Linear equations and inequalities – Year 10 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(3x=18\), so \(x=6\) (check: \(18+7=25\) ✓) Subtract 7 from both sides, then divide both sides by 3. The solution is \(x=6\).
   2. \(-2x=8\), so \(x=-4\) (check: \(5+8=13\) ✓) Subtract 5 from both sides, then divide by \(-2\); a positive divided by a negative is negative. The solution is \(x=-4\).
   3. \(\dfrac x4=5\), so \(x=20\) Add 3 to both sides, then multiply by 4 to undo the division. Check: \(20\div4-3=2\). The solution is \(x=20\).
   4. \(x-3=5\), so \(x=8\) (or \(4x-12=20\), \(4x=32\)) Divide both sides by 4 first, then add 3. Check: \(4\times(8-3)=20\). The solution is \(x=8\).

2. Answer to Question 2

   1. \(3x=15\), so \(x=5\) Subtract \(2x\) from both sides and add 3 to both sides. Check: \(25-3=22\) and \(10+12=22\). The solution is \(x=5\).
   2. \(16=4x\), so \(x=4\) The \(x\) coefficient is bigger on the right, so add \(3x\) and add 9 to both sides. Check: \(7-12=-5\) and \(4-9=-5\). The solution is \(x=4\).
   3. \(8x-4=3x+21\), \(5x=25\), so \(x=5\) Expand both brackets first, then subtract \(3x\) and add 4. Check: \(4\times9=36\) and \(3\times12=36\). The solution is \(x=5\).
   4. \(6-2x+2=3x-7\), \(8-2x=3x-7\), \(15=5x\), so \(x=3\) Be careful: \(-2\times(-1)=+2\). Then add \(2x\) and 7 to both sides. Check: \(6-4=2\) and \(9-7=2\). The solution is \(x=3\).

3. Answer to Question 3

   1. \(x+3=14\), so \(x=11\) Multiply both sides by 2 to clear the fraction, then subtract 3. Check: \(\tfrac{14}{2}=7\). The solution is \(x=11\).
   2. \(2x-1=15\), \(2x=16\), so \(x=8\) Multiply both sides by 3, add 1, then divide by 2. Check: \(\tfrac{15}{3}=5\). The solution is \(x=8\).
   3. Multiply by 6: \(3x+2x=60\), \(5x=60\), so \(x=12\) The common denominator of 2 and 3 is 6, so every term is multiplied by 6. Check: \(6+4=10\). The solution is \(x=12\).
   4. Multiply by 4: \(3x+1=2(x+9)=2x+18\), so \(x=17\) Multiply both sides by 4; the right side becomes \(2(x+9)\). Then subtract \(2x\) and 1. Check: \(\tfrac{52}{4}=13\) and \(\tfrac{26}{2}=13\). The solution is \(x=17\).

4. Answer to Question 4

   1. \(3x>15\), so \(x>5\) Solve it like an equation: add 4 to both sides, then divide by 3. The solution is \(x>5\).
   2. \(-2x\ge4\), divide by \(-2\) and reverse: \(x\le-2\) Subtract 5 from both sides. Dividing by a negative number reverses the sign, so \(\ge\) becomes \(\le\). The solution is \(x\le-2\).
   3. Subtract 1: \(-4\lt 2x\le8\). Divide by 2: \(-2\lt x\le4\) Do the same thing to all three parts of a double inequality. The solution is \(-2\lt x\le4\).
   4. \(4x-4\lt 3x+2\), \(x\lt 6\): \(x\lt 6\) Expand the bracket, then subtract \(3x\) and add 4 to both sides. The solution is \(x\lt 6\).

5. Answer to Question 5

   1. −1, 0, 1, 2, 3, 4 (−2 is not included because the sign is \(\lt \))
   2. \(3x\lt 18\), so \(x\lt 6\). The largest integer is 5.
   3. \(2x\ge8\), so \(x\ge4\). The smallest integer is 4.
   4. Draw a number line, put a filled (closed) circle at \(-2\) because \(-2\) is included, and draw an arrow pointing to the left.

6. Answer to Question 6

   1. \(4x-9=31\), \(4x=40\), so the number is 10. Let the number be \(x\). Add 9 to both sides, then divide by 4. Check: \(4\times10-9=31\).
   2. \(x+(x+3)+4x=63\), \(6x+3=63\), \(6x=60\), so Ben is 10 (sister 13, mother 40). Add the three ages and collect like terms. Check: \(10+13+40=63\).
   3. \(n+(n+1)+(n+2)=96\), \(3n+3=96\), \(n=31\): the numbers are 31, 32, 33. Consecutive numbers differ by 1, so we call them \(n\), \(n+1\) and \(n+2\). Check: \(31+32+33=96\).
   4. \(2(x+5)+2x=38\), \(4x+10=38\), \(x=7\): 7 cm by 12 cm. The perimeter is two lengths plus two widths. Check: \(2\times12+2\times7=38\).

