$T_1=K\left(l_1-l\right)$

$T_2=K(l_2-l)$

So, $\frac{T_1}{T_2}=\frac{l_1-l}{l_2-l}$

$\therefore T_1{l}_2-T_{1}l=T_2{l}_1-T_{2}l$

$\left(T_2-T_1\right)l=T_2{l}_1-T_1{l}_2$

$l= \frac{T_2{l}_1-T_1{l}_2}{(T_2-T_1)}$

$l=\left(5a-4b\right)$        .......(i)

$5-4=k(b-a)$

$k=\frac{1}{b-a }$           ...(ii)

$9=k{l}^{\prime } (l^{\prime }=$ change in length)

$9=\frac{1}{\left(b-a\right)} \times l^{\text{'}}\Rightarrow l^{\text{'}}= 9b-9a$

Hence, final length

$l_{\mathrm{final}}=l+l^{\text{'}}$

$l_{\mathrm{final}}=5a-4b+9b-9a$

$l_{\mathrm{final}}=5b-4a$