(4+z-(3+2z))/6=(-z-3(5-2))/7

Solution for (4+z-(3+2z))/6=(-z-3(5-2))/7 equation:

(4+z-(3+2z))/6=(-z-3(5-2))/7

We move all terms to the left:

(4+z-(3+2z))/6-((-z-3(5-2))/7)=0

We add all the numbers together, and all the variables

(4+z-(2z+3))/6-((-z-33)/7)=0

We calculate fractions


((z-(2z+3))*7)/()+(-((-1z-33)*6)/()=0


We calculate terms in parentheses: +((z-(2z+3))*7)/(), so:

(z-(2z+3))*7)/(

We multiply all the terms by the denominator


(z-(2z+3))*7)


We calculate terms in parentheses: +(z-(2z+3))*7), so:

z-(2z+3))*7

We multiply parentheses

z-14z-

We add all the numbers together, and all the variables

-13z

Back to the equation:

+(-13z)


We get rid of parentheses

-13z

Back to the equation:

+(-13z)



We calculate terms in parentheses: +(-((-1z-33)*6)/(), so:

-((-1z-33)*6)/(

We multiply all the terms by the denominator


-((-1z-33)*6)


We calculate terms in parentheses: -((-1z-33)*6), so:

(-1z-33)*6

We multiply parentheses

-6z-198

Back to the equation:

-(-6z-198)


We get rid of parentheses

6z+198

Back to the equation:

+(6z+198)


We get rid of parentheses

-13z+6z+198=0

We add all the numbers together, and all the variables

-7z+198=0

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

-7z=-198

z=-198/-7

z=28+2/7