Solving linear equations and rearranging formulae – Year 9 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(3x=15\), so \(x=\) 5. Check: \(15+5=20\). Subtract 5 from both sides, then divide both sides by 3.
   2. \(4x=24\), so \(x=\) 6. Check: \(24-7=17\). Add 7 to both sides, then divide both sides by 4.
   3. \(-2x=6\), so \(x=\) \(-3\). Check: \(5+6=11\). Subtract 5 from both sides; dividing by \(-2\) gives a negative answer.
   4. \(\dfrac x3=5\), so \(x=\) 15. Check: \(5+4=9\). Subtract 4 from both sides, then multiply both sides by 3 to undo the division.

2. Answer to Question 2

   1. \(3x+12=27\), \(3x=15\), \(x=\) 5 Multiply 3 into both terms of the bracket, subtract 12, then divide by 3. Check: \(3\times9=27\).
   2. \(6x-2=22\), \(6x=24\), \(x=\) 4 Expand to \(6x-2\), add 2 to both sides, then divide by 6. Check: \(2\times11=22\).
   3. \(4x-8+3=19\), \(4x-5=19\), \(4x=24\), \(x=\) 6 Expand and collect the numbers \(-8+3=-5\) before solving. Check: \(4\times4+3=19\).
   4. \(10x-15=4x+12\), \(6x=27\), \(x=\) 4.5 Expand both sides, subtract \(4x\) and add 15. Check: \(5\times6=30\) and \(4\times7.5=30\).

3. Answer to Question 3

   1. \(3x=15\), so \(x=\) 5 Subtract \(2x\) from both sides and subtract 3. Check: \(28=28\).
   2. \(4x=16\), so \(x=\) 4 Subtract \(3x\) and add 4 to both sides. Check: \(24=24\).
   3. \(10=2x\), so \(x=\) 5 Move the smaller \(x\) term: subtract \(4x\) and add 1 to both sides. Check: \(29=29\).
   4. \(3x+6=x+14\), \(2x=8\), \(x=\) 4 Expand the bracket first, then subtract \(x\) and 6 from both sides. Check: \(3\times6=18=4+14\).

4. Answer to Question 4

   1. \(x=7\times4=\) 28 Multiply both sides by 4 to undo the division.
   2. \(x+3=16\), \(x=\) 13 Multiply both sides by 2, then subtract 3. Check: \(16\div2=8\).
   3. \(2x-1=15\), \(2x=16\), \(x=\) 8 Multiply both sides by 3, add 1, then divide by 2. Check: \(15\div3=5\).
   4. Multiply by 6: \(3x+2x=60\), \(5x=60\), \(x=\) 12 The lowest common multiple of 2 and 3 is 6, so multiply every term by 6. Check: \(6+4=10\).

5. Answer to Question 5

   1. Subtract 5: \(x=y-5\) Undo adding 5 by subtracting 5 from both sides.
   2. Divide by 3: \(x=\dfrac y3\) Undo multiplying by 3 by dividing both sides by 3.
   3. Add 4 then divide by 2: \(x=\dfrac{y+4}2\) Undo the operations in reverse order: first add 4, then divide by 2.
   4. Divide by \(w\): \(l=\dfrac Aw\) \(l\) is multiplied by \(w\), so divide both sides by \(w\).

6. Answer to Question 6

   1. \(3+2\times5=\) 13 Multiply \(a\) by \(t\) before adding \(u\).
   2. \(\frac12\times10\times3=\) 15 Add the parallel sides first: \(4+6=10\), then multiply by \(\frac12\) and by 3.
   3. \(2\times\pi\times5=\) 31.4 Use the \(\pi\) key: \(10\pi=31.415\ldots\), which rounds to 31.4.
   4. \(1.8\times15+32=27+32=\) 59 Multiply 15 by 1.8 first, then add 32.

7. Answer to Question 7

   1. \(3x-7=20\), \(3x=27\), \(x=\) 9 Turn the words into an equation, add 7 to both sides and divide by 3.
   2. \(2x+5=x+12\), \(x=\) 7 The answer is 12 more than her number, so the right-hand side is \(x+12\). Subtract \(x\) and 5 from both sides.
   3. \(5(x+4)=60\), \(x+4=12\), \(x=\) 8 Adding 4 comes before multiplying, so put \(x+4\) in a bracket. Divide both sides by 5 first.
   4. Ali: \(27-7=20\). Bea: \(14+5=19=7+12\). Cal: \((8+4)\times5=60\). All correct. Substituting each answer back into the words gives the stated result.

