Algebraic notation and simplifying expressions – Year 7 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(a\times b\) is written without the multiplication sign, letters in alphabetical order: \(ab\).
   2. The number is written before the letter and the sign is left out: 5 times n is \(5n\).
   3. A letter multiplied by itself is a square, so \(p\times p\) is \(p^{2}\).
   4. Number first, then the letters in order, no signs: \(3xy\).
   5. In algebra we write division as a fraction: \(m\div4\) is \(\dfrac{m}{4}\).
   6. The whole bracket is multiplied by 5, and the number goes in front: \(5(x+2)\).

2. Answer to Question 2

   1. Each pen costs £p and there are 7 pens, so we multiply: the cost is £7p.
   2. In 6 years she will be 6 years older, so we add 6 to her age: \(n+6\).
   3. 4 bags of s sweets give \(4\times s=4s\) sweets, then we add the 3 loose ones: \(4s+3\).
   4. The width is 5 cm less than l, so it is \(l-5\) cm. The perimeter is two lengths plus two widths: \(2l+2(l-5)=2l+2l-10=\)\(4l-10\) cm.

3. Answer to Question 3

   1. \(3a\) and \(5a\) are like terms: \(3+5=8\), so the answer is \(8a\).
   2. All the terms are in x: \(7-2+4=9\), giving \(9x\).
   3. Collect the a terms: \(4a+2a=6a\). Collect the b terms: \(3b-b=2b\). Answer: \(6a+2b\).
   4. p terms: \(5p-3p=2p\). Numbers: \(2+7=9\). Answer: \(2p+9\).
   5. \(x^{2}\) terms: \(6x^{2}-2x^{2}=4x^{2}\). x terms: \(3x+x=4x\). They are unlike terms, so the answer is \(4x^{2}+4x\).
   6. \(ba\) is the same as \(ab\), so all three are like terms: \(3+4-2=5\). Answer: \(5ab\).

4. Answer to Question 4

   1. Multiply the numbers \(3\times4=12\), then the letters \(a\times b=ab\): \(12ab\).
   2. \(2\times5=10\) and \(x\times x=x^{2}\): \(10x^{2}\).
   3. A positive times a negative is negative: \(6\times(-2)=-12\), letters \(pq\): \(-12pq\).
   4. Divide the numbers \(12\div3=4\) and cancel the common x: \(4y\).
   5. \(20\div5=4\) and \(a^{2}\div a=a\): \(4a\).
   6. A negative times a negative is positive: \(4\times3=12\) and \(m\times m=m^{2}\), so \(12m^{2}\).

5. Answer to Question 5

   1. The terms are separated by + and − signs: \(5x^{2}\), \(-3x\) and \(7\). There are 3 terms.
   2. The number in front of x is the coefficient, together with its sign: \(-3\).
   3. The constant term is the number with no letter: 7.
   4. Like terms have exactly the same letter part. Only \(4x\) and \(-x\) are both in x, so they are the like terms.
   5. Any three terms that collect to \(3x+2y\) work, for example \(4x-x+2y\): \(4x-x=3x\), giving \(3x+2y\).

6. Answer to Question 6

   1. y is multiplied by itself three times, so we use the power 3: \(y^{3}\).
   2. There are two a’s, so \(a\times a=a^{2}\), and the 2 goes in front: \(2a^{2}b\).
   3. The bracket is squared: \(3x\times3x=9x^{2}\), so the answer is \(9x^{2}\).
   4. Only x is squared: \(4^{2}=16\), then \(3\times16=\)48.
   5. Work out the bracket first: \(3\times4=12\), then \(12^{2}=\)144. This is different from part (d), because the bracket changes the order.

7. Answer to Question 7

   1. 3 child tickets cost \(3c\) and 2 adult tickets cost \(2a\). Total: \(3c+2a\) pounds.
   2. The second family pays \(2c+3a\). Both families together: \((3c+2a)+(2c+3a)\).
   3. Collect like terms: \(3c+2c=5c\) and \(2a+3a=5a\), giving \(5c+5a\).
   4. Substitute \(c=6\), \(a=9\): \(5\times6+5\times9=30+45=\)£75. Both families pay £75 in total.

8. Answer to Question 8

   1. Change pence to pounds: 5p is £0.05 and 8p is £0.08. The bill is the fixed £12 plus the extras: \(12+0.05t+0.08m\).
   2. Substitute: \(12+0.05\times40+0.08\times25=12+2+2=\)£16.
   3. His sister only pays for texts: \(12+0.05\times60=12+3=\)£15.
   4. A formula starts with the subject and an equals sign: \(B=12+0.05t+0.08m\).

9. Answer to Question 9

   1. Perimeter = two lengths + two widths: \(2x+2(x-3)=2x+2x-6=\)\(4x-6\) metres.
   2. Area = length × width: \(x(x-3)=x^{2}-3x\) square metres.
   3. Substitute \(x=10\) into \(4x-6\): \(40-6=\)34 m.
   4. Cost = 8 × perimeter: \(8(4x-6)=32x-48\). For \(x=10\): \(320-48=\)£272, which is also \(8\times34\).

10. Answer to Question 10

    1. £150 fixed plus £4 for each of the n pupils: \(150+4n\).
    2. Each pupil pays £7 entry, so n pupils cost \(7n\).
    3. Add the two costs: \(150+4n+7n\). Collect \(4n+7n=11n\): \(150+11n\).
    4. Substitute \(n=30\): \(150+11\times30=150+330=\)£480.
    5. We need \(150+11n\le500\), so \(11n\le350\) and \(n\le31.8\). Pupils must be whole, and 32 would cost £502, so the largest number is 31 pupils.

