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1-1/5x=-4/5(5+x
We move all terms to the left:
1-1/5x-(-4/5(5+x)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 5(5+x)+1!=0We add all the numbers together, and all the variables
We move all terms containing x to the left, all other terms to the right
5(5+x)!=-1
x∈R
-1/5x-(-4/5(x+5)+1=0
We calculate fractions
(-5x(x+1)/(25x^2(x+1)+(-(-4*5x)/(25x^2(x+1)=0
We calculate terms in parentheses: +(-5x(x+1)/(25x^2(x+1)+(-(-4*5x)/(25x^2(x+1), so:
-5x(x+1)/(25x^2(x+1)+(-(-4*5x)/(25x^2(x+1
We can not solve this equation
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