(10x-15)+(4x-10)+(3x+5)=270(10x-15)+(4x-10)+(3x+5)=270

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Solution for (10x-15)+(4x-10)+(3x+5)=270(10x-15)+(4x-10)+(3x+5)=270 equation: 



(10x-15)+(4x-10)+(3x+5)=270(10x-15)+(4x-10)+(3x+5)=270

We move all terms to the left:

(10x-15)+(4x-10)+(3x+5)-(270(10x-15)+(4x-10)+(3x+5))=0

We get rid of parentheses

10x+4x+3x-(270(10x-15)+(4x-10)+(3x+5))-15-10+5=0



We calculate terms in parentheses: -(270(10x-15)+(4x-10)+(3x+5)), so:

270(10x-15)+(4x-10)+(3x+5)

We multiply parentheses

2700x+(4x-10)+(3x+5)-4050

We get rid of parentheses

2700x+4x+3x-10+5-4050

We add all the numbers together, and all the variables

2707x-4055

Back to the equation:

-(2707x-4055)



We add all the numbers together, and all the variables

17x-(2707x-4055)-20=0

We get rid of parentheses

17x-2707x+4055-20=0

We add all the numbers together, and all the variables

-2690x+4035=0

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

-2690x=-4035

x=-4035/-2690

x=1+1/2

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