Question 1109280
You start out with L inches of string.
You use {{{(2/9)L}}} and then use another {{{(1/4)L}}} of the string. 
So then let's say you have {{{X}}} left.
{{{(2/9)L+(1/4)L+X=L}}}
{{{X=L-(2/9)L-(1/4)L}}}
Use a common denominator,
{{{X=L-(2/9)(4/4)L-(1/4)(9/9)L}}}
{{{X=L-(8/36)L-(9/36)L}}}
{{{X=L(36/36-17/36)}}}
{{{X=(19/36)L}}}
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