Answer:
[tex]\frac{22x+3}{(3x+2)(2x-1)}[/tex]
Step-by-step explanation:
given
[tex]\frac{5}{3x+2}[/tex] + [tex]\frac{4}{2x-1}[/tex]
Before we can add the fractions, we require them to have a common denominator
multiply the numerator/denominator of the first fraction by (2x - 1 ) and
multiply the numerator/denominator of the second fraction by (3x + 2)
= [tex]\frac{5(2x-1)}{(3x+2)(2x-1)}[/tex] + [tex]\frac{4(3x+2)}{(3x+2)(2x-1)}[/tex]
distribute and simplify the numerators , over the common denominator
= [tex]\frac{10x-5+12x+8}{(3x+2)(2x-1)}[/tex]
= [tex]\frac{22x+3}{(3x+2)(2x-1)}[/tex] ← in simplest form
