Solving one-step and two-step equations – Year 7 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Undo “add 7” by subtracting 7 from both sides: \(x=15-7=\)8. Check: \(8+7=15\).
   2. Undo “subtract 9” by adding 9: \(y=4+9=\)13. Check: \(13-9=4\).
   3. Undo “multiply by 5” by dividing by 5: \(a=45\div5=\)9.
   4. Undo “divide by 4” by multiplying by 4: \(m=6\times4=\)24.
   5. Add 3.5 to both sides: \(x=8+3.5=\)11.5.
   6. Multiply both sides by 3: \(n=-5\times3=\)\(-15\). Check: \(-15\div3=-5\).

2. Answer to Question 2

   1. Subtract 4: \(3x=15\). Divide by 3: \(x=5\). Check: \(15+4=19\).
   2. Add 2: \(5y=25\). Divide by 5: \(y=5\). Check: \(25-2=23\).
   3. Subtract 9: \(2a=-6\). Divide by 2: \(a=-3\). Check: \(-6+9=3\).
   4. Subtract 6: \(\dfrac{m}{2}=5\). Multiply by 2: \(m=10\). Check: \(5+6=11\).
   5. Add 7: \(4n=-8\). Divide by 4: \(n=-2\). Check: \(-8-7=-15\).
   6. Multiply by 4: \(x-3=20\). Add 3: \(x=23\). Check: \(20\div4=5\).

3. Answer to Question 3

   1. Expand: \(3x+6=21\). Subtract 6: \(3x=15\), so \(x=5\).
   2. Expand: \(5y-5=30\). Add 5: \(5y=35\), so \(y=7\).
   3. Expand: \(6a+8=26\). Subtract 8: \(6a=18\), so \(a=3\).
   4. Expand: \(8m-12=20\). Add 12: \(8m=32\), so \(m=4\).
   5. Divide both sides by \(-2\): \(x+5=9\), so \(x=4\). Check: \(-2\times9=-18\).

   Expanding first turns each equation into a two-step equation that you already know how to solve.

4. Answer to Question 4

   1. Subtract \(2x\): \(3x+3=15\). Subtract 3: \(3x=12\), so \(x=4\). Check: \(23=23\).
   2. Subtract \(3y\): \(4y-4=12\). Add 4: \(4y=16\), so \(y=4\). Check: \(24=24\).
   3. There are more a’s on the right, so subtract \(4a\): \(1=2a-9\). Add 9: \(10=2a\), so \(a=5\). Check: \(21=21\).
   4. Expand: \(3m+6=m+14\). Subtract m: \(2m+6=14\). Subtract 6: \(2m=8\), so \(m=4\). Check: \(18=18\).

5. Answer to Question 5

   1. Let the number be n: \(4n+5=37\). Subtract 5: \(4n=32\). Divide by 4: \(n=8\).
   2. \(\dfrac{n-6}{3}=8\). Multiply by 3: \(n-6=24\). Add 6: \(n=30\).
   3. The numbers are n, \(n+1\), \(n+2\): \(3n+3=66\), so \(3n=63\) and \(n=21\). The numbers are 21, 22 and 23.
   4. \(2n-9=n+4\). Subtract n: \(n-9=4\). Add 9: \(n=13\). Check: \(26-9=17=13+4\).

6. Answer to Question 6

   1. \(3\times7-5=21-5=16\). The left side equals 16, so Zara is correct.
   2. \(3\times3.67-5=11.01-5=6.01\), not 16, so Ben is wrong.
   3. Add 5 to both sides: \(3x=21\). Divide by 3: \(x=7\).
   4. Ben seems to have worked out \(16\div3-5\) or similar: he undid the operations in the wrong order. The “subtract 5” must be undone first by adding 5, then the “multiply by 3” by dividing.

7. Answer to Question 7

   1. 2 adult tickets cost \(2\times5=£10\) and c child tickets cost \(£3c\): \(10+3c=19\).
   2. Subtract 10: \(3c=9\). Divide by 3: \(c=3\) child tickets.
   3. \(4\times5+3c=35\), so \(20+3c=35\), \(3c=15\) and \(c=5\) child tickets.
   4. With n of each: \(5n+3n=56\), so \(8n=56\) and \(n=7\): 7 adults and 7 children.

8. Answer to Question 8

   1. Fixed £9 plus £0.06 per text: \(9+0.06t=15\).
   2. Subtract 9: \(0.06t=6\). Divide by 0.06: \(t=100\) texts.
   3. \(9+0.06t=12\) gives \(0.06t=3\) and \(t=50\). She can send at most 50 texts.
   4. \(9+0.06t=5+0.10t\). Subtract 5 and \(0.06t\): \(4=0.04t\), so \(t=100\) texts (both pay £15).

9. Answer to Question 9

   1. The box weighs 500 g = 0.5 kg: \(a+0.5=2.3\). Subtract 0.5: \(a=1.8\) kg.
   2. Second bag: \(2\times1.8+0.5=3.6+0.5=4.1\) kg, which matches the given weight, so the answer is consistent.
   3. Each bag has 1.8 kg of apples: \(5\times1.8=\)9 kg.

   Writing every mass in kilograms before forming the equation avoids mixing grams and kilograms.

10. Answer to Question 10

    1. He starts with £18 and adds £4 a week: \(18+4w=62\).
    2. Subtract 18: \(4w=44\). Divide by 4: \(w=11\) weeks.
    3. \(18+4w=30+2.5w\). Subtract \(2.5w\) and 18: \(1.5w=12\), so \(w=8\) weeks.
    4. Sam: \(18+4\times8=50\); Mia: \(30+2.5\times8=50\). Each has £50.

