Simultaneous equations – Year 10 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Add: \(2x=14\), \(x=7\), \(y=3\). \(x=7\), \(y=3\) The \(y\) terms have opposite signs, so adding the equations eliminates \(y\). Check: \(7-3=4\).
   2. Add: \(4x=16\), \(x=4\), then \(4-2y=0\), \(y=2\). \(x=4\), \(y=2\) The \(-2y\) and \(+2y\) cancel when we add. Check in the first equation: \(12+4=16\).
   3. Subtract: \(3x=9\), \(x=3\), then \(15+y=17\), \(y=2\). \(x=3\), \(y=2\) Both equations contain \(+y\), so subtracting eliminates \(y\). Check: \(6+2=8\).
   4. Subtract: \(4y=8\), \(y=2\). Then \(4x-2=5\), \(4x=7\), \(x=1.75\). \(x=1.75\), \(y=2\) The \(4x\) terms are the same, so subtracting the second equation from the first removes \(x\). Check: \(7+6=13\).

2. Answer to Question 2

   1. Double the second: \(2x+2y=10\). Subtract: \(y=3\), \(x=2\). \(x=2\), \(y=3\)
   2. Double the second: \(4x+10y=32\). Subtract: \(7y=14\), \(y=2\), \(4x=12\), \(x=3\). \(x=3\), \(y=2\)
   3. Double the first: \(6x-4y=14\). Add \(x+4y=7\): \(7x=21\), \(x=3\), then \(y=1\). \(x=3\), \(y=1\)
   4. Double the second: \(2x-2y=4\). Add to the first: \(7x=28\), \(x=4\), \(y=2\). \(x=4\), \(y=2\)

3. Answer to Question 3

   1. Multiply by 3 and 2: \(9x+6y=36\) and \(10x-6y=2\). Add: \(19x=38\), \(x=2\), \(y=3\). \(x=2\), \(y=3\)
   2. Multiply by 2 and 5: \(4x+10y=32\) and \(15x-10y=25\). Add: \(19x=57\), \(x=3\), \(y=2\). \(x=3\), \(y=2\)
   3. Multiply by 3 and 4: \(12x+9y=3\) and \(12x+8y=0\). Subtract: \(y=3\), then \(4x+9=1\), \(x=-2\). \(x=-2\), \(y=3\)
   4. Multiply by 5 and 2: \(35x-10y=85\) and \(6x+10y=38\). Add: \(41x=123\), \(x=3\), then \(21-2y=17\), \(y=2\). \(x=3\), \(y=2\)

4. Answer to Question 4

   1. \(3x+(2x+1)=11\), \(5x=10\), \(x=2\), \(y=5\). \(x=2\), \(y=5\) Replace \(y\) by \(2x+1\) in the second equation, so only \(x\) is left. Check: \(6+5=11\).
   2. \(2(3y-2)+y=10\), \(7y=14\), \(y=2\), \(x=4\). \(x=4\), \(y=2\) Replace \(x\) by \(3y-2\) and expand the bracket: \(6y-4+y=10\). Check: \(8+2=10\).
   3. \(2x+3(x-4)=3\), \(5x=15\), \(x=3\), \(y=-1\). \(x=3\), \(y=-1\) Expand \(3(x-4)=3x-12\), then add 12 to both sides. Check: \(6-3=3\).
   4. \(5-x=2x-1\), \(6=3x\), \(x=2\), \(y=3\). \(x=2\), \(y=3\) Both expressions equal \(y\), so they are equal to each other. Check: \(2\times2-1=3\).

5. Answer to Question 5

   1. Substitute: \(x+x+1=7\), \(x=3\), \(y=4\). \((3,4)\) The point where two lines cross lies on both lines, so it is the solution of both equations.
   2. \(2x-1=5-x\), \(3x=6\), \(x=2\), \(y=3\). \((2,3)\) At the crossing point the two \(y\) values are equal. Check: \(5-2=3\).
   3. Both lines have gradient 2 but different intercepts (3 and \(-1\)), so they are parallel and never meet. Parallel lines never cross, so the simultaneous equations have no solution.
   4. \(4=3+1\) ✓ and \(3+4=7\) ✓. The point fits both equations, so \((3,4)\) is the point of intersection.

