The length of an elastic string is a meter when the longitudinal tension is 4 N and b metre when the tension is 5 N. The length of the string (in metre) when the longitudinal tension is 9 N is 

If L is the initial length, then the increase in length by tension P is given by

                                      $l=\frac{FL}{{\mathrm{\pi r}}^2\mathrm{Y}}$

Here  $a=L+l=L+\frac{4L}{{\mathrm{\pi r}}^2\mathrm{Y}}=L+4c \left(1\right)$

and    $b=L+\frac{5L}{{\mathrm{\pi r}}^2\mathrm{Y}}=L+5c \left(2\right)$

where $c=\frac{L}{{\mathrm{\pi r}}^2\mathrm{Y}}$.Solving (1) and (2) for Land c, we get L =5a $-$ 4b and c = b $-$ a. For F = 9 N, we have 

                    $$\begin{gathered} x=L+\frac{9L}{{\mathrm{\pi r}}^2\mathrm{Y}} \\ =L+9c=(5a - 4b) + 9(b - a) \\ =5b - 4a \end{gathered}$$