11. Answer to Question 11

    1. The brother gets £2 more than Priya, so we add 2: \(x+2\).
    2. The cousin gets twice Priya’s amount: \(2x\).
    3. Total \(=x+(x+2)+2x\). Collect the x terms \(1+1+2=4\): \(4x+2\).
    4. Substitute \(x=5\): \(4\times5+2=20+2=\)£22.
    5. Yes. From part (d), when \(x=5\) the total is exactly £22, so Priya gets £5, her brother £7 and her cousin £10.

12. Answer to Question 12

    1. Start with \(n\). Multiply by 3: \(3n\). Add 6: \(3n+6\). Divide by 3: \(\dfrac{3n+6}{3}=n+2\). Take away n: \(n+2-n\).
    2. In \(n+2-n\) the n terms cancel, because \(n-n=0\). What is left is 2, whatever number you start with.
    3. Any trick where the starting number cancels works, for example: think of n, double it to get \(2n\), add 10 to get \(2n+10\), halve to get \(n+5\), subtract n to get 5.
    4. Start with 7: double 14, add 10 gives 24, halve 12, subtract 7 gives 5. The trick works.

13. Answer to Question 13

    1. Each new square needs 3 more matchsticks: 4, 7, 10, so the next is \(10+3=\)13.
    2. The pattern goes up by 3, so it starts with \(3n\). For \(n=1\), \(3n=3\) and we need 4, so add 1: \(3n+1\).
    3. Substitute \(n=20\): \(3\times20+1=\)61 matchsticks.
    4. We need \(3n+1\le100\), so \(3n\le99\) and \(n\le33\). Check: \(3\times33+1=100\). She can make 33 squares.

14. Answer to Question 14

    1. Add all six sides: x terms \(3x+2x+x+x+2x+x=10x\), numbers \(2-2+2=2\). Perimeter: \(10x+2\) cm.
    2. Substitute \(x=5\): \(10\times5+2=50+2=\)52 cm.
    3. Form the equation \(10x+2=74\). Subtract 2: \(10x=72\). Divide by 10: \(x=\)7.2. Check: \(72+2=74\).

15. Answer to Question 15

    1. Area = base × height. A: \(2a\times h=\)\(2ah\), B: \(3a\times h=\)\(3ah\), C: \(a\times h=\)\(ah\).
    2. The three areas are like terms: \(2ah+3ah+ah=\)\(6ah\).
    3. The big rectangle has base \(2a+3a+a=6a\) and height h. Perimeter: \(2(6a)+2h=\)\(12a+2h\).
    4. Area: \(6\times4\times5=\)120 cm². Perimeter: \(12\times4+2\times5=48+10=\)58 cm.

16. Answer to Question 16

    1. Company A: £3 plus £2 per mile, \(3+2m\). Company B: £1 plus £2.50 per mile, \(1+2.5m\).
    2. Substitute 2, 4, 6, 8. A: 7, 11, 15, 19. B: 6, 11, 16, 21.
    3. Set the costs equal: \(3+2m=1+2.5m\). Subtract \(2m\) and 1: \(2=0.5m\), so \(m=4\). Both cost £11 at 4 miles.
    4. A: \(3+20=23\), B: \(1+25=26\). Company A is cheaper by \(26-23=\)£3.

17. Answer to Question 17

    1. 3 kg of apples \(3a\), 2 kg of bananas \(2b\), 1 kg of grapes \(g\): \(3a+2b+g\).
    2. 1 kg apples, 4 kg bananas, 2 kg grapes: \(a+4b+2g\).
    3. Collect like terms: \(3a+a=4a\), \(2b+4b=6b\), \(g+2g=3g\): \(4a+6b+3g\).
    4. Mrs Clarke: \(6+3+4=\)£13. Mr Dean: \(2+6+8=\)£16. Total £29. Check with (c): \(8+9+12=29\), so it agrees.

18. Answer to Question 18

    1. Zoe added unlike terms. \(3x\) and \(4y\) have different letters, so \(3x+4y\) cannot be simplified.
    2. Adam treated \(5a\) and \(2a^{2}\) as like terms. They are not, so \(5a-2a^{2}\) stays as it is.
    3. Beth forgot that \(x\times x=x^{2}\). Correct: \(2\times3=6\), so \(6x^{2}\).
    4. Carl only changed one sign. The minus changes both terms: \(-x+3\).

19. Answer to Question 19

    1. Twice packet A is \(2x\), and 3 more gives \(2x+3\).
    2. Half of packet B: \(\dfrac{2x+3}{2}=x+1.5\). Packet C has \(x+1.5\) sweets.
    3. Total \(=x+(2x+3)+(x+1.5)\). x terms: \(4x\); numbers: \(4.5\). Answer: \(4x+4.5\).
    4. For \(x=5\): \(20+4.5=\)24.5. No: \(x+1.5\) always ends in .5, so packet C is never whole and the situation is impossible for real sweets.

20. Answer to Question 20

    1. Add neighbours: \(a+2a=3a\) and \(2a+3a=5a\). Middle row: 3a, 5a.
    2. The top block is the sum of the middle row: \(3a+5a=\)8a.
    3. Middle row: \(x+y\) and \(y+x\). Top: \((x+y)+(y+x)=\)\(2x+2y\).
    4. Middle: \(x+4\) and \(4+x\). Top: \(2x+8=24\). Subtract 8: \(2x=16\), so \(x=\)8. Check: \(16+8=24\).