10(3-x)+6(x+3)=19+7(2x-1)

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Solution for 10(3-x)+6(x+3)=19+7(2x-1) equation:



10(3-x)+6(x+3)=19+7(2x-1)
We move all terms to the left:
10(3-x)+6(x+3)-(19+7(2x-1))=0
We add all the numbers together, and all the variables
10(-1x+3)+6(x+3)-(19+7(2x-1))=0
We multiply parentheses
-10x+6x-(19+7(2x-1))+30+18=0
We calculate terms in parentheses: -(19+7(2x-1)), so:
19+7(2x-1)
determiningTheFunctionDomain 7(2x-1)+19
We multiply parentheses
14x-7+19
We add all the numbers together, and all the variables
14x+12
Back to the equation:
-(14x+12)
We add all the numbers together, and all the variables
-4x-(14x+12)+48=0
We get rid of parentheses
-4x-14x-12+48=0
We add all the numbers together, and all the variables
-18x+36=0
We move all terms containing x to the left, all other terms to the right
-18x=-36
x=-36/-18
x=+2

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