10x+(2+3x)=4x+(7x-20)+50

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Solution for 10x+(2+3x)=4x+(7x-20)+50 equation: 



10x+(2+3x)=4x+(7x-20)+50

We move all terms to the left:

10x+(2+3x)-(4x+(7x-20)+50)=0

We add all the numbers together, and all the variables

10x+(3x+2)-(4x+(7x-20)+50)=0

We get rid of parentheses

10x+3x-(4x+(7x-20)+50)+2=0



We calculate terms in parentheses: -(4x+(7x-20)+50), so:

4x+(7x-20)+50

We get rid of parentheses

4x+7x-20+50

We add all the numbers together, and all the variables

11x+30

Back to the equation:

-(11x+30)



We add all the numbers together, and all the variables

13x-(11x+30)+2=0

We get rid of parentheses

13x-11x-30+2=0

We add all the numbers together, and all the variables

2x-28=0

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

2x=28

x=28/2

x=14

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