Printable · GCSE Foundation · ages 14-16

Algebra worksheet — GCSE Foundation

Fifteen questions across the algebra statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.

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Algebra worksheet — GCSE Foundation

Maths · GCSE Foundation · Algebra · 15 questions

MathsUKwww.geekhero.co.uk

1.Solve the inequality 4x + 1 > 2x + 9.
(a)x > 5
(b)x < 5
(c)x < 4
(d)x > 4
2.Solve 3x + 14 = 8x − 6
(a)x = 15
(b)x = 2.5
(c)x = 4
(d)x = −1.6
3.Make x the subject of the formula y = (x − 4)/5.
(a)x = 5(y + 4)
(b)x = 5y − 4
(c)x = y/5 + 4
(d)x = 5y + 4
4.A rule turns each input into an output. An input of 0 gives an output of −1, an input of 1 gives an output of 1, and an input of 2 gives an output of 3. Work out the rule, writing the input as x and the output as y.
(a)y = x − 1
(b)y = 2x − 1
(c)y = 3x − 1
(d)y = 2x + 1
5.Expand 2(x − 13)
(a)2x + 26
(b)2x − 11
(c)2x − 13
(d)2x − 26
6.Explain why the equation 4x + 3 = 4x + 10 has no solution for x.
(a)Subtracting 4x gives 3 = 10, which is never true.
(b)x would have to be negative, but x must be positive.
(c)It's true for every x, so there are infinite solutions.
(d)x = 7, since 10 − 3 = 7.
7.A quadratic graph has equation y = (x − 4)². A student says this graph crosses the x-axis at two different points. Explain why the student is wrong.
(a)It touches the x-axis once, only at x = 4.
(b)It never touches the x-axis at all.
(c)It crosses twice, at x = 4 and x = −4.
(d)It crosses twice, at x = 2 and x = −2.
8.Work out the gradient of the straight line with equation 3y = 12 − 6x.
(a)2
(b)4
(c)−2
(d)−6
9.The point A lies on the x-axis. Which statement about A must be true?
(a)y = 0
(b)x = 0
(c)x > 0
(d)x = y
10.A stack of firewood has 3 logs in the top layer. Each layer below has 4 more logs than the layer above it. Work out the number of logs in the 6th layer from the top.
(a)23
(b)27
(c)24
(d)20
11.The diagram shows the graph of the cost, in pounds, of a taxi journey plotted against the distance travelled, in miles, for journeys of up to 4 miles. The same fixed charge and the same cost per mile apply to longer journeys. Work out the cost of a 6-mile journey.
(a)£24
(b)£14
(c)£12
(d)£8
12.Which of these points lies in the first quadrant?
(a)(−1, −4)
(b)(4, −1)
(c)(−2, 3)
(d)(3, 5)
13.Which expression means the same as 3 × a × b?
(a)3ab
(b)3a + b
(c)3 + a + b
(d)a³b
14.Write down the coefficient of p in the expression 7 − 4p + 9q.
(a)7
(b)4
(c)9
(d)−4
15.The line y = 3x + 6 meets the x-axis at one point. Work out the coordinates of that point.
y = 3x + 6
(a)(2, 0)
(b)(−2, 0)
(c)(6, 0)
(d)(0, 6)

