Pre-U Pre-U 9794/1 2011 June — Question 3 3 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2011
SessionJune
Marks3
TopicModulus function
TypeSolve |linear| = linear (non-modulus)
DifficultyModerate -0.5 This is a straightforward absolute value equation requiring case analysis (7-4x ≥ 0 and 7-4x < 0), leading to two linear equations to solve. It's slightly easier than average as it only requires basic algebraic manipulation and understanding of absolute value definition, with no complex problem-solving or multi-step reasoning beyond the standard technique.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function
Solve the equation \(3 + 2x = |7 - 4x|\). [3]
AnswerMarks Guidance
Method for modulus eqn, maybe implied.M1 e.g graphical or if algebraic, must consider \(3 + 2x = 7 - 4x\) and \(3 + 2x = 4x - 7\)
State \(x = 5\)B1
State \(x = \frac{2}{3}\)B1 [3] 
Method for modulus eqn, maybe implied. | M1 | e.g graphical or if algebraic, must consider $3 + 2x = 7 - 4x$ and $3 + 2x = 4x - 7$

State $x = 5$ | B1 |

State $x = \frac{2}{3}$ | B1 | [3] | Ignore $y$ co-ordinates. Accept unsimplified
Solve the equation $3 + 2x = |7 - 4x|$. [3]

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q3 [3]}}
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