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

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

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

We move all terms to the left:

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


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

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

z!=-4

z∈R



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

z∈R


We add all the numbers together, and all the variables

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

We calculate fractions


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


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

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

We add all the numbers together, and all the variables

3z0/((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: +(-(2*(z+4))/((z+4)*3(-1z+5))), so:

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

We multiply all the terms by the denominator


-(2*(z+4))


We calculate terms in parentheses: -(2*(z+4)), so:

2*(z+4)

We multiply parentheses

2z+8

Back to the equation:

-(2z+8)


We get rid of parentheses

-2z-8

Back to the equation:

+(-2z-8)


We get rid of parentheses

3z-2z-8-6=0

We add all the numbers together, and all the variables

z-14=0

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

z=14