A shop sells notebooks for \($n\) each and folders for \($f\) each. \(4\) notebooks and \(3\) folders cost \($23.40\). \(6\) notebooks and \(5\) folders cos...

Two unknown prices need two equations, one for each purchase described. Here \(n\) is the cost of a notebook and \(f\) the cost of a folder, both in dollars.

(a) Four notebooks and three folders cost \(\$23.40\), so \(4n+3f=23.40\) [B1]. Six notebooks and five folders cost \(\$37.00\), so \(6n+5f=37.00\) [B1].

(b) To eliminate \(f\), match its coefficients using the lowest common multiple of \(3\) and \(5\), which is \(15\). Multiply the first equation by \(5\) and the second by \(3\):

\[20n+15f=117\qquad\text{and}\qquad 18n+15f=111\]

which is the method mark [M1]. The \(f\) terms are identical and both positive, so subtract: \(2n=6\), giving \(n=3.00\), so a notebook costs \(\$3.00\) [A1].

Substituting into \(4n+3f=23.40\) gives \(12+3f=23.40\), so \(3f=11.40\) and a folder costs \(\$3.80\) [A1].

Check the second purchase: \(6(3.00)+5(3.80)=18.00+19.00=37.00\), as stated.

Multiply every term, including the money value on the right hand side, when scaling an equation. Leaving \(23.40\) unchanged while multiplying the left by \(5\) is the error that most often derails this question.