Answer key

(d) x > 4 — Method: collect the x terms on one side and the numbers on the other, then divide by the coefficient of x; dividing by a positive number leaves the sign as it is. Working: subtracting 2x from both sides of 4x + 1 > 2x + 9 gives 2x + 1 > 9; subtracting 1 from both sides gives 2x > 8; dividing both sides by 2 gives x > 4. Answer: x > 4. The distractors: x < 4 comes from turning the sign round while dividing by 2; x > 5 comes from adding the 1 to the 9 instead of subtracting it, giving 2x > 10; x < 5 comes from making both of those mistakes together.
(c) x = 4 — Method: with an unknown on both sides, subtract the smaller x term from both sides so that all the x is on one side, then collect the numbers on the other. Working: subtracting 3x from both sides gives 14 = 5x − 6; adding 6 to both sides gives 20 = 5x; dividing both sides by 5 gives x = 4. Checking: 3 × 4 + 14 = 26 and 8 × 4 − 6 = 26. Answer: x = 4. The distractors: x = 2.5 comes from taking 3x off the left-hand side only, leaving 14 = 8x − 6 and so 8x = 20; x = −1.6 comes from changing the sign of the 8x when it is moved across but leaving the sign of the 14 unchanged, giving 3x − 8x = −6 + 14 and so −5x = 8; x = 15 comes from reaching 5x = 20 correctly and then subtracting 5 instead of dividing by 5.
(d) x = 5y + 4 — To make x the subject of y = (x − 4)/5, first multiply both sides by 5 to clear the fraction: 5y = x − 4, then add 4 to both sides: x = 5y + 4. Writing x = 5y − 4 multiplies correctly but keeps the minus sign on the 4 instead of changing it to a plus when moving it across. Writing x = y/5 + 4 divides by 5 instead of multiplying, the wrong inverse of the fraction. Writing x = 5(y + 4) adds 4 before multiplying by 5, reversing the correct order of the two steps. The correct rearrangement is x = 5y + 4.
(b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
(d) 2x − 26 — Method: multiply each term inside the bracket by the 2 in front of it, and keep the sign that belongs to each term. Working: 2 × x = 2x and 2 × 13 = 26; the bracket contains a subtraction, so the second term is subtracted. Answer: 2x − 26. The distractors: 2x − 13 comes from multiplying only the x by 2 and copying the 13 across unchanged; 2x + 26 comes from multiplying both terms correctly but losing the minus sign that belongs to the second term; 2x − 11 comes from subtracting 2 from 13 instead of multiplying 13 by 2.
(a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
(a) It touches the x-axis once, only at x = 4. — (x − 4)² is a square, so it equals zero only when x − 4 = 0, that is at x = 4 — the curve just touches the x-axis there rather than crossing it, since a square cannot be negative on either side to cross through. Saying it crosses at x = 4 and x = −4 wrongly introduces a plus-or-minus, as if taking a square root of x, rather than recognising the bracket is already squared and only zero once. Saying it never touches the x-axis forgets that a squared term CAN equal zero, even though it can never be negative. Saying it crosses at x = 2 and x = −2 confuses (x − 4)² with the different expression x² − 4.
(c) −2 — Method: rearrange the equation into the form y = mx + c, then read off the gradient. Working: 3y = 12 − 6x, so dividing every term by 3 gives y = 4 − 2x, so the gradient is −2. Answer: the gradient is −2. 2 comes from dropping the negative sign after dividing by 3. 4 comes from using the y-intercept, 4, instead of the gradient. −6 comes from reading off the coefficient of x before dividing the whole equation by 3.
(a) y = 0 — Method: in the pair (x, y) the first coordinate measures how far left or right of the origin a point is and the second how far above or below the x-axis it is, so a point on an axis has one of those measurements equal to zero. Working: the x-axis is the horizontal line through the origin, so a point sitting on it is neither above nor below that line and its second coordinate is zero, while its first coordinate may be positive, negative or zero. Answer: y = 0. The distractors: x = 0 is the condition for lying on the y-axis, the other axis; x > 0 comes from assuming a point on the x-axis must be to the right of the origin, which is true only of part of that axis; x = y holds only at the origin, which is one point of the x-axis rather than a property shared by all of them.
(a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
(b) £14 — Method: read the fixed charge (the cost at 0 miles) and the rate (the cost per extra mile) from the graph, then use them to work out the cost for a distance beyond the part that is plotted. Working: the graph shows a fixed charge of £2 at 0 miles, and the cost rises by £2 for every extra mile, so for 6 miles the cost is £2 + (£2 × 6) = £2 + £12 = £14. Answer: £14. Distractor refutation: £12 comes from multiplying the rate by the distance and leaving out the £2 fixed charge. £8 comes from misreading the rate as £1 per mile instead of £2 per mile. £24 comes from adding the fixed charge to the rate first and then multiplying the total by the distance, instead of multiplying the rate by the distance and then adding the fixed charge.
(d) (3, 5) — Method: the quadrant a point lies in is decided by the signs of its two coordinates, and in the first quadrant both coordinates are positive. Working: (−2, 3) has a negative x-coordinate, so it sits to the left of the y-axis; (4, −1) has a negative y-coordinate, so it sits below the x-axis; (−1, −4) has both coordinates negative; only (3, 5) has a positive x-coordinate and a positive y-coordinate. Answer: (3, 5). The distractors: (−2, 3) is chosen by candidates who check only the y-coordinate and take a positive height as enough; (4, −1) is chosen by those who check only the x-coordinate; (−1, −4) is chosen by those who number the quadrants from the bottom left, so that the region with two negative coordinates is called the first.
(a) 3ab — 3 × a × b means 3, a and b are all multiplied together, and in algebraic notation this is written with no multiplication signs: 3ab. Writing 3 + a + b turns every multiplication into an addition, giving a completely different expression. Writing a³b misreads the 3 as a power on a rather than as a coefficient in front of both letters. Writing 3a + b multiplies the 3 by a correctly but then adds b instead of also multiplying it in. The expression that means 3 × a × b is 3ab.
(d) −4 — The coefficient of a letter is the number multiplying it, taken with the sign written in front of that number. The term containing p is being subtracted, so the term is −4p and the number multiplying p is −4. Quoting 4 drops the sign; 7 is a constant term with no letter attached to it; 9 is the number multiplying q, which is a different letter.
(b) (−2, 0) — Method: a graph meets the x-axis where the y-value is 0, so setting y = 0 turns the equation into a linear equation in x. Working: 0 = 3x + 6 gives 3x = −6, so x = (−6) ÷ 3 = −2 and the meeting point is (−2, 0). Answer: (−2, 0). The distractors: (2, 0) comes from solving 3x = −6 and then dropping the minus sign from the result; (0, 6) is the y-axis crossing, found by substituting x = 0 instead of y = 0; (6, 0) comes from reading the constant 6 straight off as the x-coordinate, without dividing by 3 and without changing its sign.

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