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Either paper 2 marks
The graphs of \(y = -x + 7\) and \(y = x + 1\) cross at the point \((3, 4)\).
(a)Write down the solution of the simultaneous equations
\(y = -x + 7\)
\(y = x + 1\)
(b)The graph of \(y = -3x + 13\) also passes through \((3, 4)\).
Write down the solution of the simultaneous equations
\(y = -x + 7\)
\(y = -3x + 13\)
Show answer
(a) \(x = 3\), \(y = 4\)
(b) \(x = 3\), \(y = 4\)The crossing point is the one point that is on both graphs, so its coordinates are the one pair of values that fits both equations.
\(x = 3\), \(y = 4\)
The coordinates are written \((3, 4)\), but the solution is written as two values — that is the only difference between the two answers.
All three graphs pass through \((3, 4)\), so any two of them cross there.
\(x = 3\), \(y = 4\)
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Either paper 3 marks
The graphs of \(y = 2x + 9\) and \(y = -2x - 7\) are both drawn on the grid.
(a)State the coordinates of the point where the two lines cross.
(b)Use the graph to solve the simultaneous equations \(y = 2x + 9\) and \(y = -2x - 7\)
Show answer
(a) \((-4, 1)\)
(b) \(x = -4\), \(y = 1\)Read the crossing point off the grid the same way as any other point: across to the \(x\)-axis, then down or up to the \(y\)-axis.
The lines cross at \((-4, 1)\).
A point on the first line fits the first equation, and a point on the second line fits the second equation. The crossing point is on both, so it is the one pair of values that fits both equations at once.
\(x = -4\), \(y = 1\)
Check it by putting \(x = -4\) into each equation: \(-8 + 9 = 1\) and \(8 - 7 = 1\)
- (b) M1uses the crossing point of the two lines, \((-4, 1)\)
- (b) A1\(x = -4\) and \(y = 1\)
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Either paper 3 marks
Omar is solving the simultaneous equations
\(y = -x - 4\)
\(y = 2x - 7\)
by drawing both graphs. Omar says the graphs cross at \((1, -5)\).
(a)Calculate the value of \(y\) given by \(y = -x - 4\) when \(x = 1\).
(b)Calculate the value of \(y\) given by \(y = 2x - 7\) when \(x = 1\).
(c)Is his answer right? Give a reason.
Show answer
(a) \(-5\)
(b) \(-5\)
(c) Yes — the point is on both graphsPut \(x = 1\) into \(y = -x - 4\):
\(y = -1 - 4 = -5\)
Put the same \(x = 1\) into \(y = 2x - 7\):
\(y = 2 - 7 = -5\)
A crossing point has to be on both graphs, so both equations have to give the same \(y\) at the same \(x\). At \(x = 1\) they give \(-5\) and \(-5\).
Answer: Yes — the point is on both graphs.
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Either paper 3 marks
The graphs of \(y = -x - 3\) and \(y = 2\) are both drawn on the grid.
(a)State the coordinates of the point where the two graphs cross.
(b)Use the graph to solve the simultaneous equations \(y = -x - 3\) and \(y = 2\)
Show answer
(a) \((-5, 2)\)
(b) \(x = -5\), \(y = 2\)The flat line is \(y = 2\). Follow it across until it meets the sloping line, then read down (or up) to the \(x\)-axis.
They cross at \((-5, 2)\).
The crossing point is on both graphs, so its coordinates are the solution:
\(x = -5\), \(y = 2\)
Because the second equation is just \(y = 2\), you can check it without the graph: put \(y = 2\) into the first equation and it gives \(x = -5\).
- (b) M1follows the line \(y = 2\) across to where it meets the sloping line
- (b) A1both values stated: \(x = -5\), \(y = 2\)
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Either paper 2 marks
The graphs of \(y = -2x - 10\) and \(y = x + 8\) are both drawn on the grid. Both axes are numbered in twos.
Use the graph to solve the simultaneous equations.
Show answer
\(x = -6\), \(y = 2\)The solution is the point where the two lines cross, because that is the only point that is on both graphs.
Check the scale before reading it — each square is \(2\), not \(1\).
The lines cross at \((-6, 2)\), so
\(x = -6\), \(y = 2\)
- M1reads the crossing point using the scale of 2 per square, \((-6, 2)\)
- A1\(x = -6\) and \(y = 2\)
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Either paper 2 marks
The graphs of \(y = x - 5\) and \(y = x + 4\) are both drawn on the grid.
Explain why the simultaneous equations \(y = x - 5\) and \(y = x + 4\) have no solution.
