Year 8 solving linear equations worksheet answers to download as PDF

These Year 8 solving linear equations worksheet answers give a full worked solution to all 20 questions of the solving linear equations worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
Undo the operations in reverse order with inverse operations, doing the same thing to both sides each time.
1. Subtract 9: \(x=8\). Check: \(8+9=17\).
2. Divide by 5: \(x=9\).
3. Add 4: \(3x=15\). Divide by 3: \(x=5\). Check: \(15-4=11\).
4. Subtract 2: \(\dfrac{x}{4}=4\). Multiply by 4: \(x=16\).
5. Subtract 7: \(-2x=-6\). Divide by \(-2\): \(x=3\). Check: \(7-6=1\).
Answer to Question 2
1. Expand: \(2x+6=18\). Subtract 6: \(2x=12\). Divide by 2: \(x=6\). Check: \(2\times9=18\).
2. Expand: \(4x-20=12\). Add 20: \(4x=32\). Divide by 4: \(x=8\). Check: \(4\times3=12\).
3. Expand: \(6x+3=27\). Subtract 3: \(6x=24\), so \(x=4\). Check: \(3\times9=27\).
4. Expand: \(5x-10+3=18\), so \(5x-7=18\). Add 7: \(5x=25\), so \(x=5\). Check: \(5\times3+3=18\).
In (a) you could also divide both sides by 2 first: \(x+3=9\), giving the same answer more quickly.
Answer to Question 3
1. Subtract \(2x\): \(3x+3=18\). Subtract 3: \(3x=15\), so \(x=5\). Check: \(28=28\).
2. Subtract \(3x\): \(4x-4=12\). Add 4: \(4x=16\), so \(x=4\). Check: \(24=24\).
3. Subtract x: \(3x+9=-6\). Subtract 9: \(3x=-15\), so \(x=-5\). Check: \(-11=-11\).
4. Expand: \(2x+8=x+11\). Subtract x: \(x+8=11\), so \(x=3\). Check: \(14=14\).
Subtracting the smaller x term keeps the x coefficient positive, which avoids sign mistakes.
Answer to Question 4
1. Multiply both sides by 3: \(x=21\). Check: \(21\div3=7\).
2. Multiply by 5: \(x+4=15\). Subtract 4: \(x=11\). Check: \(15\div5=3\).
3. Add 1: \(\dfrac{2x}{3}=6\). Multiply by 3: \(2x=18\), so \(x=9\).
4. The LCM of 2 and 3 is 6. Multiply every term by 6: \(3x+2x=60\), so \(5x=60\) and \(x=12\). Check: \(6+4=10\).
Multiplying every term by the denominator removes the fractions in one step.
Answer to Question 5
1. Subtract x: \(x+3=-4\). Subtract 3: \(x=-7\). Check: \(-11=-11\).
2. Subtract 5: \(3x=-3\). Divide by 3: \(x=-1\). Check: \(-3+5=2\).
3. Subtract 5: \(-4x=12\). Divide by \(-4\): \(x=-3\). Check: \(5+12=17\).
4. Subtract \(3x\): \(5x+1=4\). Subtract 1: \(5x=3\), so \(x=\dfrac{3}{5}\). Check: both sides equal 5.8.
Negative and fractional answers are perfectly valid; do not change the equation to make the answer whole.
Answer to Question 6
1. \(4n-7=29\). Add 7: \(4n=36\). Divide by 4: \(n=9\).
2. \(2(n+5)=24\). Divide by 2: \(n+5=12\), so \(n=7\).
3. \(n+12=3n\). Subtract n: \(12=2n\), so \(n=6\). Check: \(18=3\times6\).
4. \(n+(n+1)+(n+2)=51\), so \(3n+3=51\), \(3n=48\), \(n=16\). The numbers are 16, 17 and 18.
For consecutive numbers, calling the smallest n makes the others \(n+1\) and \(n+2\), so only one letter is needed.
Answer to Question 7
1. Two adults cost \(2\times5=£10\) and c children cost \(£3c\): \(10+3c=19\).
2. Subtract 10: \(3c=9\). Divide by 3: \(c=3\) children.
3. \(3\times5+3c=30\), so \(15+3c=30\), \(3c=15\), \(c=5\) children.
4. With a adults and a children: \(5a+3a=56\), so \(8a=56\) and \(a=7\) adults (and 7 children).
Answer to Question 8
1. A: \(20+5h\). B: \(8+8h\).
2. \(20+5h=8+8h\), so \(12=3h\), \(h=4\).
3. \(20+5\times4=\) £40 (check B: \(8+32=40\)).
4. A: \(20+30=£50\). B: \(8+48=£56\). Company A saves her £6.
Company A has the higher fixed charge but the lower hourly rate, so it becomes the cheaper choice for any hire longer than 4 hours.
Answer to Question 9
1. £150 plus £4 for each of the p extra pupils: \(150+4p=210\).
2. Subtract 150: \(4p=60\). Divide by 4: \(p=15\).
3. p counts pupils over 20, so \(20+15=\)35 pupils went.
4. Shared equally: \(7n=210\), so \(n=\)30 pupils.
Be careful to answer the question asked: p is only the number of pupils above 20, not the total.
