Year 10 simultaneous equations test answers to download as PDF

These Year 10 simultaneous equations test answers give the worked solutions and the marks for every question of the simultaneous 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. Multiply to match coefficients: \(9x+6y=36\) and \(10x-6y=2\) (1). Add: \(19x=38\) (1). \(x=2\) (1). \(y=3\) (1). Multiply the first equation by 3 and the second by 2, so the \(y\) terms are \(+6y\) and \(-6y\) and cancel when added. Substitute \(x=2\) into \(3x+2y=12\): \(6+2y=12\), so \(y=3\). Check: \(10-9=1\).
Check both equations with \(x=2\), \(y=3\): \(6+6=12\) and \(10-9=1\). A common error is to subtract \(9x+6y=36\) and \(10x-6y=2\), which leaves \(12y\) and does not eliminate anything.
Answer to Question 2 (4 marks)
1. \(2a+3c=34\) and \(3a+c=30\) (1). Multiply the second by 3: \(9a+3c=90\) (1). Subtract: \(7a=56\), \(a=8\) (1). Then \(c=30-24=6\): adult £8, child £6 (1). Multiplying by 3 makes the \(c\) terms match, so subtracting removes \(c\). Check: \(16+18=34\) and \(24+6=30\). One adult ticket costs £8 and one child ticket costs £6.
Letters must be defined: \(a\) is the price of an adult ticket in pounds and \(c\) the price of a child ticket. Substitution also works, with \(c=30-3a\).
Answer to Question 3 (4 marks)
1. \(2x+3x-4=11\) (1), \(5x=15\) (1), \(x=3\), \(y=5\) (1). \(3\times3-4=5\) ✓ but \(3+5=8\ne7\) (1). Substitute \(y=3x-4\) into the second equation and solve for \(x\). The point \((3,5)\) fits \(y=3x-4\) but not \(x+y=7\), so it is not on that line.
Check the solution in both original equations: \(y=3\times3-4=5\) and \(2\times3+5=11\). Elimination is also possible by writing the first equation as \(3x-y=4\) and adding it to \(2x+y=11\), which gives \(5x=15\) again. The point \((3,5)\) is where the two lines meet.
Answer to Question 4 (4 marks)
1. \(x+2x+1=10\) (1), \(x=3\), \(y=7\): \((3,7)\) (1) Substitute \(y=2x+1\) into \(x+y=10\). The lines cross at \((3,7)\).
2. \(2+1.5m=5+m\) (1), \(0.5m=3\), \(m=6\) miles (1) Set the two costs equal and solve. Check: \(2+9=11\) and \(5+6=11\). The charges are the same for 6 miles.
In (a), check the point in both equations: \(2\times3+1=7\) and \(3+7=10\). In (b), let \(m\) be the number of miles; for journeys longer than 6 miles the second firm is cheaper because it charges less per mile.
Answer to Question 5 (4 marks)
1. \(x^2+(x+2)^2=34\) (1), \(2x^2+4x-30=0\) so \(x^2+2x-15=0\) (1), \((x+5)(x-3)=0\), \(x=-5\) or \(3\) (1). Solutions \((3,5)\) and \((-5,-3)\) (1). Substitute \(y=x+2\) into the circle equation and expand \((x+2)^2=x^2+4x+4\). Divide by 2 and factorise, then find \(y\) from \(y=x+2\). Check: \(9+25=34\) and \(25+9=34\).
Both pairs must be given, matched correctly. A common error is to pair \(x=3\) with \(y=-3\); always use the linear equation to find \(y\). Geometrically, the line meets the circle of radius \(\sqrt{34}\) at two points.

Total : 20 marks

How to use these Year 10 simultaneous 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 simultaneous equations lesson and redo the worksheet.

Also, the Year 10 simultaneous 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.