Question 174203
# 1


Problem: 


simplify
1. 2z+1/3z-z-6/5z




Solution:



{{{(2z+1)/(3z)-(z-6)/(5z)}}} Start with the given expression.



{{{(5/5)((2z+1)/(3z))-(z-6)/(5z)}}} Multiply the 1st fraction {{{(2z+1)/(3z)}}} by {{{5/5}}}



{{{(5(2z+1))/(5(3z))-(z-6)/(5z)}}} Combine the fractions.



{{{(10z+5)/(5(3z))-(z-6)/(5z)}}} Distribute



{{{(10z+5)/(15z)-(z-6)/(5z)}}} Multiply



{{{(10z+5)/(15z)-(3/3)((z-6)/(5z))}}}  Multiply the 2nd fraction {{{(z-6)/(5z)}}} by {{{3/3}}}





{{{(10z+5)/(15z)-(3(z-6))/(3(5z))}}} Combine the fractions.



{{{(10z+5)/(15z)-(3z-18)/(3(5z))}}} Distribute.



{{{(10z+5)/(15z)-(3z-18)/(15z)}}} Multiply.





Now that the fractions have equal denominators, we can combine them:



{{{(10z+5-(3z-18))/(15z)}}} Combine the fractions.



{{{(10z+5-3z+18)/(15z)}}} Distribute



{{{(7z+23)/(15z)}}} Combine like terms.




So {{{(2z+1)/(3z)-(z-6)/(5z)}}} simplifies to {{{(7z+23)/(15z)}}}




In other words, {{{(2z+1)/(3z)-(z-6)/(5z)=(7z+23)/(15z)}}} where {{{z<>0}}}




<hr>



# 2




Problem:


solve
1. 5b^2-15b=0



Solution:



{{{5b^2-15b=0}}} Start with the given equation



{{{5b(b-3)=0}}} Factor out the GCF 5b



Now set each factor equal to zero:



{{{5b=0}}} or  {{{b-3=0}}} 



{{{b=0}}} or  {{{b=3}}}    Now solve for b in each case



So our answers are 


 {{{b=0}}} or  {{{b=3}}}