A phone shop's loyalty scheme gives a discount, in pounds, described by the equation \(\dfrac{m+4}{m-2}=3\), where m is the number of months a customer has ...
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A phone shop's loyalty scheme gives a discount, in pounds, described by the equation \(\dfrac{m+4}{m-2}=3\), where m is the number of months a customer has held an account.
- Solve the equation to find m. (3)
- State why \(m=2\) cannot be a solution of the original equation. (2)
Answer Details
(a) Multiplying both sides of \(\dfrac{m+4}{m-2}=3\) by \((m-2)\) to clear the fraction gives \(m+4=3(m-2)\). Expanding the right-hand side, \(m+4=3m-6\). Rearranging, adding \(6\) to both sides and subtracting \(m\) from both sides, gives \(10=2m\), so \(m=5\). [3 marks]
(b) Substituting \(m=2\) into the original equation would make the denominator, \(m-2\), equal to zero, and division by zero is undefined, so \(m=2\) cannot be a solution of the original equation, regardless of what the numerator equals. [2 marks]
Multiplying through by \((m-2)\) in part (a) is only valid provided \(m\neq2\); checking this restriction afterwards, as part (b) does, is essential whenever an equation contains a variable in the denominator.
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