7. Answer to Question 7

   1. Plan A: \(12+0.03m\). Plan B: \(5+0.10m\).
   2. \(12+0.03m=5+0.10m\), so \(7=0.07m\) and \(m=\) 100 minutes. Subtract \(0.03m\) and 5 from both sides, then divide by 0.07.
   3. \(12+0.03\times100=\) £15 (and \(5+10=15\)).
   4. For 150 minutes: A costs \(12+4.50=£16.50\) and B costs \(5+15=£20\). Plan A is cheaper, because above 100 minutes the lower rate per minute outweighs the higher fee.

8. Answer to Question 8

   1. \(2(3x-2)+2(x+4)=60\), so \(6x-4+2x+8=60\), so \(8x+4=60\). The perimeter is two lengths plus two widths; expand the brackets and collect like terms.
   2. \(8x=56\), so \(x=7\). Subtract 4 from both sides, then divide by 8.
   3. Length \(=3\times7-2=\) 19 m, width \(=7+4=\) 11 m, area \(=19\times11=\) 209 m². Substitute \(x=7\) into each expression. Check: \(2\times19+2\times11=60\). The area is 209 m².
   4. \(60\times14.50=\) £870. Multiply the perimeter by the price per metre. The fencing costs £870.

9. Answer to Question 9

   1. City Cabs: \(2.40+1.60d\). Quick Cars: \(3.80+1.40d\).
   2. \(2.40+1.60d=3.80+1.40d\), so \(0.2d=1.4\) and \(d=\) 7 miles. The cost is \(2.40+11.20=\) £13.60.
   3. 3 miles: City Cabs \(=£7.20\), Quick Cars \(=£8.00\). City Cabs is cheaper by 80p.
   4. 12 miles: City Cabs \(=£21.60\), Quick Cars \(=£20.60\). Quick Cars is cheaper by £1.00.

10. Answer to Question 10

    1. Child tickets \(=120-a\). Every ticket is either adult or child, so the child tickets are the remaining \(120-a\).
    2. \(25a+15(120-a)=2400\), so \(25a+1800-15a=2400\), so \(10a+1800=2400\). Multiply each number of tickets by its price and add. Expand the bracket and collect the \(a\) terms.
    3. \(10a=600\), so \(a=60\) adult tickets and 60 child tickets. Subtract 1800 from both sides, then divide by 10. Check: \(25\times60+15\times60=1500+900=2400\).
    4. \(10a+1800\ge2700\), so \(a\ge90\): at least 90 adult tickets. Write the takings as an inequality and solve it like an equation. The school must sell at least 90 adult tickets.

11. Answer to Question 11

    1. \(48n\ge212+8=220\). Everyone needs a seat: 212 pupils plus 8 teachers make 220 people, and \(n\) coaches seat \(48n\) people.
    2. \(n\ge\dfrac{220}{48}=4.58\), so the school needs 5 coaches. Divide both sides by 48. We round up because 4 coaches seat only 192 people. The school needs 5 coaches.
    3. \(5\times350=\) £1750. Multiply the number of coaches by the cost of one coach. The coaches cost £1750.
    4. \(350n\le1500\), so \(n\le4.29\), so at most 4 coaches, carrying \(4\times48=192\) people, which is not enough for everyone. We round down because 5 coaches would cost more than £1500.

12. Answer to Question 12

    1. Savings \(=35+8w\). She starts with £35 and adds £8 each week, so after \(w\) weeks she has \(35+8w\) pounds.
    2. \(35+8w\ge150\). She can afford the console when her savings are at least £150, so we use \(\ge\).
    3. \(8w\ge115\), \(w\ge14.375\), so 15 weeks. Subtract 35 from both sides, then divide by 8. We round up because after 14 weeks she has only £147. She needs 15 weeks.
    4. With £55: \(55+8w\ge150\), \(8w\ge95\), \(w\ge11.875\), so 12 weeks. She now starts with \(35+20=55\) pounds. After 11 weeks she has £143, which is not enough, so she needs 12 weeks.

13. Answer to Question 13

    1. Fit4U: \(30+18m\). Gymzone: \(25m\). Fit4U adds £18 each month to the £30 fee; Gymzone charges £25 each month.
    2. \(30\lt 7m\), so \(m>4.29\). Subtract \(18m\) from both sides, then divide by 7: \(m>\tfrac{30}{7}\).
    3. 5 months or more (after 5 months: Fit4U £120, Gymzone £125). The number of months must be a whole number greater than 4.29, so Fit4U is cheaper from 5 months onwards.
    4. 12 months: Fit4U \(=30+216=£246\), Gymzone \(=£300\). She saves £54. Twelve months is more than 5 months, so Fit4U is cheaper and Alisha saves £54.

14. Answer to Question 14

    1. \((x+30)+2(2x-5)=180\), so \(x+30+4x-10=180\), so \(5x+20=180\). An isosceles triangle has two equal base angles, so we use \(2(2x-5)\) for both of them.
    2. \(5x=160\), so \(x=32\). Subtract 20 from both sides, then divide by 5.
    3. Apex \(=32+30=\) 62°. Each base angle \(=2\times32-5=\) 59°. Substitute \(x=32\) into each expression.
    4. \(62+59+59=180\) ✓ The angles add up to \(180^\circ\), so the answer is correct.

