03x+2(x-1)+04(2x+3)=25(x+3)+73

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Solution for 03x+2(x-1)+04(2x+3)=25(x+3)+73 equation:



03x+2(x-1)+04(2x+3)=25(x+3)+73
We move all terms to the left:
03x+2(x-1)+04(2x+3)-(25(x+3)+73)=0
We multiply parentheses
03x+2x+8x-(25(x+3)+73)-2+12=0
We calculate terms in parentheses: -(25(x+3)+73), so:
25(x+3)+73
We multiply parentheses
25x+75+73
We add all the numbers together, and all the variables
25x+148
Back to the equation:
-(25x+148)
We add all the numbers together, and all the variables
13x-(25x+148)+10=0
We get rid of parentheses
13x-25x-148+10=0
We add all the numbers together, and all the variables
-12x-138=0
We move all terms containing x to the left, all other terms to the right
-12x=138
x=138/-12
x=-11+1/2

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