2-5(2x+1)=3/4(16x-8)-22x

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Solution for 2-5(2x+1)=3/4(16x-8)-22x equation: 



2-5(2x+1)=3/4(16x-8)-22x

We move all terms to the left:

2-5(2x+1)-(3/4(16x-8)-22x)=0



Domain of the equation: 4(16x-8)-22x)!=0

x∈R



We multiply parentheses

-10x-(3/4(16x-8)-22x)-5+2=0

We multiply all the terms by the denominator


-10x*4(16x-8)-22x)-(3-5*4(16x-8)-22x)+2*4(16x-8)-22x)=0

Wy multiply elements

-40x^2(1-22x)-(3-20x(1-22x)+2*4(16x-8)-22x)=0



We calculate terms in parentheses: -(3-20x(1-22x)+2*4(16x-8)-22x), so:

3-20x(1-22x)+2*4(16x-8)-22x

determiningTheFunctionDomain
-20x(1-22x)+2*4(16x-8)-22x+3

We add all the numbers together, and all the variables

-20x(-22x+1)+2*4(16x-8)-22x+3

We add all the numbers together, and all the variables

-22x-20x(-22x+1)+2*4(16x-8)+3

We multiply parentheses

440x^2-22x-20x+2*4(16x-8)+3

Wy multiply elements

440x^2-22x-20x+8x(1+3

We add all the numbers together, and all the variables

440x^2-42x+8x(1+3

Back to the equation:

-(440x^2-42x+8x(1+3)



We add all the numbers together, and all the variables

-40x^2(-22x+1)-(440x^2-42x+8x4=0

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