Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
x∈(−∞,0)∪(35,+∞)
Evaluate
x1×31<51
Find the domain
x1×31<51,x=0
Multiply the terms
More Steps

Multiply the terms
x1×31
Multiply the terms
x×31
Use the commutative property to reorder the terms
3x1
3x1<51
Move the expression to the left side
3x1−51<0
Subtract the terms
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Evaluate
3x1−51
Reduce fractions to a common denominator
3x×55−5×3x3x
Multiply the numbers
15x5−5×3x3x
Multiply the numbers
15x5−15x3x
Write all numerators above the common denominator
15x5−3x
15x5−3x<0
Set the numerator and denominator of 15x5−3x equal to 0 to find the values of x where sign changes may occur
5−3x=015x=0
Calculate
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Evaluate
5−3x=0
Move the constant to the right-hand side and change its sign
−3x=0−5
Removing 0 doesn't change the value,so remove it from the expression
−3x=−5
Change the signs on both sides of the equation
3x=5
Divide both sides
33x=35
Divide the numbers
x=35
x=3515x=0
Calculate
x=35x=0
Determine the test intervals using the critical values
x<00<x<35x>35
Choose a value form each interval
x1=−1x2=1x3=3
To determine if x<0 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
More Steps

Evaluate
3(−1)1<51
Simplify
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Evaluate
3(−1)1
Simplify
−31
Use b−a=−ba=−ba to rewrite the fraction
−31
−31<51
Calculate
−0.3˙<51
Calculate
−0.3˙<0.2
Check the inequality
true
x<0 is the solutionx2=1x3=3
To determine if 0<x<35 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
More Steps

Evaluate
3×11<51
Reduce the fraction
31<51
Calculate
0.3˙<51
Calculate
0.3˙<0.2
Check the inequality
false
x<0 is the solution0<x<35 is not a solutionx3=3
To determine if x>35 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
More Steps

Evaluate
3×31<51
Multiply the numbers
91<51
Calculate
0.1˙<51
Calculate
0.1˙<0.2
Check the inequality
true
x<0 is the solution0<x<35 is not a solutionx>35 is the solution
The original inequality is a strict inequality,so does not include the critical value ,the final solution is x∈(−∞,0)∪(35,+∞)
x∈(−∞,0)∪(35,+∞)
Check if the solution is in the defined range
x∈(−∞,0)∪(35,+∞),x=0
Solution
x∈(−∞,0)∪(35,+∞)
Show Solution