11. Answer to Question 11

    1. Perimeter = 2 lengths + 2 widths: \(2(x+5)+2(x-1)=2x+10+2x-2=\)\(4x+8=36\).
    2. Subtract 8: \(4x=28\). Divide by 4: \(x=7\).
    3. Length \(7+5=\)12 cm, width \(7-1=\)6 cm. Check: \(2(12+6)=36\).
    4. Area = length × width: \(12\times6=\)72 cm².

12. Answer to Question 12

    1. 12 minutes is \(12\div60=0.2\) hours, and \(40\times0.2=8\) km. Then \(3+8=11\) km, so the numbers add up.
    2. 15 minutes is 0.25 hours. Multiply both sides of \(\dfrac{3}{v}=0.25\) by v: \(3=0.25v\). Divide by 0.25: \(v=12\).
    3. His cycling speed is 12 km/h: he covers 3 km in a quarter of an hour.

13. Answer to Question 13

    1. Jumper £22 plus 3 T-shirts at £t: \(22+3t=43\).
    2. Subtract 22: \(3t=21\). Divide by 3: \(t=7\), so each T-shirt costs £7.
    3. \(50-43=\)£7 is left on the card.
    4. Yes: one more T-shirt costs £7, exactly the balance left, so she can buy it and the card will then be empty.

14. Answer to Question 14

    1. 5 trays of n plus 7 extra: \(5n+7=47\).
    2. Subtract 7: \(5n=40\). Divide by 5: \(n=8\) cupcakes per tray.
    3. \(47\times0.60=\)£28.20.
    4. \(0.6c=30\), so \(c=30\div0.6=\)50 cupcakes, three more than she has baked.

    Checking part (b): \(5\times8+7=40+7=47\), the total number of cupcakes.

15. Answer to Question 15

    1. \(8+4h=28\), \(4h=20\), \(h=\)5 hours.
    2. \(8+4h=36\), \(4h=28\), \(h=\)7 hours.
    3. Ruby, by 2 hours.
    4. \(8+4h=4+5h\), \(4=h\). They are the same for 4 hours (£24).

    In each case the fixed charge is subtracted first, then the total is divided by the price per hour; checking with the original prices confirms the answers.

16. Answer to Question 16

    1. Add the sides: \(2x+1+x+6+3x-2=6x+5=41\). Subtract 5: \(6x=36\), so \(x=6\).
    2. \(2\times6+1=\)13 cm, \(6+6=\)12 cm, \(3\times6-2=\)16 cm. Check: \(13+12+16=41\).
    3. No: all three sides are different, so the triangle is scalene.
    4. Set the equal sides equal: \(3x-2=2x+1\), so \(x=3\). Both sides are \(3\times3-2=\)7 cm.

17. Answer to Question 17

    1. 4 rides cost \(4\times2.50=£10\), plus \(£1.80s\) for snacks: \(10+1.8s=17.2\).
    2. Subtract 10: \(1.8s=7.2\), so \(s=4\). A whole number of snacks makes sense: 4 snacks.
    3. \(3\times2.50+5\times1.80=7.50+9.00=\)£16.50.
    4. \(2\times2.50+1.8s=14\), so \(1.8s=9\) and \(s=5\) snacks.

    Check part (b): \(10+1.8\times4=10+7.2=17.2\), which is Eva’s total.

18. Answer to Question 18

    1. Anna: \(2n-7=15\), \(2n=22\), \(n=11\). Ben: \(3(n-4)=18\), \(n-4=6\), \(n=10\). Cara: \(\dfrac{n+5}{2}=9\), \(n+5=18\), \(n=13\).
    2. Comparing 11, 10 and 13, Cara thought of the biggest number.
    3. \(11+10+13=\)34.
    4. \(34\div4=8.5\), not a whole number, so 34 is not a multiple of 4 (\(4\times8=32\) and \(4\times9=36\)).

19. Answer to Question 19

    1. Ali added 6 instead of subtracting: \(3x=18\), \(x=6\).
    2. Beth did not multiply the 3 by 2: \(2x-6=10\), \(x=8\).
    3. Carl multiplied only the first term by 5: \(x-10=15\), \(x=25\).
    4. Dina added 3x to the left. She should subtract: \(2x=8\), \(x=4\).

    In every case, substituting the wrong answer back into the original equation would have shown that it does not work.

20. Answer to Question 20

    1. \(\dfrac{3x-4}{2}\).
    2. \(\dfrac{3x-4}{2}=13\), \(3x-4=26\), \(x=\)10.
    3. Reverse: multiply by 2, add 4, divide by 3. Output 5: \(5\times2=10\), \(10+4=14\), \(14\div3=\)4.67 (4 2/3).
    4. \(\dfrac{3x-4}{2}=x\), \(3x-4=2x\), \(x=\)4. Check: \(3\times4=12\), \(12-4=8\), \(8\div2=4\).

    The reverse machine undoes each step with the inverse operation, in the opposite order.

21. Answer to Question 21

    1. Maya: \(x\). Brother: \(x+4\). Mother: \(3x\).
    2. \((x+5)+(x+9)+(3x+5)=64\), \(5x+19=64\).
    3. \(5x=45\), \(x=9\). Maya is 9, her brother is 13 and her mother is 27.
    4. Maya is 18 in 9 years' time. The mother will then be \(27+9=\)36.

    Check part (c): in 5 years they will be 14, 18 and 32, and \(14+18+32=64\).