Answer:
[tex]l<7\dfrac{21}{44}[/tex]
Step-by-step explanation:
Given the inequality
[tex]5\dfrac{1}{6}+7\dfrac{1}{3}l<60[/tex]
First, convert mixed numbers to improper fractions:
[tex]5\dfrac{1}{6}=\dfrac{5\cdot 6+1}{6}=\dfrac{31}{6}\\ \\7\dfrac{1}{3}=\dfrac{7\cdot 3+1}{3}=\dfrac{22}{3}[/tex]
Now the inequality looks like
[tex]\dfrac{31}{6}+\dfrac{22}{3}l<60[/tex]
Multiply it by 6:
[tex]\dfrac{31}{6}\cdot 6+\dfrac{22}{3}l\cdot 6<60\cdot 6\\ \\31+44l<360\\ \\44l<360-31\\ \\44l<329\\ \\l<\dfrac{329}{44}\\ \\l<7\dfrac{21}{44}[/tex]
