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Either paper 3 marks(a)
Make \(x\) the subject of \(y = x - 13\)
(b)Make \(t\) the subject of \(w = 5t\)
(c)Make \(k\) the subject of \(p = \frac{k}{ 4 }\)
Show answer
(a) \(x = y + 13\)
(b) \(t = \frac{w}{ 5 }\)
(c) \(k = 4p\)Add 13 to both sides:
\(x = y + 13\)
\(t\) has been multiplied by 5, so divide both sides by 5:
\(t = \frac{w}{ 5 }\)
\(k\) has been divided by 4, so multiply both sides by 4:
\(k = 4p\)
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Either paper 1 mark\(ab^{3} = c\) Circle the correct expression for \(a\).
\(cb^{3}\) \(3bc\) \(\frac{ c }{ b^{3} }\) \(\frac{ c }{ 3b }\) Show answer
\(\frac{ c }{ b^{3} }\)\(ab^{3} = c\)
In this formula \(a\) is multiplied by the whole of \(\;b^{3}\;\), so undo it by dividing both sides by all of \(\;b^{3}\;\):
\(a = \frac{ c }{ b^{3} }\)
The other three expressions each undo something else. One uses the opposite operation, one treats the power as a multiple of \(b\) or squares instead of square-rooting, and one does both, so none of them leaves \(a\) on its own.
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Either paper 2 marks
Make \(x\) the subject of the formula \(y = 3x + 2\)
Show answer
\(x = \frac{y - 2 }{ 3 }\)Undo the 2 first. Subtract 2 from both sides:
\(3x = y - 2\)
Now divide both sides by 3:
\(x = \frac{y - 2 }{ 3 }\)
- M1Subtract 2 from both sides: \(3x = y - 2\)
- A1divides by 3: \(x = \frac{y - 2 }{ 3 }\)
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Either paper 2 marks
Rearrange the formula \(y = \frac{x}{ 3 } + 5\) to make \(x\) the subject.
Show answer
\(x = 3(y - 5)\)Subtract 5 from both sides:
\(\frac{x}{ 3 } = y - 5\)
Multiply both sides by 3. The whole right-hand side is multiplied, so it needs a bracket:
\(x = 3(y - 5)\)
- M1Subtract 5 from both sides: \(\frac{x}{ 3 } = y - 5\)
- A1multiplies the whole right-hand side by 3: \(x = 3(y - 5)\)
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Either paper 4 marks
A hire shop uses the formula
\(C = 4d + 15\)
to work out the cost, \(C\) pounds, of hiring a carpet cleaner for \(d\) days.
(a)Make \(d\) the subject of the formula.
(b)Priya pays £43 to hire the carpet cleaner.
Find the number of days she hires it for.
Show answer
(a) \(d = \frac{C - 15 }{ 4 }\)
(b) \(7\ \text{days}\)Subtract 15 from both sides:
\(C - 15 = 4d\)
Divide both sides by 4:
\(d = \frac{C - 15 }{ 4 }\)
\(d = \frac{ 43 - 15 }{ 4 } = \frac{ 28 }{ 4 } = 7\)
So Priya hires the carpet cleaner for 7 days.
- (a) M1subtract 15 from both sides: \(C - 15 = 4d\)
- (a) A1divide by 4: \(d = \frac{C - 15 }{ 4 }\)
- (b) M1substitute and subtract 15: \(43 - 15 = 28\)
- (b) A1divide by 4: \(7\) days
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Either paper 2 marks
Make \(x\) the subject of the formula \(y = \frac{ x - 17 }{ 6 }\)
Show answer
\(x = 6y + 17\)The dividing line runs under the whole of \(x - 17\), so the 17 is inside the division and cannot be dealt with first. Undo the division by multiplying both sides by 6:
\(6y = x - 17\)
Add 17 to both sides:
\(x = 6y + 17\)
This is a different formula from \(y = \frac{x}{ 6 } - 17\), where only the \(x\) is divided by 6. There the 17 is outside the division, so the two undo the same pair of operations in opposite orders.
- M1multiplies both sides by 6: \(6y = x - 17\)
- A1Add 17 to both sides: \(x = 6y + 17\)
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Either paper 2 marks
Rearrange the formula \(y = 5(x + 8)\) to make \(x\) the subject.
Show answer
\(x = \frac{ y }{ 5 } - 8\)Everything inside the bracket has been multiplied by 5, so divide both sides by 5 and the bracket goes:
\(\frac{ y }{ 5 } = x + 8\)
Subtract 8 from both sides:
\(x = \frac{ y }{ 5 } - 8\)
Expanding first reaches the same place. \(5(x + 8) = 5x + 40\), so the formula is \(y = 5x + 40\). Subtract 40 from both sides:
\(5x = y - 40\)
and dividing both sides by 5 gives
\(x = \frac{ y - 40 }{ 5 }\)
Both forms are correct. Dividing the top of the second one by 5 term by term turns it into the first.
- M1divides both sides by 5: \(\frac{ y }{ 5 } = x + 8\)
- A1Subtract 8 from both sides: \(x = \frac{ y }{ 5 } - 8\)
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Either paper 2 marks
Make \(x\) the subject of the formula \(y = 40 - 2x\)
Show answer
\(x = \frac{ 40 - y }{ 2 }\)Add \(2x\) to both sides so the \(x\) term becomes positive:
\(y + 2x = 40\)
Subtract \(y\) from both sides:
\(2x = 40 - y\)
Divide both sides by 2:
\(x = \frac{ 40 - y }{ 2 }\)
- M1makes the \(x\) term positive and isolates it: \(2x = 40 - y\)
- A1divides by 2: \(x = \frac{ 40 - y }{ 2 }\)
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Either paper 3 marks
Make \(x\) the subject of the formula \(y = \frac{ 3x }{ 8 } + 20\)
Show answer
\(x = \frac{ 8(y - 20) }{ 3 }\)Three things have been done to \(x\): it has been multiplied by 3, divided by 8, and then 20 has been added. Undo them in the opposite order.
Subtract 20 from both sides:
\(\frac{ 3x }{ 8 } = y - 20\)
Multiply both sides by 8. The whole of the right-hand side is multiplied, so it needs a bracket:
\(3x = 8(y - 20)\)
Divide both sides by 3:
\(x = \frac{ 8(y - 20) }{ 3 }\)
- M1Subtract 20 from both sides: \(\frac{ 3x }{ 8 } = y - 20\)
- M1multiply the whole right-hand side by 8: \(3x = 8(y - 20)\)
- A1divide by 3: \(x = \frac{ 8(y - 20) }{ 3 }\)
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Either paper 2 marks
Rearrange the formula \(y = 82 - \frac{x}{ 4 }\) to make \(x\) the subject.
Show answer
\(x = 4(82 - y)\)The \(x\) term is being taken away, so add \(\frac{x}{ 4 }\) to both sides to make it positive:
\(y + \frac{x}{ 4 } = 82\)
Subtract \(y\) from both sides:
\(\frac{x}{ 4 } = 82 - y\)
Multiply both sides by 4. The whole right-hand side is multiplied, so it needs a bracket:
\(x = 4(82 - y)\)
- M1isolates the \(x\) term: \(\frac{x}{ 4 } = 82 - y\)
- A1multiplies the whole right-hand side by 4: \(x = 4(82 - y)\)