Show answer
Both lines have gradient 1, so they are parallel and never cross — there is no point on both graphs, so no pair of values fits both equations.Look at the two equations. The number in front of \(x\) is \(1\) in both, so both lines have gradient \(1\) and they are parallel.
Parallel lines never meet, so there is no point that lies on both graphs.
Solving a pair of equations graphically means finding the crossing point, and here there is not one — so the equations have no solution.
You can see the same thing without the graph: if \(y\) were equal to both \(x - 5\) and \(x + 4\), then \(-5\) would have to equal \(4\), and it does not.
- C1both lines have gradient \(1\), so they are parallel
- C1parallel lines never cross, so there is no point on both graphs and no solution
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Either paper 3 marks
The graph of \(y = -x + 2\) has been drawn on the grid.
(a)On the same grid, draw the graph of \(y = x - 6\).
(b)Use the two graphs to solve the simultaneous equations
\(y = -x + 2\)
\(y = x - 6\)
You cannot draw on this page. Draw (a) on paper, then show the answer and check your drawing against the model.
Show answer
(a) A straight line through \((0, -6)\) and \((4, -2)\), with gradient \(1\)
(b) \(x = 4\), \(y = -2\)(a)The line crosses the \(y\)-axis at \(-6\), and its gradient is \(1\), so from that point go one square right and \(1\) up each time, and the same back the other way.
Plot two or three of those points and rule the line through them, right across the grid.
The two lines cross at \((4, -2)\), and that crossing point is the only pair of values on both graphs.
\(x = 4\), \(y = -2\)
- (a) B1a straight line with gradient \(1\)
- (a) B1the correct line, through \((0, -6)\) and \((4, -2)\)
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Either paper 2 marks
The graphs of \(y = 3x - 9\) and \(y = -2x - 4\) cross at the point \((1, -6)\).
(a)Write down the solution of the simultaneous equations
\(y = 3x - 9\)
\(y = -2x - 4\)
(b)The graph of \(y = -3x - 3\) also passes through \((1, -6)\).
Write down the solution of the simultaneous equations
\(y = 3x - 9\)
\(y = -3x - 3\)
Show answer
(a) \(x = 1\), \(y = -6\)
(b) \(x = 1\), \(y = -6\)The crossing point is the one point that is on both graphs, so its coordinates are the one pair of values that fits both equations.
\(x = 1\), \(y = -6\)
The coordinates are written \((1, -6)\), but the solution is written as two values — that is the only difference between the two answers.
All three graphs pass through \((1, -6)\), so any two of them cross there.
\(x = 1\), \(y = -6\)
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Either paper 3 marks
The graphs of \(y = -x - 2\) and \(y = x + 4\) are both drawn on the grid.
(a)State the coordinates of the point where the two lines cross.
(b)Use the graph to solve the simultaneous equations \(y = -x - 2\) and \(y = x + 4\)
Show answer
(a) \((-3, 1)\)
(b) \(x = -3\), \(y = 1\)Read the crossing point off the grid the same way as any other point: across to the \(x\)-axis, then down or up to the \(y\)-axis.
The lines cross at \((-3, 1)\).
A point on the first line fits the first equation, and a point on the second line fits the second equation. The crossing point is on both, so it is the one pair of values that fits both equations at once.
\(x = -3\), \(y = 1\)
Check it by putting \(x = -3\) into each equation: \(3 - 2 = 1\) and \(-3 + 4 = 1\)
- (b) M1uses the crossing point of the two lines, \((-3, 1)\)
- (b) A1\(x = -3\) and \(y = 1\)
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Either paper 3 marks
Tomas is solving the simultaneous equations
\(y = 3x - 7\)
\(y = -x - 3\)
by drawing both graphs. Tomas says the graphs cross at \((-1, -10)\).
(a)Work out the value of \(y\) given by \(y = 3x - 7\) when \(x = -1\).
(b)Work out the value of \(y\) given by \(y = -x - 3\) when \(x = -1\).
(c)Is his answer right? Give a reason.
Show answer
(a) \(-10\)
(b) \(-2\)
(c) No — the point is on the first graph but not on the secondPut \(x = -1\) into \(y = 3x - 7\):
\(y = -3 - 7 = -10\)
Put the same \(x = -1\) into \(y = -x - 3\):
\(y = 1 - 3 = -2\)
A crossing point has to be on both graphs, so both equations have to give the same \(y\) at the same \(x\). At \(x = -1\) they give \(-10\) and \(-2\).
Answer: No — the point is on the first graph but not on the second.