8. Answer to Question 8

   1. \(2(2x+3)+2x=36\), so \(4x+6+2x=36\), \(6x=30\), \(x=\) 5. The perimeter is two lengths plus two widths. Expand, collect like terms and solve.
   2. Width \(=5\) m. Length \(=2\times5+3=\) 13 m. Substitute \(x=5\) into the expressions for the length and the width.
   3. \(13\times5=\) 65 m² Area of a rectangle = length × width.
   4. \(65\times4.50=\) £292.50 Multiply the area by the price per square metre.

9. Answer to Question 9

   1. \(x+2x+10+3x-10=180\), so \(6x=180\) and \(x=\) 30. The angles in a triangle add up to \(180^\circ\). The numbers \(+10\) and \(-10\) cancel.
   2. \(30^\circ\), \(70^\circ\), \(80^\circ\) Substitute \(x=30\) into each expression.
   3. \(30+70+80=180\). The three angles add up to exactly \(180^\circ\), so the answer is correct.
   4. Scalene: all three angles are different, so no sides are equal. In a triangle, equal angles are opposite equal sides, so different angles mean different sides.

10. Answer to Question 10

    1. A: \(3+1.5m\). B: \(1+2m\). Each cost is a fixed charge plus the price per mile multiplied by the number of miles.
    2. \(3+1.5m=1+2m\), so \(2=0.5m\) and \(m=\) 4 miles. Set the two expressions equal, then subtract \(1.5m\) and 1 from both sides.
    3. \(3+1.5\times4=\) £9 Substitute \(m=4\) into either expression; taxi B gives \(1+8=9\) too.
    4. A: \(3+15=£18\); B: \(1+20=£21\). Taxi A is cheaper by £3. Substitute \(m=10\) into both expressions and compare.

11. Answer to Question 11

    1. Child tickets: \(40-a\) There are 40 tickets altogether, so the rest of them are child tickets.
    2. \(8a+5(40-a)=260\), so \(8a+200-5a=260\), \(3a=60\), \(a=\) 20. Adult tickets raise \(8a\) pounds and child tickets raise \(5(40-a)\) pounds; expand and solve.
    3. 20 adult tickets and 20 child tickets. Check: \(160+100=260\). Substitute \(a=20\) into \(40-a\) and check the total money.
    4. The most it could be is 40 adult tickets: \(40\times8=£320\), so £300 is possible, with \(3a+200=300\), \(a=33.3\), which is not a whole number, so not exactly £300. The total \(3a+200\) can only be 299 or 302 near £300, because \(a\) must be a whole number.

12. Answer to Question 12

    1. \(40\times5+20=220\) min \(=\) 3 hours 40 minutes Substitute \(w=5\), then change minutes into hours and minutes.
    2. \(T-20=40w\), so \(w=\dfrac{T-20}{40}\) Subtract 20 from both sides, then divide both sides by 40.
    3. \(w=(300-20)\div40=\) 7 kg Substitute \(T=300\) into the rearranged formula.
    4. \(T=260\) min \(=4\) h 20 min. \(17{:}30-4{:}20=\) 13:10. Find the cooking time for 6 kg, then count back from 17:30.

13. Answer to Question 13

    1. \(t=\dfrac ds\) and \(s=\dfrac dt\) Divide both sides of \(d=st\) by \(s\) to find \(t\), or by \(t\) to find \(s\).
    2. \(1800\div600=\) 3 hours Substitute into \(t=\dfrac ds\).
    3. \(1800\div2.5=\) 720 km/h Substitute into \(s=\dfrac dt\).
    4. \(3\times3000=\) 9 000 litres The first flight takes 3 hours, so multiply the fuel per hour by 3.

14. Answer to Question 14

    1. Alex gets £\((x+3)\). Alex gets 3 pounds more than Sam.
    2. \(x+x+3=17\), \(2x=14\), \(x=\) 7. Add the two amounts and set the total equal to 17, then subtract 3 and divide by 2.
    3. Sam £7, Alex £10. Sam gets \(x=7\) pounds and Alex gets \(7+3\) pounds.
    4. \(8\times17=\) £136 Together they receive £17 a week, so multiply by 8.