6. Answer to Question 6

   1. From the second, \(x=-3y\). Then \(-6y-y=7\), \(y=-1\), \(x=3\). \(x=3\), \(y=-1\)
   2. Add: \(2x=3\), \(x=1.5\), \(y=-0.5\). \(x=1.5\), \(y=-0.5\) Check: \(1.5+(-0.5)=1\) and \(1.5-(-0.5)=2\).
   3. Add: \(3x=2\), \(x=\tfrac23\), \(y=-\tfrac13\). \(x=\tfrac23\), \(y=-\tfrac13\) Check: \(\tfrac43-\tfrac13=1\) and \(\tfrac23+\tfrac13=1\).
   4. The second equation is twice the first, so they are the same line: there are infinitely many solutions.

7. Answer to Question 7

   1. \(2a+3c=28\) and \(a+4c=24\). Each equation turns one sentence into algebra: the number of tickets times the price, added together.
   2. Double the second: \(2a+8c=48\). Subtract: \(5c=20\), \(c=4\), then \(a=8\). Adult £8, child £4. Doubling the second equation makes the \(a\) terms match, so subtracting removes \(a\). Check: \(16+12=28\) and \(8+16=24\).
   3. \(2\times8+2\times4=\) £24. Substitute the two prices. A family ticket costs £24.
   4. \(5\times8+5\times4=\) £60. Five adults and five children cost \(5\times(8+4)=5\times12\). The group pays £60.

8. Answer to Question 8

   1. \(3c+2t=11.10\) and \(2c+4t=11.80\). Write one equation for each order: the number of drinks times the price, added together.
   2. Double the first: \(6c+4t=22.20\). Subtract the second: \(4c=10.40\), \(c=2.60\). Then \(2t=11.10-7.80=3.30\), \(t=1.65\). Coffee £2.60, tea £1.65. Doubling the first equation makes the \(t\) terms match. Check: \(5.20+6.60=11.80\).
   3. \(4\times2.60+1.65=10.40+1.65=\) £12.05. Substitute the two prices. Four coffees and a tea cost £12.05.
   4. \(20-1.65=18.35\), \(18.35\div2.60=7.06\), so 7 coffees. Pay for the tea first, then see how many whole coffees fit in the money left. We round down because she cannot buy part of a coffee.

9. Answer to Question 9

   1. \(3a+2r=29\) and \(a+r=11\). One equation uses the 3 kg and 2 kg mix, the other uses 1 kg of each.
   2. From the second, \(r=11-a\): \(3a+22-2a=29\), so \(a=7\) and \(r=4\). Almonds £7/kg, raisins £4/kg. Substitute \(r=11-a\) into the first equation. Check: \(21+8=29\).
   3. \(2\times7+3\times4=14+12=\) £26. Multiply each mass by its price per kilogram and add.
   4. \(26\div5=\) £5.20 per kg. The mix has a mass of \(2+3=5\) kg, so divide the cost by 5.

10. Answer to Question 10

    1. \(2l+2w=34\) (so \(l+w=17\)) and \(l=w+5\). The perimeter is two lengths plus two widths, and the second sentence gives the link between \(l\) and \(w\).
    2. \(w+5+w=17\), \(2w=12\), \(w=6\), \(l=11\). Length 11 m, width 6 m. Substitute \(l=w+5\) into \(l+w=17\). Check: \(2\times11+2\times6=34\).
    3. \(11\times6=\) 66 m². The area of a rectangle is length times width.
    4. \(66\times4.50=\) £297. Multiply the area by the price per square metre. The turf costs £297.

11. Answer to Question 11

    1. \(a+b=30\) and \(20a+50b=1020\).
    2. \(b=30-a\): \(20a+1500-50a=1020\), \(-30a=-480\), \(a=16\), \(b=14\). 16 twenty-pence and 14 fifty-pence coins.
    3. \(16\times20=320\) and \(14\times50=700\), total \(1020\)p \(=\) £10.20 ✓
    4. \(20a+50(28-a)=1020\) gives \(-30a=-380\), \(a=12.67\), not a whole number, so it is impossible.

12. Answer to Question 12

    1. \(f=3s\) and \(f+12=2(s+12)\). The father's age now is three times his son's. In 12 years both ages increase by 12.
    2. \(3s+12=2s+24\), \(s=12\), \(f=36\). Son 12, father 36. Substitute \(f=3s\) into the second equation, then solve for \(s\).
    3. Son 24, father 48 (\(48=2\times24\) ✓). Add 12 to each age now. The father will be twice as old, as the question says.
    4. \(36-12=\) 24 years old. The age difference never changes, so the father was 24 when his son was born.

