SOLUTION: Will the expression 5 over(fraction)x always be less than 5?

Question 346974: Will the expression 5 over(fraction)x always be less than 5?

Answer by haileytucki(390) (Show Source): You can put this solution on YOUR website!
((5)/(x))<5
Remove the parentheses around the expression (5)/(x).
(5)/(x)<5
To set the left-hand side of the inequality equal to 0, move all the expressions to the left-hand side.
(5)/(x)-5<0
To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is x. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
(5)/(x)-5*(x)/(x)<0
Complete the multiplication to produce a denominator of x in each expression.
(5)/(x)-(5x)/(x)<0
Combine the numerators of all expressions that have common denominators.
(5-5x)/(x)<0
Reorder the polynomial 5-5x alphabetically from left to right, starting with the highest order term.
(-5x+5)/(x)<0
Factor out the GCF of -5 from each term in the polynomial.
(-5(x)-5(-1))/(x)<0
Factor out the GCF of -5 from -5x+5.
(-5(x-1))/(x)<0
Move the -1 to the front of the fraction.
-(5(x-1))/(x)<0
Multiply each term in the inequality by -1.
-(5(x-1))/(x)*-1>0*-1
Simplify the left-hand side of the inequality by canceling the common factors.
(5(x-1))/(x)>0*-1
Multiply 0 by -1 to get 0.
(5(x-1))/(x)>0
Find all the values where the expression switches from negative to positive by setting each factor equal to 0 and solving.
x=0_(x-1)=0
Solve for each factor to find the values where the absolute value expression goes from negative to positive.
x=0_x=1
To find the solution set that makes the expression greater than 0, break the set into real number intervals based on the values found earlier.
x<0_0 Determine if the given interval makes each factor positive or negative. If the number of negative factors is odd, then the entire expression over this interval is negative. If the number of negative factors is even, then the entire expression over this interval is positive.
x<0 makes the expression positive_0 Since this is a 'greater than 0' inequality, all intervals that make the expression positive are part of the solution.
x<0 or x>1
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