Year 8 solving linear equations test answers to download as PDF

These Year 8 solving linear equations test answers give the worked solutions and the marks for every question of the solving linear equations test. The test is marked out of 20, so you can work out your score.

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Answer to Question 1 (4 marks)
1. Add 7 to both sides to undo the subtraction: \(5x=25\) (1 mark). Divide both sides by 5: \(x=5\) (1 mark). Check: \(5\times5-7=25-7=18\).
2. Expand the bracket: \(3x+6=2x+11\) (1 mark). Subtract \(2x\) from both sides so that x is only on the left: \(x+6=11\). Subtract 6: \(x=5\) (1 mark).
Check (b): the left side is \(3(5+2)=21\) and the right side is \(2\times5+11=21\), so both sides balance. It is a coincidence that both equations have the same solution; always substitute back into the original equation, not into a later line of working, so that an early slip is caught.
Answer to Question 2 (4 marks)
1. Subtract \(3x\) from both sides: \(4x+2=18\). Subtract 2: \(4x=16\) (1 mark). Divide by 4: \(x=4\) (1 mark). Check: \(7\times4+2=30\) and \(3\times4+18=30\).
2. Multiply both sides by 3 to clear the fraction: \(x+5=12\). Subtract 5: \(x=7\). Check: \((7+5)\div3=4\). (1 mark)
3. Subtract 4 from both sides: \(-3x=15\). Divide both sides by \(-3\): \(x=-5\). Check: \(4-3\times(-5)=4+15=19\). (1 mark)
In (c) the sign of x is easy to lose. Another route is to add \(3x\) to both sides first, \(4=19+3x\), which gives \(3x=-15\) and the same answer.
Answer to Question 3 (4 marks)
1. The perimeter is the sum of the three sides: \(x+(x+3)+(2x-1)=30\). Collect like terms: \(4x+2=30\). (1 mark)
2. Subtract 2 from both sides: \(4x=28\). Divide by 4: \(x=7\). (1 mark)
3. Substitute \(x=7\): the sides are 7 cm, \(7+3=10\) cm and \(2\times7-1=13\) cm. The longest side is 13 cm. (1 mark)
4. \(7+10+13=30\) cm, which matches the perimeter given. (1 mark)
It is also worth checking that the triangle can exist: the two shorter sides give \(7+10=17\), which is greater than 13, so the three lengths do form a real triangle.
Answer to Question 4 (4 marks)
1. Let m be the number of miles. Taxi A costs \(2.5+1.6m\) and taxi B costs \(4+1.1m\). The fares are equal when \(2.5+1.6m=4+1.1m\). (1 mark)
2. Subtract \(1.1m\) from both sides: \(2.5+0.5m=4\). Subtract 2.5: \(0.5m=1.5\) (1 mark). Divide by 0.5: \(m=3\) miles (1 mark).
3. For 5 miles, taxi A costs \(2.5+1.6\times5=£10.50\) and taxi B costs \(4+1.1\times5=£9.50\). Taxi B is cheaper, by £1. (1 mark)
Check (b): at 3 miles both cost £7.30. Beyond 3 miles taxi B's lower rate per mile makes it the better choice.
Answer to Question 5 (4 marks)
1. Consecutive even numbers differ by 2, so they are n, \(n+2\) and \(n+4\). The equation is \(n+(n+2)+(n+4)=114\). (1 mark)
2. Collect terms: \(3n+6=114\). Subtract 6: \(3n=108\). Divide by 3: \(n=36\) (1 mark). The three numbers are 36, 38, 40 (1 mark). Check: \(36+38+40=114\).
3. \(3n+6=100\) gives \(3n=94\) and \(n=31.33\ldots\), which is not a whole number, let alone an even one, so no. (1 mark)
A neat reason for (c): the sum \(3n+6=3(n+2)\) is always a multiple of 3, and 100 is not divisible by 3.

Total : 20 marks

How to use these Year 8 solving linear equations test answers

First, sit the test under timed conditions before you read anything here. Then mark each answer and give yourself the marks shown, including method marks. When you lose marks, read the worked solution and find the step you missed. After that, revise the solving linear equations lesson and redo the worksheet.

Also, the Year 8 solving linear equations test answers are available as a PDF, so a teacher can mark printed tests 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.