13. Answer to Question 13

    1. \(C=30+0.2m\) and \(C=20+0.3m\). Each company charges a fixed amount plus a cost for each mile.
    2. \(30+0.2m=20+0.3m\), \(10=0.1m\), \(m=100\) miles. The costs are equal when the two expressions for \(C\) are equal. Subtract \(0.2m\) and 20 from both sides.
    3. \(30+0.2\times100=\) £50. Substitute \(m=100\) into either equation. Both companies charge £50.
    4. 250 miles: A \(=30+50=£80\), B \(=20+75=£95\). A is cheaper by £15. After 100 miles company A becomes cheaper because its cost per mile is lower.

14. Answer to Question 14

    1. \(3c+2m=190\) and \(2c+3m=160\).
    2. Multiply by 3 and 2: \(9c+6m=570\) and \(4c+6m=320\). Subtract: \(5c=250\), \(c=50\), then \(m=20\). Coach 50, minibus 20.
    3. \(4\times50+20=\) 220 pupils. Four coaches seat 200 pupils and one minibus seats 20.
    4. For example 4 coaches and 2 minibuses: \(200+40=240\ge230\). Or 3 coaches and 4 minibuses: \(150+80=230\) exactly.

15. Answer to Question 15

    1. \(C_A=50+4n\) and \(C_B=20+6n\).
    2. \(50+4n=20+6n\), \(30=2n\), \(n=15\), cost \(=50+60=\) £110.
    3. 10 shirts: A \(=£90\), B \(=£80\), so B is cheaper. 25 shirts: A \(=£150\), B \(=£170\), so A is cheaper. The lines cross at (15, 110).
    4. 40 shirts: A \(=210\), B \(=260\). A saves £50.

16. Answer to Question 16

    1. \(x+y=47\) and \(x-y=13\). Add: \(2x=60\), \(x=30\), \(y=17\). 30 and 17.
    2. \(t+u=11\) and \((10u+t)-(10t+u)=27\), so \(9u-9t=27\) and \(u-t=3\). Add to \(t+u=11\): \(2u=14\), \(u=7\), \(t=4\). The number is 47.
    3. \(x+y=20\) and \(2x+3y=48\). \(y=20-x\): \(2x+60-3x=48\), \(x=12\), \(y=8\). 12 and 8.
    4. Reversed number \(74\). \(74-47=27\) ✓. Digit sum \(4+7=11\) ✓.

17. Answer to Question 17

    1. \(x^2+(x-1)^2=25\), \(2x^2-2x+1=25\), \(2x^2-2x-24=0\), so \(x^2-x-12=0\).
    2. \((x-4)(x+3)=0\), so \(x=4\) or \(x=-3\).
    3. \(x=4\): \(y=3\). \(x=-3\): \(y=-4\). \((4,3)\) and \((-3,-4)\)
    4. The line cuts the circle (radius 5, centre the origin) at two points; both pairs satisfy \(x^2+y^2=25\) (\(16+9\) and \(9+16\)).

18. Answer to Question 18

    1. \(x^2-4=3x\), so \(x^2-3x-4=0\).
    2. \((x-4)(x+1)=0\), so \(x=4\) or \(x=-1\).
    3. \(x=4\): \(y=12\). \(x=-1\): \(y=-3\). \((4,12)\) and \((-1,-3)\)
    4. \(x^2-4=-5\) gives \(x^2=-1\), which has no real solution, so no: the lowest point of the parabola is \(y=-4\).

19. Answer to Question 19

    1. \(a+2b=7\) and \(3a-b=7\). The line passes through each point, so substitute \(x=1\), \(y=2\) and then \(x=3\), \(y=-1\) into \(ax+by=7\).
    2. \(b=3a-7\): \(a+6a-14=7\), \(7a=21\), \(a=3\), \(b=2\). Make \(b\) the subject of the second equation and substitute it into the first. Check: \(3+4=7\) and \(9-2=7\).
    3. \(3x+2y=7\) Substitute \(a=3\) and \(b=2\) into \(ax+by=7\).
    4. \(x=0\): \(2y=7\), so \((0,3.5)\). Every point on the \(y\)-axis has \(x=0\). The line crosses the \(y\)-axis at \((0,3.5)\).

20. Answer to Question 20

    1. The second equals twice the first when \(k=\) 10.
    2. They have the same gradient, so for any other \(k\) (\(k\ne10\)) the lines are parallel and never meet.
    3. \(k=10\): one line (infinitely many solutions). \(k\ne10\): two parallel lines (no solutions).
    4. Double the first: \(4x+6y=10\). Subtract from the second: \(3y=k-10\), \(y=\dfrac{k-10}{3}\). Then \(2x=5-(k-10)=15-k\), \(x=\dfrac{15-k}{2}\).