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]
Answer:
A) -8
Step-by-step explanation:
You have to make the whole equation equal to zero. so rewrite the equation and equal it to zero. Write all numerators above the common denominator
3(b-5)+1-(5b+2) over 3b equal zero
Then remove parentheses
3b-15+1-5b-2 over 3b equals zero
Then collect the like terms
Calculate
-2b-16 over 3b equals zero
Then set the numerator equal to zero
-2b-16=0
Then move constant to the right
-2b=16
Then divide both sides by -2
b=-8