15. Answer to Question 15

    1. \(5x=25\), \(x=\) 5. Check: \(32=32\). Subtract \(2x\) from both sides and add 3.
    2. \(6x+3=4x+15\), \(2x=12\), \(x=\) 6. Check: \(39=39\). Expand the bracket, then subtract \(4x\) and 3 from both sides.
    3. \(x+5=3x-3\), \(8=2x\), \(x=\) 4. Check: \(3=3\). Multiply both sides by 3 first, then collect the \(x\) terms on the right.
    4. \(2x-6-3x+12=5\), \(-x+6=5\), \(x=\) 1. Check: \(-4+9=5\). Take care with \(-3(x-4)=-3x+12\); then collect like terms.
    5. Equation (c). Equation (c) has a denominator of 3, which must be cleared first.

16. Answer to Question 16

    1. \(a=\dfrac{v-u}t\); \(l=\dfrac P2-w\); \(h=\dfrac{2A}b\); \(x=\dfrac{y-c}m\). Undo each operation in reverse order. For example, in \(P=2(l+w)\) divide by 2 and then subtract \(w\).
    2. \(a=20\div4=\) 5; \(l=15-6=\) 9; \(h=60\div12=\) 5; \(x=12\div4=\) 3. Substitute the given values into each rearranged formula.
    3. Check \(A\): \(\frac12\times12\times5=30\). Correct. Substituting the answer back into the original formula gives the given value.
    4. If \(b=0\) then \(A\) would always be 0, and division by zero is impossible. The formula \(h=\dfrac{2A}{b}\) would need a division by zero.

17. Answer to Question 17

    1. A: \(15+0.02m\). B: \(5+0.05m\). Change pence into pounds: 2p is £0.02 and 5p is £0.05.
    2. \(15+0.02m=5+0.05m\), so \(10=0.03m\) and \(m=333.3\), so about 333 minutes. Set the costs equal, subtract \(0.02m\) and 5 from both sides, then divide by 0.03.
    3. 400 minutes: A is £23 and B is £25, so A is cheaper. 100 minutes: B is cheaper (£10 against £17). Above about 333 minutes Plan A is cheaper; below it Plan B is cheaper.
    4. 250 minutes: A \(=15+5=£20\); B \(=5+12.5=£17.50\). Plan B saves \(2.50\times12=\) £30 a year. Plan B is £2.50 cheaper each month, and there are 12 months in a year.

18. Answer to Question 18

    1. \(212=1.8C+32\), \(180=1.8C\), \(C=\) 100 Subtract 32 from both sides, then divide by 1.8.
    2. \(-4-32=-36\), \(C=-36\div1.8=\) \(-20\) Subtract 32 from \(-4\) carefully, then divide by 1.8.
    3. \(F-32=1.8C\), so \(C=\dfrac{F-32}{1.8}\) Subtract 32 from both sides, then divide both sides by 1.8.
    4. Put \(F=C\): \(C=1.8C+32\), so \(-0.8C=32\) and \(C=\) \(-40\). Replace \(F\) by \(C\), collect the \(C\) terms on one side and solve.
    5. At \(-40\), \(1.8\times(-40)+32=-72+32=-40\), so both scales read \(-40\). Substituting \(-40\) into the formula gives \(-40\) back.

19. Answer to Question 19

    1. Length \(=x+20\) The length is 20 m more than the width.
    2. \(2(x+x+20)=140\), so \(4x+40=140\), \(x=\) 25. The perimeter is twice the length plus twice the width; expand and solve.
    3. Width 25 m, length 45 m, area \(25\times45=\) 1 125 m². Substitute \(x=25\), then multiply the length by the width.
    4. \(140\times6.50=\) £910 Multiply the perimeter by the price per metre.
    5. \(137\times6.50=\) £890.50 The gate means 3 m less fence: \(140-3=137\) m.

20. Answer to Question 20

    1. \(4b+6=3(b+6)\), so \(4b+6=3b+18\) and \(b=\) 12. In 6 years both people are 6 years older, so write \(4b+6\) and \(b+6\).
    2. Mother is \(4\times12=\) 48 now and 54 in 6 years, which is \(3\times18\). Substitute \(b=12\); in 6 years Ben is 18 and his mother is 54.
    3. Multiply by 6: \(3(3x+1)-2(x-2)=21\), so \(9x+3-2x+4=21\), \(7x=14\), \(x=\) 2. The lowest common multiple of 2 and 3 is 6, so multiply every term by 6 and expand carefully.
    4. \(\dfrac72-0=\dfrac72\). Correct. Substituting \(x=2\) gives \(\dfrac72-\dfrac03=\dfrac72\).
    5. \(2y=3x+1\), \(3x=2y-1\), so \(x=\dfrac{2y-1}3\). Multiply both sides by 2, subtract 1 and divide by 3.