Exam questions · Maths · Further Algebra
Simultaneous Equations
- 6 exam questions
- 20 marks
- 9 quick checks
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1 Use [2 marks]
The graph shows the straight lines \(A\) and \(B\). Use the graph to solve the simultaneous equations \(y = x - 1\) and \(2x + y = 11\).
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Model answer
The lines cross at \((4, 3)\), so \(x = 4\) and \(y = 3\).
Mark scheme
- \(x = 4\) — B1
- \(y = 3\) — B1
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2 Solve [3 marks]
Solve the simultaneous equations \(x + 2y = 11\) and \(x - y = 2\).
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Model answer
Subtracting gives \(3y = 9\), so \(y = 3\). Then \(x = 5\). Check: \(5 - 3 = 2\).
Mark scheme
- \(3y = 9\) or another correct elimination — M1
- \(y = 3\) — A1
- \(x = 5\) — A1
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3 Solve [3 marks]
Solve the simultaneous equations \(4x + y = 19\) and \(3x - y = 9\).
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Model answer
Adding gives \(7x = 28\), so \(x = 4\). Then \(16 + y = 19\), so \(y = 3\). Check: \(12 - 3 = 9\).
Mark scheme
- \(7x = 28\) — M1
- \(x = 4\) — A1
- \(y = 3\) — A1
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4 Solve [4 marks]
Solve the simultaneous equations \(4x + 3y = 17\) and \(2x - y = 1\).
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Model answer
From the second equation \(y = 2x - 1\). Substituting gives \(4x + 3(2x - 1) = 17\), so \(10x - 3 = 17\) and \(x = 2\). Then \(y = 3\).
Mark scheme
- \(y = 2x - 1\), or the equations made to match — M1
- \(4x + 3(2x - 1) = 17\) — M1
- \(x = 2\) — A1
- \(y = 3\) — A1
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5 Calculate [4 marks]
2 large boxes and 3 small boxes have a total mass of 29 kg. 1 large box and 2 small boxes have a total mass of 17 kg. Calculate the mass of a large box and the mass of a small box.
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Model answer
\(2l + 3s = 29\) and \(l + 2s = 17\). Doubling the second gives \(2l + 4s = 34\). Subtracting the first gives \(s = 5\). Then \(l = 17 - 10 = 7\). A large box is 7 kg and a small box is 5 kg.
Mark scheme
- Two correct equations — M1
- A correct elimination step — M1
- Small box 5 kg — A1
- Large box 7 kg — A1
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6 Solve [4 marks]
Solve the simultaneous equations \(y = x^2\) and \(y = 3x + 4\).
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Model answer
Setting them equal gives \(x^2 = 3x + 4\), so \(x^2 - 3x - 4 = 0\) and \((x - 4)(x + 1) = 0\). So \(x = 4\) or \(x = -1\), and \(y = 16\) or \(y = 1\). The solutions are \((4, 16)\) and \((-1, 1)\).
Mark scheme
- \(x^2 = 3x + 4\) — M1
- \((x - 4)(x + 1)\) — M1
- \(x = 4\) and \(x = -1\) — A1
- \((4, 16)\) and \((-1, 1)\) — A1
Quick check
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1
Solve \(x + y = 9\) and \(x - y = 1\).
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C: \(x = 5\), \(y = 4\)
Adding gives \(2x = 10\), so \(x = 5\) and \(y = 4\).
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2
When do you add the two equations in elimination?
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B: When the terms in one letter are opposites
Opposites such as \(+3y\) and \(-3y\) cancel when added.
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3
Solve \(3x + 2y = 16\) and \(x + 2y = 8\).
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A: \(x = 4\), \(y = 2\)
Subtract: \(2x = 8\), so \(x = 4\). Then \(4 + 2y = 8\) gives \(y = 2\).
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4
Solve \(2x + 3y = 13\) and \(3x - y = 3\).
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D: \(x = 2\), \(y = 3\)
Multiply the second equation by 3 and add: \(11x = 22\), so \(x = 2\), \(y = 3\).
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5
Solve \(y = 2x + 1\) and \(3x + y = 16\).
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C: \(x = 3\), \(y = 7\)
\(3x + 2x + 1 = 16\), so \(x = 3\) and \(y = 7\).
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6
Two straight lines are parallel. How many solutions do their simultaneous equations have?
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B: None
Parallel lines never meet, so there is no point on both.
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7
2 adult and 3 child tickets cost \(\pounds 19\). 3 adult and 1 child ticket cost \(\pounds 18\). What does an adult ticket cost?
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A: \(\pounds 5\)
\(2a + 3c = 19\) and \(3a + c = 18\) give \(a = 5\), \(c = 3\).
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8
Solve \(y = x^2\) and \(y = x + 6\).
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D: \((3, 9)\) and \((-2, 4)\)
\(x^2 = x + 6\) gives \((x - 3)(x + 2) = 0\).
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9
Solve \(x^2 + y^2 = 25\) and \(y = x + 1\).
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C: \((3, 4)\) and \((-4, -3)\)
\(x^2 + (x + 1)^2 = 25\) gives \(x^2 + x - 12 = 0\), so \(x = 3\) or \(x = -4\).