6(z+2)=4(13-z)

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Solution for 6(z+2)=4(13-z) equation: 



6(z+2)=4(13-z)

We move all terms to the left:

6(z+2)-(4(13-z))=0

We add all the numbers together, and all the variables

6(z+2)-(4(-1z+13))=0

We multiply parentheses

6z-(4(-1z+13))+12=0



We calculate terms in parentheses: -(4(-1z+13)), so:

4(-1z+13)

We multiply parentheses

-4z+52

Back to the equation:

-(-4z+52)



We get rid of parentheses

6z+4z-52+12=0

We add all the numbers together, and all the variables

10z-40=0

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

10z=40

z=40/10

z=4

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