Solve for x x/(1-x)>1

Math
x1-x>1
Move 1 to the left side of the equation by subtracting it from both sides.
x1-x-1>0
Simplify x1-x-1.
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To write -1 as a fraction with a common denominator, multiply by 1-x1-x.
x1-x-1⋅1-x1-x>0
Simplify terms.
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Combine -1 and 1-x1-x.
x1-x+-(1-x)1-x>0
Combine the numerators over the common denominator.
x-(1-x)1-x>0
x-(1-x)1-x>0
Simplify the numerator.
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Apply the distributive property.
x-1⋅1–x1-x>0
Multiply -1 by 1.
x-1–x1-x>0
Multiply –x.
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Multiply -1 by -1.
x-1+1×1-x>0
Multiply x by 1.
x-1+x1-x>0
x-1+x1-x>0
Add x and x.
2x-11-x>0
2x-11-x>0
2x-11-x>0
Find all the values where the expression switches from negative to positive by setting each factor equal to 0 and solving.
2x-1=0
1-x=0
Add 1 to both sides of the equation.
2x=1
Divide each term by 2 and simplify.
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Divide each term in 2x=1 by 2.
2×2=12
Cancel the common factor of 2.
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Cancel the common factor.
2×2=12
Divide x by 1.
x=12
x=12
x=12
Subtract 1 from both sides of the equation.
-x=-1
Multiply each term in -x=-1 by -1
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Multiply each term in -x=-1 by -1.
(-x)⋅-1=(-1)⋅-1
Multiply (-x)⋅-1.
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Multiply -1 by -1.
1x=(-1)⋅-1
Multiply x by 1.
x=(-1)⋅-1
x=(-1)⋅-1
Multiply -1 by -1.
x=1
x=1
Solve for each factor to find the values where the absolute value expression goes from negative to positive.
x=12
x=1
Consolidate the solutions.
x=12,1
Find the domain of 2x-11-x.
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Set the denominator in 2x-11-x equal to 0 to find where the expression is undefined.
1-x=0
Solve for x.
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Subtract 1 from both sides of the equation.
-x=-1
Multiply each term in -x=-1 by -1
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Multiply each term in -x=-1 by -1.
(-x)⋅-1=(-1)⋅-1
Multiply (-x)⋅-1.
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Multiply -1 by -1.
1x=(-1)⋅-1
Multiply x by 1.
x=(-1)⋅-1
x=(-1)⋅-1
Multiply -1 by -1.
x=1
x=1
x=1
The domain is all values of x that make the expression defined.
(-∞,1)∪(1,∞)
(-∞,1)∪(1,∞)
Use each root to create test intervals.
x<12
12<x<1
x>1
Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.
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Test a value on the interval x<12 to see if it makes the inequality true.
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Choose a value on the interval x<12 and see if this value makes the original inequality true.
x=0
Replace x with 0 in the original inequality.
01-(0)>1
The left side 0 is not greater than the right side 1, which means that the given statement is false.
False
False
Test a value on the interval 12<x<1 to see if it makes the inequality true.
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Choose a value on the interval 12<x<1 and see if this value makes the original inequality true.
x=0.75
Replace x with 0.75 in the original inequality.
0.751-(0.75)>1
The left side 3 is greater than the right side 1, which means that the given statement is always true.
True
True
Test a value on the interval x>1 to see if it makes the inequality true.
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Choose a value on the interval x>1 and see if this value makes the original inequality true.
x=4
Replace x with 4 in the original inequality.
41-(4)>1
The left side -1.3‾ is not greater than the right side 1, which means that the given statement is false.
False
False
Compare the intervals to determine which ones satisfy the original inequality.
x<12 False
12<x<1 True
x>1 False
x<12 False
12<x<1 True
x>1 False
The solution consists of all of the true intervals.
12<x<1
The result can be shown in multiple forms.
Inequality Form:
12<x<1
Interval Notation:
(12,1)
<div data-graph-input="{"graphs":[{"ascii":"(1)/(2)<x
Solve for x x/(1-x)>1

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