1/3(z+4)=2/3(7-z)

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



1/3(z+4)=2/3(7-z)

We move all terms to the left:

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



Domain of the equation: 3(z+4)!=0

z∈R





Domain of the equation: 3(7-z))!=0

z∈R



We add all the numbers together, and all the variables

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

We calculate fractions


(3z(-)/(3(z+4)*3(-1z+7)))+(-6zz/(3(z+4)*3(-1z+7)))=0



We calculate terms in parentheses: +(3z(-)/(3(z+4)*3(-1z+7))), so:

3z(-)/(3(z+4)*3(-1z+7))

We add all the numbers together, and all the variables

3z0/(3(z+4)*3(-1z+7))

We multiply all the terms by the denominator


3z0

We add all the numbers together, and all the variables

3z

Back to the equation:

+(3z)





We calculate terms in parentheses: +(-6zz/(3(z+4)*3(-1z+7))), so:

-6zz/(3(z+4)*3(-1z+7))

We multiply all the terms by the denominator


-6zz

Back to the equation:

+(-6zz)



We get rid of parentheses

3z-6zz=0

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