4(q-3)=3(4-q)+32

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Solution for 4(q-3)=3(4-q)+32 equation: 



4(q-3)=3(4-q)+32

We move all terms to the left:

4(q-3)-(3(4-q)+32)=0

We add all the numbers together, and all the variables

4(q-3)-(3(-1q+4)+32)=0

We multiply parentheses

4q-(3(-1q+4)+32)-12=0



We calculate terms in parentheses: -(3(-1q+4)+32), so:

3(-1q+4)+32

We multiply parentheses

-3q+12+32

We add all the numbers together, and all the variables

-3q+44

Back to the equation:

-(-3q+44)



We get rid of parentheses

4q+3q-44-12=0

We add all the numbers together, and all the variables

7q-56=0

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

7q=56

q=56/7

q=8

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