-(1)/(40)(r+40)-(1)/(20)(2r+22)=(2)/(5)

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Solution for -(1)/(40)(r+40)-(1)/(20)(2r+22)=(2)/(5) equation: 



-(1)/(40)(r+40)-(1)/(20)(2r+22)=(2)/(5)

We move all terms to the left:

-(1)/(40)(r+40)-(1)/(20)(2r+22)-((2)/(5))=0



Domain of the equation: 40(r+40)!=0

r∈R





Domain of the equation: 20(2r+22)!=0

r∈R



We add all the numbers together, and all the variables

-1/40(r+40)-1/20(2r+22)-(+2/5)=0

We get rid of parentheses

-1/40(r+40)-1/20(2r+22)-2/5=0

We calculate fractions


(-100r2/(40(r+40)*20(2r+22)*5)+(-200rr/(40(r+40)*20(2r+22)*5)+(-1600r^2r/(40(r+40)*20(2r+22)*5)=0



We calculate terms in parentheses: +(-100r2/(40(r+40)*20(2r+22)*5)+(-200rr/(40(r+40)*20(2r+22)*5)+(-1600r^2r/(40(r+40)*20(2r+22)*5), so:

-100r2/(40(r+40)*20(2r+22)*5)+(-200rr/(40(r+40)*20(2r+22)*5)+(-1600r^2r/(40(r+40)*20(2r+22)*5

We can not solve this equation

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