Answer to Question 10
1. Perimeter = two lengths + two widths: \(2(2x+3)+2x=4x+6+2x=\)\(6x+6=36\).
2. Subtract 6: \(6x=30\). Divide by 6: \(x=5\).
3. Width 5 m, length \(2\times5+3=\)13 m. Check: \(2(13+5)=36\).
4. Area = length × width: \(13\times5=\)65 m².
5. The fence goes round the 36 m perimeter: \(36\times9.50=\)£342.
Answer to Question 11
1. \(60+15m=240\). Subtract 60: \(15m=180\). Divide by 15: \(m=12\) minutes.
2. \(240-20m=100\). Subtract 100 and add \(20m\): \(140=20m\), so \(m=7\) minutes.
3. \(60+15m=100+10m\). Subtract \(10m\) and 60: \(5m=40\), so \(m=8\) minutes.
4. \(60+15\times8=60+120=\)180°C; check the other oven: \(100+80=180\).
Each part is a linear equation: write the temperature rule, set it equal to the target value, then solve.
Answer to Question 12
1. Use pounds: 8p is £0.08. \(12+0.08t=16.40\).
2. Subtract 12: \(0.08t=4.40\). Divide by 0.08: \(t=55\) extra texts.
3. \(55-30=25\) extra texts: \(12+0.08\times25=12+2=\)£14.
4. \(12+0.08t=10+0.12t\). Subtract 10 and \(0.08t\): \(2=0.04t\), so \(t=50\); each bill is \(12+4=\)£16.
Answer to Question 13
1. Tom gains £6 a week: \(35+6w\). Jess loses £3 a week: \(80-3w\).
2. \(35+6w=80-3w\). Add \(3w\) and subtract 35: \(9w=45\), so \(w=5\) weeks.
3. \(35+6\times5=\)£65 each (Jess: \(80-15=65\)).
4. \(80-3w<20\) gives \(3w>60\), so \(w>20\). At 20 weeks she has exactly £20, so she first has less in week 21.
Answer to Question 14
1. Weekly saving is \(12-3=9\), so \(45+9w\)
2. \(45+9w=189\), \(9w=144\), \(w=16\) weeks
3. \(75+9w=189\), \(9w=114\), \(w=12.67\), so 13 weeks (the money passes £189 during week 13).
4. No: after 15 weeks he has \(45+135=£180\), which is £9 short.
Each week adds £9, so the total after w weeks is the starting amount plus \(9w\); rounding up in part (c) is needed because he must reach the full price.
Answer to Question 15
1. \(3n+8=5n-6\). Subtract \(3n\) and add 6: \(14=2n\), so \(n=7\). Check: \(29=29\).
2. \(5(n+4)=3n+30\). Expand: \(5n+20=3n+30\). Then \(2n=10\), so \(n=5\). Check: \(45=45\).
3. \(n+\dfrac{n}{2}=27\) means \(\dfrac{3n}{2}=27\). Multiply by 2 and divide by 3: \(n=18\).
4. No: the solution of Ali’s puzzle is 7, which is positive.
Answer to Question 16
1. \(4x-12=2x+6\), \(2x=18\), \(x=9\)
2. Left: \(4(9-3)=24\). Right: \(2\times9+6=24\). They match, so Ali is correct.
3. Bea: forgot the minus when expanding (\(4x+12\)). Cal: divided \(18\) by \(6\) instead of \(2\) (or wrong subtraction). Dev: expanded to \(4x-12\) but then subtracted the wrong terms to get \(2x=9\).
Substituting each wrong answer into the original equation would have shown at once that the two sides are not equal.
Answer to Question 17
1. \(x+2x+10+3x-10=180\), so \(6x=180\)
2. \(x=30\)
3. \(30^\circ\), \(2\times30+10=\) \(70^\circ\), \(3\times30-10=\) \(80^\circ\)
4. Check \(30+70+80=180\). All angles differ: scalene.
5. \(4y+20=180\), \(y=40\): angles \(40^\circ\), \(40^\circ\), \(100^\circ\) (isosceles).
The angle sum of 180° gives the equation; checking that the three angles really add up to 180° confirms the solution.
Answer to Question 18
1. Rectangle perimeter \(2(x+4)+2\times5\); square perimeter \(4(x-1)\): \(2(x+4)+10=4(x-1)\).
2. Expand: \(2x+18=4x-4\). Subtract \(2x\) and add 4: \(22=2x\), so \(x=11\).
3. Square side \(11-1=\)10 cm; rectangle 15 cm by 5 cm. Both perimeters are 40 cm.
4. Square \(10\times10=100\) cm², rectangle \(15\times5=75\) cm². Equal perimeters do not mean equal areas: the square is larger.
Answer to Question 19
1. Every ticket that is not adult is a child ticket: \(240-a\).
2. Income from adults \(6a\) plus children \(4(240-a)\): \(6a+4(240-a)=1180\).
3. Expand: \(6a+960-4a=1180\). Then \(2a=220\), so \(a=110\) adult tickets.
4. \(240-110=\)130 child tickets. Check: \(110\times6+130\times4=660+520=1180\).
Answer to Question 20
1. In 6 years both are 6 years older: Ben \(x+6\), mother \(4x+6\).
2. The mother will be three times Ben’s age: \(4x+6=3(x+6)\).
3. Expand: \(4x+6=3x+18\). Subtract \(3x\) and 6: \(x=12\). Ben is 12.
4. Mother now \(4\times12=\)48; in 6 years 54, when Ben is 18, and \(3\times18=54\).

How to use these Year 8 solving linear equations worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the solving linear equations lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 8 solving linear equations worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 8 solving linear equations worksheet answers show every step, so nothing is left unexplained.