1/3(z+4)-6=2/3(5-z)=

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Solution for 1/3(z+4)-6=2/3(5-z)= equation: 



1/3(z+4)-6=2/3(5-z)=

We move all terms to the left:

1/3(z+4)-6-(2/3(5-z))=0



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

z∈R





Domain of the equation: 3(5-z))!=0

z∈R



We add all the numbers together, and all the variables

1/3(z+4)-(2/3(-1z+5))-6=0

We calculate fractions


(3z(-)/(3(z+4)*3(-1z+5)))+(-6zz/(3(z+4)*3(-1z+5)))-6=0



We calculate terms in parentheses: +(3z(-)/(3(z+4)*3(-1z+5))), so:

3z(-)/(3(z+4)*3(-1z+5))

We add all the numbers together, and all the variables

3z0/(3(z+4)*3(-1z+5))

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+5))), so:

-6zz/(3(z+4)*3(-1z+5))

We multiply all the terms by the denominator


-6zz

Back to the equation:

+(-6zz)



We get rid of parentheses

3z-6zz-6=0

We move all terms containing z to the left, all other terms to the right

3z-6zz=6

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