Answer:
[tex]\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{13a}{20} +\frac{b}{20}[/tex]
Step-by-step explanation:
We have:
[tex]\frac{(a+b)}{4} +\frac{(2a-b)}{5}[/tex]
We can use common denominator.
Observation:
If you have, [tex]\frac{a}{b} +\frac{c}{d}=\frac{(a*d)+(c*b)}{b*d}[/tex]
Then,
[tex]\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{5(a+b)+4(2a-b)}{4*5}[/tex]
Using distributive property:
Observation:
c(a+b)=ca+cb
[tex]\frac{5(a+b)+4(2a-b)}{4*5}=\frac{5a+5b+8a-4b}{20}=\frac{(5a+8a)+(5b-4b)}{20}[/tex]
Finally,
[tex]\frac{(5a+8a)+(5b-4b)}{20} =\frac{13a+b}{20}=\frac{13a}{20} +\frac{b}{20}[/tex]
The answer then is:
[tex]\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{13a}{20} +\frac{b}{20}[/tex]
