[tex]\dfrac7{24}=\dfrac6{24}+\dfrac1{24}=\dfrac14+\dfrac1{24}[/tex]
[tex]\dfrac1{24}=\dfrac1{8\time3}=\dfrac a8+\dfrac b3=\dfrac{3a+8b}{24}[/tex]
[tex]\implies3a+8b=1[/tex]
We can choose [tex]a=-1[/tex] and [tex]b=\dfrac12[/tex], so that
[tex]\dfrac1{24}=-\dfrac18+\dfrac16[/tex]
Recalling that [tex]\dfrac13=0.333\ldots[/tex], it follows that [tex]\dfrac16=0.166\ldots[/tex]. So,
[tex]\dfrac1{24}=0.125+0.166\ldots=0.04166\ldots[/tex]
[tex]\implies\dfrac7{24}=0.25+0.4166\ldots=0.29166\ldots[/tex]
