1/5(z-25)=2/5(z+30)

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Solution for 1/5(z-25)=2/5(z+30) equation: 



1/5(z-25)=2/5(z+30)

We move all terms to the left:

1/5(z-25)-(2/5(z+30))=0



Domain of the equation: 5(z-25)!=0

z∈R





Domain of the equation: 5(z+30))!=0

z∈R



We calculate fractions


(5zz/(5(z-25)*5(z+30)))+(-10zz/(5(z-25)*5(z+30)))=0



We calculate terms in parentheses: +(5zz/(5(z-25)*5(z+30))), so:

5zz/(5(z-25)*5(z+30))

We multiply all the terms by the denominator


5zz

Back to the equation:

+(5zz)





We calculate terms in parentheses: +(-10zz/(5(z-25)*5(z+30))), so:

-10zz/(5(z-25)*5(z+30))

We multiply all the terms by the denominator


-10zz

Back to the equation:

+(-10zz)



We get rid of parentheses

5zz-10zz=0

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