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Either paper 3 marks
Solve these simultaneous equations.
\(4x + 5y = 39\)
\(4x - 5y = -31\)
Show answer
\(x = 1\), \(y = 7\)The \(y\) terms are \(+5y\) and \(-5y\), so adding the two equations removes \(y\) altogether:
\(8x = 8\)
\(x = 1\)
Substitute \(x = 1\) into the first equation:
\(4 \times (1) + 5y = 39\)
\(5y = 35\)
\(y = 7\)
- M1adds the equations to remove \(y\): \(8x = 8\)
- M1substitutes \(x = 1\) to get \(5y = 35\)
- A1\(x = 1\) and \(y = 7\)
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Either paper 3 marks
Solve the simultaneous equations
\(y = 2x + 6\)
\(4x + 3y = 48\)
Show answer
\(x = 3\), \(y = 12\)The first equation already gives \(y\) in terms of \(x\), so put that expression in place of \(y\) in the second equation:
\(4x + 3(2x + 6) = 48\)
Multiply out the bracket:
\(4x + 6x + 18 = 48\)
Collect the \(x\) terms:
\(10x + 18 = 48\)
\(10x = 30\)
\(x = 3\)
Substitute back into \(y = 2x + 6\):
\(y = 2 \times (3) + 6 = 6 + 6 = 12\)
- M1substitutes for \(y\): \(4x + 3(2x + 6) = 48\)
- M1collects and solves: \(10x = 30\)
- A1\(x = 3\) and \(y = 12\)
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Either paper 3 marks
\(x = 3\) and \(y = 8\) is the solution of the simultaneous equations
\(6x + ay = 34\)
\(x - 5y = -37\)
Calculate the value of \(a\).
Show answer
\(2\)The solution makes both equations true, so put \(x = 3\) and \(y = 8\) into the first one. Only that equation contains \(a\).
\(6 \times 3 + a \times 8 = 34\)
\(18 + 8a = 34\)
\(8a = 34 - 18 = 16\)
\(a = 2\)
- M1substitutes into the first equation: \(6 \times 3 + a \times 8 = 34\)
- M1isolates the term in \(a\): \(8a = 16\)
- A1\(a = 2\)
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Either paper 3 marks
Solve these simultaneous equations.
\(5x + 6y = 18\)
\(5x + 2y = 26\)
Show answer
\(x = 6\), \(y = -2\)Both equations have \(5x\), and the two \(x\) terms have the same sign, so it is subtracting one equation from the other that removes \(x\).
Taking the second equation from the first:
\(4y = -8\)
\(y = -2\)
Substitute \(y = -2\) into the first equation:
\(5x + 6 \times (-2) = 18\)
\(5x = 30\)
\(x = 6\)
- M1subtracts the equations to remove \(x\): \(4y = -8\)
- M1substitutes \(y = -2\) to get \(5x = 30\)
- A1\(x = 6\) and \(y = -2\)
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Either paper 4 marks
Solve the simultaneous equations
\(5x + 4y = -11\)
\(7x + 12y = -9\)
Show answer
\(x = -3\), \(y = 1\)Neither letter has matching coefficients yet. The \(y\) coefficients are 4 and 12, and \(4 \times 3 = 12\), so multiply every term of the first equation by 3:
\(15x + 12y = -33\)
Now subtract the second equation from this one. The \(y\) terms cancel:
\(8x = -24\)
\(x = -3\)
Substitute \(x = -3\) into the original first equation:
\(5 \times (-3) + 4y = -11\)
\(4y = 4\)
\(y = 1\)
- M1multiplying the first equation by 3: \(15x + 12y = -33\)
- M1subtracting to eliminate \(y\): \(8x = -24\)
- A1\(x = -3\)
- A1\(y = 1\) from substituting into an original equation
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Either paper 4 marks
A stationery shop sells folders and notepads.
3 folders and 5 notepads cost £52 altogether.
9 folders and 2 notepads cost £91 altogether.
Calculate the cost of one folder and the cost of one notepad.
Show answer
One folder costs £9 and one notepad costs £5.Let \(a\) stand for the cost of one folder in pounds, and \(b\) for the cost of one notepad in pounds.
\(3a + 5b = 52\)
\(9a + 2b = 91\)
Multiplying the first equation by 3 makes both \(a\) terms \(9a\):
\(9a + 15b = 156\)
Subtract the second equation. The \(a\) terms cancel:
\(13b = 65\)
\(b = 5\)
Substitute back into the first equation:
\(3a + 5 \times 5 = 52\)
\(3a = 27\)
\(a = 9\)
One folder costs £9 and one notepad costs £5.
- M1equations \(3a + 5b = 52\) and \(9a + 2b = 91\) multiplied to match one letter: \(9a + 15b = 156\)
- M1subtracting to eliminate \(a\): \(13b = 65\)
- A1one notepad costs £5
- A1one folder costs £9
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Either paper 4 marks
Here are two simultaneous equations.
\(3x = 12 - 4y\)
\(5x - 4y = -44\)
(a)Write the first equation in the form \(ax + by = c\).
(b)Solve the simultaneous equations.
