1/5(x+5)+2(x-2)+(x-1)=28

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Solution for 1/5(x+5)+2(x-2)+(x-1)=28 equation: 



1/5(x+5)+2(x-2)+(x-1)=28

We move all terms to the left:

1/5(x+5)+2(x-2)+(x-1)-(28)=0



Domain of the equation: 5(x+5)!=0

x∈R



We multiply parentheses

1/5(x+5)+2x+(x-1)-4-28=0

We get rid of parentheses

1/5(x+5)+2x+x-1-4-28=0

We multiply all the terms by the denominator


2x*5(x+5)+x*5(x+5)-1*5(x+5)-4*5(x+5)-28*5(x+5)+1=0

Wy multiply elements

10x^2(x+5x^2(x-5x(x-20x(x-140x(x+1=0

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