2/3(z-2)=-4/5(z+2)

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Solution for 2/3(z-2)=-4/5(z+2) equation: 



2/3(z-2)=-4/5(z+2)

We move all terms to the left:

2/3(z-2)-(-4/5(z+2))=0



Domain of the equation: 3(z-2)!=0

z∈R





Domain of the equation: 5(z+2))!=0

z∈R



We calculate fractions


(10zz/(3(z-2)*5(z+2)))+(-(-12zz)/(3(z-2)*5(z+2)))=0



We calculate terms in parentheses: +(10zz/(3(z-2)*5(z+2))), so:

10zz/(3(z-2)*5(z+2))

We multiply all the terms by the denominator


10zz

Back to the equation:

+(10zz)





We calculate terms in parentheses: +(-(-12zz)/(3(z-2)*5(z+2))), so:

-(-12zz)/(3(z-2)*5(z+2))

We multiply all the terms by the denominator


-(-12zz)

We get rid of parentheses

12zz

Back to the equation:

+(12zz)

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