525-3(2w-3)+3w=7(3w-21)+42(3w-1)-4(w+5)+13

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Solution for 525-3(2w-3)+3w=7(3w-21)+42(3w-1)-4(w+5)+13 equation: 



525-3(2w-3)+3w=7(3w-21)+42(3w-1)-4(w+5)+13

We move all terms to the left:

525-3(2w-3)+3w-(7(3w-21)+42(3w-1)-4(w+5)+13)=0

We add all the numbers together, and all the variables

3w-3(2w-3)-(7(3w-21)+42(3w-1)-4(w+5)+13)+525=0

We multiply parentheses

3w-6w-(7(3w-21)+42(3w-1)-4(w+5)+13)+9+525=0



We calculate terms in parentheses: -(7(3w-21)+42(3w-1)-4(w+5)+13), so:

7(3w-21)+42(3w-1)-4(w+5)+13

We multiply parentheses

21w+126w-4w-147-42-20+13

We add all the numbers together, and all the variables

143w-196

Back to the equation:

-(143w-196)



We add all the numbers together, and all the variables

-3w-(143w-196)+534=0

We get rid of parentheses

-3w-143w+196+534=0

We add all the numbers together, and all the variables

-146w+730=0

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

-146w=-730

w=-730/-146

w=+5

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