15. Answer to Question 15

    1. \(4(x+3)=2(2x-1+x+2)\), so \(4x+12=2(3x+1)=6x+2\). The square has four equal sides; the rectangle has two lengths and two widths.
    2. \(10=2x\), so \(x=5\). Subtract \(4x\) and 2 from both sides, then divide by 2.
    3. Square side \(=\) 8 cm. Rectangle \(=9\) cm by 7 cm (perimeter \(2(9+7)=32=4\times8\) ✓).
    4. Square area \(=64\) cm². Rectangle area \(=63\) cm². The square is bigger by 1 cm². Equal perimeters do not mean equal areas.

16. Answer to Question 16

    1. Profit \(=3d-1.2d-60=1.8d-60\). Each drink makes \(3-1.20=1.80\) pounds before costs, and the daily costs of £60 are subtracted.
    2. \(1.8d-60>0\), \(d>33.3\), so at least 34 drinks. A profit means the expression is greater than 0. We round up because 33 drinks give a small loss. The café needs at least 34 drinks.
    3. \(1.8d-60\ge100\), \(1.8d\ge160\), \(d\ge88.9\), so at least 89 drinks. Add 60 to both sides, then divide by 1.8. We round up because 88 drinks give only £98.40.
    4. \(1.8\times80-60=144-60=\) £84. The target of £100 is not reached. So 80 drinks give a profit of £84, which is £16 short of the target.

17. Answer to Question 17

    1. \((x-2)(x-3)\lt 0\). The graph is below the axis between the roots: \(2\lt x\lt 3\). The graph of \(y=(x-2)(x-3)\) is a U-shape crossing the axis at 2 and 3, and it is below the axis between them. The solution is \(2\lt x\lt 3\).
    2. \(x^2\ge16\) gives \(x\le-4\) or \(x\ge4\). The roots of \(x^2=16\) are \(-4\) and \(4\). The squares are at least 16 outside these values, so \(x\le-4\) or \(x\ge4\).
    3. \((x-4)(x+2)>0\). The graph is above the axis outside the roots: \(x\lt -2\) or \(x>4\). Factorise to find the roots \(-2\) and \(4\). We want the parts of the U-shaped graph above the axis.
    4. \(-3.16\lt x\lt 3.16\), so the integers are −3, −2, −1, 0, 1, 2, 3 (7 integers). The roots of \(x^2=10\) are \(\pm\sqrt{10}=\pm3.16\), so \(-3.16\lt x\lt 3.16\) and there are 7 integers.

18. Answer to Question 18

    1. \(x=1\) (solid), \(y=5\) (solid) and \(y=x\) (dashed, because the inequality is \(>\) not \(\ge\)).
    2. \(y>1\) and \(y\le5\): \((1,2),(1,3),(1,4),(1,5)\).
    3. \(x=1\): 4 points. \(x=2\): \((2,3),(2,4),(2,5)\). \(x=3\): \((3,4),(3,5)\). \(x=4\): \((4,5)\).
    4. There are \(4+3+2+1=\) 10 points. \((3,3)\) is not in \(R\) because \(y>x\) needs \(3>3\), which is false (it lies on the dashed line).

19. Answer to Question 19

    1. Multiply by 6: \(2(2x+1)=3(x-4)\), \(4x+2=3x-12\), so \(x=-14\). The common denominator of 3 and 2 is 6. Check: \(\tfrac{-27}{3}=-9\) and \(\tfrac{-18}{2}=-9\).
    2. Multiply by 15: \(5x+3(x-2)=30\), \(8x-6=30\), \(8x=36\), so \(x=4.5\). The common denominator of 3 and 5 is 15. Check: \(1.5+0.5=2\).
    3. \(\dfrac{n+7}{3}=n-5\), \(n+7=3n-15\), \(22=2n\), so \(n=11\). Let the number be \(n\) and write the sentence as an equation, then multiply both sides by 3.
    4. \(\dfrac{11+7}{3}=\dfrac{18}{3}=6\) and \(11-5=6\) ✓ Both sides give 6, so the number is 11.

20. Answer to Question 20

    1. \((n+2)+(n+4)=n+22\), so \(2n+6=n+22\), \(n=16\): the numbers are 16, 18, 20.
    2. \(n+(n+2)+(n+4)=3n+6=3(n+2)\), and since \(n+2\) is even, \(3(n+2)=6\times\dfrac{n+2}{2}\), a multiple of 6.
    3. \(x=\dfrac{12-k}{3}>0\), so \(k\lt 12\). Make \(x\) the subject, then the solution is positive when \(12-k>0\), which gives \(k\lt 12\).
    4. For \(x=2\) (the smallest positive even number): \(k=12-6=6\). For \(x=4\): \(k=0\); for larger \(x\), \(k\) is smaller. So the largest is \(k=6\).