Show answer
(a) \(3x + 4y = 12\)
(b) \(x = -4\), \(y = 6\)Move the term that is on the wrong side of the equals sign across, changing its sign as it goes, so that both letters are on the left and the number is on the right.
\(3x + 4y = 12\)
With the first equation written as \(3x + 4y = 12\), the \(y\) terms are \(+4y\) and \(-4y\), so adding the two equations removes \(y\):
\(8x = -32\)
\(x = -4\)
Substitute \(x = -4\) into the rearranged first equation:
\(3 \times (-4) + 4y = 12\)
\(4y = 24\)
\(y = 6\)
Rearranging first is not the only route — adding the two equations as they are printed cancels the \(y\) terms too. It is written out this way because lining the equations up is what makes that visible.
- (b) M1adds the equations to remove \(y\): \(8x = -32\)
- (b) M1substitutes \(x = -4\) to get \(4y = 24\)
- (b) A1\(x = -4\) and \(y = 6\)
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Either paper 5 marks
A plumber charges a fixed call-out fee and then the same amount for every hour of work.
A job that takes 4 hours costs £203.
A job that takes 6 hours costs £291.
(a)Calculate the call-out fee and the amount charged for each hour.
(b)Work out the cost of a job that takes 9 hours.
Show answer
(a) The call-out fee is £27 and the charge for each hour is £44.
(b) \(\text{£}423\)Let \(f\) stand for the call-out fee in pounds, and \(r\) for the number of pounds charged for each hour.
\(f + 4r = 203\)
\(f + 6r = 291\)
Both equations contain exactly one \(f\), so subtracting the first from the second removes it:
\(2r = 88\)
\(r = 44\)
Substitute \(r = 44\) into the first equation:
\(f + 4 \times 44 = 203\)
\(f + 176 = 203\)
\(f = 27\)
The cost is the call-out fee plus 9 hours at £44 an hour.
\(27 + 9 \times 44 = 27 + 396 = 423\)
- (a) M1equations \(f + 4r = 203\) and \(f + 6r = 291\)
- (a) M1subtracting to eliminate \(f\): \(2r = 88\)
- (a) A1charge for each hour £44
- (a) A1call-out fee £27
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Either paper 4 marks
A car park charges £4 for a car and £6 for a van.
One day there were 38 vehicles in the car park, and the total charge was £196.
Work out how many cars and how many vans there were.
Show answer
There were 16 cars and 22 vans.Let \(x\) stand for the number of cars and \(y\) for the number of vans.
There were 38 vehicles altogether:
\(x + y = 38\)
Each of the cars was charged £4 and each of the vans £6, and the charges came to £196 altogether:
\(4x + 6y = 196\)
Multiply the first equation by 4, so that both \(x\) terms are \(4x\):
\(4x + 4y = 152\)
Subtracting that from the second equation removes \(x\):
\(2y = 44\)
\(y = 22\)
Then \(x = 38 - 22 = 16\).
So there were 16 cars and 22 vans.
- M1two equations \(x + y = 38\) and \(4x + 6y = 196\)
- M1scales to equal coefficients, \(4x + 4y = 152\)
- M1eliminates one unknown, \(2y = 44\)
- A116 cars and 22 vans
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Either paper 5 marks
In a coding club, there are \(x\) boys and \(y\) girls.
If 6 more boys and 14 more girls join, there would be half as many boys as girls.
(a)Show that \(y = 2x - 2\)
(b)If instead 12 more boys and 2 more girls join the coding club, there would be the same number of boys and girls.
Work out the value of \(x\) and the value of \(y\).
Show answer
(a) \(x + 6 = \frac{ y + 14 }{2}\), which rearranges to \(y = 2x - 2\)
(b) \(x = 12\) and \(y = 22\)After the new members join there are \(x + 6\) boys and \(y + 14\) girls.
There are half as many boys as girls:
\(x + 6 = \frac{ y + 14 }{2}\)
Multiply both sides by 2:
\(2(x + 6) = y + 14\)
\(2x + 12 = y + 14\)
Subtract 14 from both sides:
\(y = 2x + 12 - 14\)
\(y = 2x - 2\)
This is the result we were asked to show.
With 12 more boys and 2 more girls the numbers are equal:
\(x + 12 = y + 2\)
Subtract 2 from both sides to get \(y\) on its own:
\(y = x + 10\)
From part (a), \(y = 2x - 2\). Both expressions equal \(y\), so
\(x + 10 = 2x - 2\)
Subtract \(x\) from both sides:
\(10 = x - 2\)
Add 2 to both sides:
\(x = 10 + 2\)
\(x = 12\)
Substitute into \(y = x + 10\):
\(y = 12 + 10\)
\(y = 22\)
Check in the first equation: \(2 \times 12 - 2 = 22\).
- (a) M1forms the equation \(x + 6 = \frac{ y + 14 }{2}\), or \(2(x + 6) = y + 14\)
- (a) A1rearranges, showing \(2x + 12 - 14\), to reach \(y = 2x - 2\)
- (b) M1forms the equation \(x + 12 = y + 2\), or \(y = x + 10\)
- (b) M1eliminates \(y\) and solves, reaching \(x = 12\)
- (b) A1substitutes to find \(y = 22\)