Answer:
Step-by-step explanation:
[tex]\dfrac{4y+5}{3}=\dfrac{y-8}{2}\qquad\text{multiply both sides by }\ 2\cdot3=6\\\\6\!\!\!\!\diagup^2\cdot\dfrac{4y+5}{\not3_1}=6\!\!\!\!\diagup^3\cdot\dfrac{y-8}{\not2_1}\\\\2(4y+5)=3(y-8)\qquad\text{use distributive property}\\\\(2)(4y)+(2)(5)=(3)(y)+(3)(-8)\\\\8y+10=3y-24\qquad\text{subtract 10 from both sides}\\\\8y=3y-34\qquad\text{subtract 3y from both sides}\\\\5y=-34\qquad\text{divide both sides by 5}\\\\\boxed{y=-\dfrac{34}{5}}[/tex]