1/3(z+4)=62/35-z)

Solution for 1/3(z+4)=62/35-z) equation:

1/3(z+4)=62/35-z)

We move all terms to the left:

1/3(z+4)-(62/35-z))=0


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

z∈R



Domain of the equation: 35-z))!=0

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

-z))!=-35

z!=-35/1

z!=-35

z∈R


We add all the numbers together, and all the variables

1/3(z+4)-(-1z+62/35))=0

We get rid of parentheses

1/3(z+4)+1z-62/35)=0

We calculate fractions


1z+()/3z+(-186zz/3z=0

We add all the numbers together, and all the variables

z+()/3z+(-186zz/3z=0

We multiply all the terms by the denominator


z*3z+(-186zz+()=0


We calculate terms in parentheses: +(-186zz+(), so:

-186zz+(

We add all the numbers together, and all the variables

-186zz

Back to the equation:

+(-186zz)


Wy multiply elements

3z^2+(-186zz)=0

We get rid of parentheses

3z^2-186zz=0