives multiply to a positive, so the term is +10.
  • Dividing by 3 first in 3x + 4 = 19 because the 3 is written next to the x. The addition has to be undone before the multiplication.
  • Reading 4 - 2x as though the coefficient were +2, which gives x = 3 instead of x = -3 - a lost sign rather than a lost method.
  • Expanding 3(2x + 5) as 6x + 5, multiplying only the first term inside the bracket.

Undo the operations in reverse order

In 3x + 4 = 19, two things have happened to x: it was multiplied by 3, and then 4 was added. To get back to x, undo them in the opposite order to the way they were done - take the 4 away first, then divide by 3.

Students who divide by 3 first are following the layout of the page rather than the structure of the expression, and the arithmetic then goes wrong in a way that is hard to spot. The habit worth building is to say the two operations aloud before touching the equation.

The check is part of the answer, not an extra

Every solution below ends with the value substituted back into the original equation. This is not a formality. Algebra is the first topic where a wrong answer looks exactly like a right one - there is no obviously silly result, no negative length, nothing to catch the eye.

For an equation with the unknown on both sides, work out the two sides separately rather than as one line. For 5x - 7 = 2x + 8 with x = 5, that is 18 and 18. If they do not match, the error is somewhere in the working and it will be found in seconds; if they do, the answer is right and there is nothing left to worry about.

Where the negatives do the damage

Three of these twelve are built around signs rather than around method. 2p - 5q with q = -2 has a negative times a negative in the middle of it. 4 - 2x = 10 hides a negative coefficient in plain sight. 2(x - 3) = 5x + 9 has a negative answer, which many students distrust and then talk themselves out of.

None of these need a new technique. They need the term to be read with its sign attached: the coefficient in 4 - 2x is -2, not 2, and dividing by -2 is what produces x = -3. Reading -2x as 2x is the single most common cause of a lost sign at this level.

Common questions

Why do you have to do the same thing to both sides?

Because an equation is a statement that two things are equal, and the only changes that keep it true are ones that treat both sides identically. Subtract 4 from one side alone and the two sides are no longer equal, so the equation you go on to solve is a different one from the one you were given.

Can a solution really be negative?

Yes, and 4 - 2x = 10 gives x = -3. Students who expect an answer to be a positive whole number often abandon correct working at this point. The check settles it: substituting -3 back in gives 4 + 6 = 10, which is the equation, so the answer stands.

What is the difference between simplifying and solving?

Simplifying tidies an expression and the answer still contains letters: 5a + 3a - 2a simplifies to 6a. Solving finds a number that makes an equation true, so the answer is a value: 6x = 42 solves to x = 7. An expression has no equals sign, and it cannot be solved.

Does it matter which side the letters end up on?

No. Solving 5x - 7 = 2x + 8 by moving the 5x instead gives -7 = -3x + 8, then -15 = -3x, then x = 5 - the same answer with more negatives to handle. Moving the smaller x term keeps the coefficient positive, which is why it is worth doing that way round.

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