1/4(40-8x)=19x+3-5x

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Solution for 1/4(40-8x)=19x+3-5x equation:



1/4(40-8x)=19x+3-5x
We move all terms to the left:
1/4(40-8x)-(19x+3-5x)=0
Domain of the equation: 4(40-8x)!=0
x∈R
We add all the numbers together, and all the variables
1/4(-8x+40)-(14x+3)=0
We get rid of parentheses
1/4(-8x+40)-14x-3=0
We multiply all the terms by the denominator
-14x*4(-8x+40)-3*4(-8x+40)+1=0
Wy multiply elements
-56x^2(--12x(-+1=0
We use the square of the difference formula
-56x^2(+12x(-1=0

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