3(1-g)+12=4(g-4)+3

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Solution for 3(1-g)+12=4(g-4)+3 equation: 



3(1-g)+12=4(g-4)+3

We move all terms to the left:

3(1-g)+12-(4(g-4)+3)=0

We add all the numbers together, and all the variables

3(-1g+1)-(4(g-4)+3)+12=0

We multiply parentheses

-3g-(4(g-4)+3)+3+12=0



We calculate terms in parentheses: -(4(g-4)+3), so:

4(g-4)+3

We multiply parentheses

4g-16+3

We add all the numbers together, and all the variables

4g-13

Back to the equation:

-(4g-13)



We add all the numbers together, and all the variables

-3g-(4g-13)+15=0

We get rid of parentheses

-3g-4g+13+15=0

We add all the numbers together, and all the variables

-7g+28=0

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

-7g=-28

g=-28/-7

g=+4

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