Let x be the length of the call per minute;
Let n be the cost per minute (in pence);
[tex]\begin{gathered} \text{ cost = (cost/minute)}\times length \\ \end{gathered}[/tex]
That is, at 12:30pm;
[tex]\begin{gathered} 60=n\times x \\ 60=nx \\ n=\frac{60}{x} \end{gathered}[/tex]
Similarly at 1:00pm;
[tex]\begin{gathered} 60=(n-1)\times(x+2) \\ n-1=\frac{60}{x+2} \\ n=\frac{60}{x+2}+1 \end{gathered}[/tex]
Then, we have expressed n in terms of x at 12:30pm and 1:00pm, then we have;
[tex]n=\frac{60}{x+2}+1=\frac{60}{x}[/tex]
Simplifying further, we have;
[tex]\begin{gathered} \frac{60}{x+2}+1=\frac{60}{x} \\ M\text{ultiply through by x(x+2);} \\ 60x+x^2+2x=(x+2)60 \\ 60x+x^2+2x=60x+120 \\ 60x-60x+x^2+2x-120=0 \\ x^2+2x-120=0 \end{gathered}[/tex]
Thus, the equation is correct.
(iii) By simplifying the equation above, we have;
[tex]\begin{gathered} x^2+2x-120=0 \\ x^2+12x-10x-120=0 \\ x(x+12)-10(x+12)=0 \\ x-10=0\text{ or x+12=0} \\ x=10\text{ or x = -12} \end{gathered}[/tex]
Thus, the length of Jane's call is 10minutes
