First of all, observe that all fractions exist as long as b is not zero.Multiply both numerator and denominator of the first fraction by 3:[tex]\dfrac{3b-15}{3b}+\dfrac{1}{3b}=\dfrac{5b+2}{3b}[/tex]Now all fractions have the same denominator. Under the assumption that b is not zero, multiply both sides by 3b:[tex]3b-15+1=5b+2 \iff 3b-14=5b+2[/tex]Subtract 3b from both sides:[tex]-14=2b+2[/tex]Subtract 2 from both sides:[tex]-16=2b[/tex]Divide both sides by 2:[tex]b=-8[/tex]