2/3(p+2)+1/5(p-4)=7/5

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Solution for 2/3(p+2)+1/5(p-4)=7/5 equation: 



2/3(p+2)+1/5(p-4)=7/5

We move all terms to the left:

2/3(p+2)+1/5(p-4)-(7/5)=0



Domain of the equation: 3(p+2)!=0

p∈R





Domain of the equation: 5(p-4)!=0

p∈R



We add all the numbers together, and all the variables

2/3(p+2)+1/5(p-4)-(+7/5)=0

We get rid of parentheses

2/3(p+2)+1/5(p-4)-7/5=0

We calculate fractions


(50pp/(3(p+2)*5(p-4)*5)+(3pp/(3(p+2)*5(p-4)*5)+(-21pp/(3(p+2)*5(p-4)*5)=0



We calculate terms in parentheses: +(50pp/(3(p+2)*5(p-4)*5)+(3pp/(3(p+2)*5(p-4)*5)+(-21pp/(3(p+2)*5(p-4)*5), so:

50pp/(3(p+2)*5(p-4)*5)+(3pp/(3(p+2)*5(p-4)*5)+(-21pp/(3(p+2)*5(p-4)*5

We can not solve this equation

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