The length of an elastic string is am when the tension is 4N , and...

{T}_{1} = K\left({l}_{1} - l\right)$<br /><br />${T}_{2} = K ( {l}_{2} - l )$<br /><br />So,$\dfrac{{T}_{1}}{{T}_{2}} = \dfrac{{l}_{1} - l}{{l}_{2} - l}$<br /><br />$∴ {T}_{1}{l}_{2} - {T}_{1}l = {T}_{2}{l}_{1} - {T}_{2}l$<br /><br />\left({T}_{2} - {T}_{1}\right)l = {T}_{2}{l}_{1} - {T}_{1}{l}_{2}$

$l = \dfrac{{T}_{2}{l}_{1} - {T}_{1}{l}_{2}}{( {T}_{2} - {T}_{1} )}$

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

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

$k = \dfrac{1}{b - a}$ ...(ii)

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

$9 = \dfrac{1}{\left(b - a\right)} \times {l}^{'} ⇒ {l}^{'} = 9b - 9a$

Hence, final length

${l}_{final} = l + {l}^{'}$

${l}_{final} = 5a - 4b + 9b - 9a$

${l}_{final} = 5b - 4a$