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Either paper 2 marks
\(6x - 5 = 31\)
Work out the value of \(x\).
Show answer
\(6\)Add 5 to both sides:
\(6x = 36\)
Now divide both sides by 6:
\(x = 6\)
- M1isolates the x term: \(6x = 36\)
- A1\(x = 6\)
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Either paper 2 marks
Solve the equation \(\frac{x}5 + 7 = 17\)
Show answer
\(50\)Subtract 7 from both sides:
\(\frac{x}5 = 10\)
Now multiply both sides by 5:
\(x = 5 \times 10 = 50\)
- M1isolates the x term: \(\frac{x}5 = 10\)
- A1\(x = 50\)
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Either paper 3 marks
\(3x + 1 + 3x = 67\)
Work out the value of \(x\).
Show answer
\(11\)Combine the \(x\) terms on the left-hand side:
\(6x + 1 = 67\)
Subtract 1 from both sides:
\(6x = 66\)
Now divide both sides by 6:
\(x = 11\)
- M1combines the \(x\) terms: \(6x + 1 = 67\)
- M1clears the constant: \(6x = 66\)
- A1\(x = 11\)
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Either paper 3 marks
Solve the equation \(7(x + 5) = 98\)
Show answer
\(9\)Multiply out the bracket first:
\(7x + 35 = 98\)
Subtract 35 from both sides, then divide by 7:
\(x = 9\)
- M1expands the bracket: \(7x + 35 = 98\)
- M1Subtract 35 from both sides to isolate the \(x\) term
- A1\(x = 9\)
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Either paper 2 marks
The solution of the equation
\(9x + b = -84\)
is \(x = -8\).
Find the value of \(b\).
Show answer
\(-12\)The solution is the value of \(x\) that makes the equation true, so substitute \(x = -8\).
\(9 \times -8 + b = -84\)
\(-72 + b = -84\)
\(b = -84 - -72 = -12\)
- M1substitutes \(x = -8\) to get \(-72 + b = -84\)
- A1\(b = -12\)
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Either paper 2 marks
Solve \(25 - 6x = -35\)
Show answer
\(10\)Add \(6x\) to both sides, then add \(35\) to both sides:
\(6x = 60\)
Now divide both sides by 6:
\(x = 10\)
- M1rearranges to \(6x = 60\)
- A1\(x = 10\)
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Either paper 3 marks
Isla is trying to solve \(9x - 36 = 45\). Here is her working, with each line numbered.
Line 1: \(9x + 36 = 45\)
Line 2: \(9x = 9\)
Line 3: \(x = 1\)
(a)State the number of the line on which Isla first makes a mistake.
(b)Calculate the correct value of \(x\).
Show answer
(a) \(1\)
(b) \(9\)The mistake is on line 1: the constant was copied down with the wrong sign.
Starting again from the original equation, Add 36 to both sides:
\(9x = 81\)
Now divide both sides by 9:
\(x = 9\)
- (b) M1removes the constant from both sides: \(9x = 81\)
- (b) A1divides by 9 to get \(x = 9\)
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Either paper 3 marks
Solve \(8x + 8 = 3x - 7\)
Show answer
\(-3\)Take \(3x\) from both sides:
\(5x + 8 = -7\)
Then solve as usual:
\(x = -3\)
- M1collects the \(x\) terms: \(5x + 8 = -7\)
- M1removes the constant \(8\) from the left-hand side to isolate the \(x\) term
- A1\(x = -3\)
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Either paper 2 marks
The solution of the equation
\(a(x + 7) = 40\)
is \(x = 1\).
Calculate the value of \(a\).
Show answer
\(5\)Substitute \(x = 1\) and work out what is inside the bracket first.
\(a(1 + 7) = 40\)
\(8a = 40\)
\(a = 40 \div 8 = 5\)
- M1substitutes \(x = 1\) to get \(8a = 40\)
- A1\(a = 5\)
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Either paper 3 marks
Solve the equation \(9(x + 4) = 6(x + 9)\)
Show answer
\(6\)Expand both brackets:
\(9x + 36 = 6x + 54\)
Collect the \(x\) terms on the left-hand side, since it has the larger coefficient:
\(3x = 18\)
Now divide both sides by 3:
\(x = 6\)
- M1expands both brackets: \(9x + 36 = 6x + 54\)
- M1collects terms: \(3x = 18\)
- A1\(x = 6\)