Solve $7x - 9 = 26$ [2]
- for adding 9 to both sides, $7x = 35$, or for $(26+9)\div 7$M1
- $x = 5$A1
- Answer: $x = 5$
- Examiner tip: Do the inverse operations in reverse order: undo the −9 before dividing by 7.
(a) Solve $\dfrac{x}{4} + 3 = 10$ [1]
(b) Solve $5(y - 2) = 35$ [2]
- (a) $x = 28$B1
- (b) for $5y - 10 = 35$ or $y - 2 = 7$M1
- (b) $y = 9$A1
- Answer: (a) $x = 28$ (b) $y = 9$
- Examiner tip: In (b) dividing both sides by 5 first ($y - 2 = 7$) avoids the common error $5y - 2 = 35$.
Solve $8p + 5 = 3p - 20$ [3]
- for collecting the $p$ terms on one side, e.g. $8p - 3p + 5 = -20$ or $5p + 5 = -20$M1
- for collecting the number terms on the other side, e.g. $5p = -25$M1
- $p = -5$A1
- Answer: $p = -5$
- Examiner tip: Moving $+5$ across gives $-25$, not $-15$; the sign slip loses the accuracy mark.
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Lena is solving the equation $5 - 2(x - 4) = 3x$
Here is her working.
$5 - 2x - 8 = 3x$
$-3 = 5x$
$x = -\tfrac{3}{5}$
(a) Describe the mistake Lena has made. [1]
(b) Solve $5 - 2(x - 4) = 3x$ correctly. [2]
- (a) she wrote $-8$ instead of $+8$ when expanding; accept $-2 \times -4 = +8$ (not $-8$) or 'she got the sign of the 8 wrong'C1
- (b) for $5 - 2x + 8 = 3x$ or $13 = 5x$M1
- (b) $x = \dfrac{13}{5}$ oe, e.g. $2.6$ or $2\tfrac{3}{5}$A1
- Answer: (a) $-2 \times -4$ should be $+8$ (b) $x = \tfrac{13}{5} = 2.6$
- Examiner tip: When a bracket follows a minus sign, multiply every term inside by the negative number.
The diagram shows triangle $ABC$.
The angles, in degrees, are $2x + 15$, $3x - 10$ and $x + 25$.
Show that triangle $ABC$ is isosceles. [4]
- for forming the equation $(2x + 15) + (3x - 10) + (x + 25) = 180$M1
- for simplifying to $6x + 30 = 180$ or $6x = 150$M1
- $x = 25$A1
- for working out all three angles $65°$, $65°$, $50°$ and concluding two angles are equal, so the triangle is isoscelesC1
- Answer: $x = 25$; angles $65°, 65°, 50°$: two equal angles so isosceles
- Examiner tip: For 'show that', finding $x$ is not enough: you must work out the angles and state that two are equal.
The table shows the monthly charges for two mobile data plans.
| Plan | Fixed charge per month | Cost per GB of data |
|---|---|---|
| A | \$15 | \$0.20 |
| B | \$6 | \$0.50 |
(a) Write an equation and solve it to find the number of GB of data, $n$, for which both plans cost the same in a month. [3]
(b) Jo uses 45 GB of data a month. Which plan is cheaper for Jo, and by how much? [2]
- (a) for $15 + 0.2n = 6 + 0.5n$ oe (allow in cents, e.g. $1500 + 20n = 600 + 50n$)M1
- (a) for $9 = 0.3n$ oeM1
- (a) $n = 30$A1
- (b) for working out the cost of either plan for 45 GB, e.g. $15 + 0.2 \times 45 = 24$ or $6 + 0.5 \times 45 = 28.5$M1
- (b) Plan A, cheaper by \$4.50 (accept 4.5)A1
- Answer: (a) $n = 30$ GB (b) Plan A, by \$4.50
- Examiner tip: Set up ONE equation with both plans equal; trial values of $n$ do not earn the method marks when an equation is asked for.
The diagram shows a rectangle and a square. All lengths are in centimetres.
The perimeter of the rectangle is equal to the perimeter of the square.
Work out the area of the rectangle. [4]
- for a correct expression for either perimeter, e.g. $2(2x + 3) + 2(x - 1)$ or $6x + 4$, or $4(x + 4)$ or $4x + 16$M1
- for forming the equation $6x + 4 = 4x + 16$ oeM1
- $x = 6$A1
- $75$ (cm²)A1
- Answer: $75\text{ cm}^2$
- Examiner tip: After finding $x$, go back and answer the question: the mark is for the area, not for $x = 6$.
Solve $\dfrac{x + 3}{2} - \dfrac{2x - 1}{5} = 3$
Show clear algebraic working. [3]
- for multiplying every term by 10 (or a common multiple), e.g. $5(x + 3) - 2(2x - 1) = 30$M1
- for correctly expanding and simplifying, e.g. $5x + 15 - 4x + 2 = 30$ or $x + 17 = 30$M1
- $x = 13$A1
- Answer: $x = 13$
- Examiner tip: The second fraction's whole numerator is subtracted: $-2(2x - 1) = -4x + 2$, not $-4x - 2$.
Sami thinks of a number.
One third of his number, plus 14, is equal to three quarters of his number, minus 1.
Use algebra to find Sami's number. [3]
- for setting up the equation $\dfrac{n}{3} + 14 = \dfrac{3n}{4} - 1$ oeM1
- for clearing fractions and collecting terms correctly, e.g. $4n + 168 = 9n - 12$ or $15 = \dfrac{5n}{12}$M1
- $36$A1
- Answer: $36$
- Examiner tip: 'Use algebra' means a trial-and-improvement answer of 36 with no equation scores no marks.
Priya drives from town $P$ to town $Q$ at an average speed of 60 km/h.
She drives back from $Q$ to $P$ along the same road at an average speed of 40 km/h.
The total driving time for the two journeys is 5 hours.
(a) Work out the distance from $P$ to $Q$. [3]
(b) Priya says, 'My average speed for the whole journey was 50 km/h.' Show that Priya is wrong. [2]
- (a) for an expression for the time of either journey, e.g. $\dfrac{d}{60}$ or $\dfrac{d}{40}$M1
- (a) for forming and starting to solve $\dfrac{d}{60} + \dfrac{d}{40} = 5$ oe, e.g. $2d + 3d = 600$M1
- (a) $120$ (km)A1
- (b) for total distance $\div$ total time, e.g. $240 \div 5$ (ft their distance)M1
- (b) $48$ km/h, which is not 50, so Priya is wrongC1
- Answer: (a) 120 km (b) average speed $= 240 \div 5 = 48$ km/h, not 50
- Examiner tip: Average speed is total distance over total time; averaging the two speeds (50) is the classic wrong answer.
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