Algebra
Calculus
Trigonometry
Matrix
Question
Solve the inequality
x∈(−∞,−2)∪(−2,−23]
Evaluate
∣x+2∣∣x+1∣≥1
Find the domain
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Evaluate
∣x+2∣=0
Rewrite the expression
x+2=0
Move the constant to the right side
x=0−2
Removing 0 doesn't change the value,so remove it from the expression
x=−2
∣x+2∣∣x+1∣≥1,x=−2
Move the expression to the left side
∣x+2∣∣x+1∣−1≥0
Subtract the terms
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Evaluate
∣x+2∣∣x+1∣−1
Reduce fractions to a common denominator
∣x+2∣∣x+1∣−∣x+2∣∣x+2∣
Write all numerators above the common denominator
∣x+2∣∣x+1∣−∣x+2∣
∣x+2∣∣x+1∣−∣x+2∣≥0
Separate the inequality into 2 possible cases
{∣x+1∣−∣x+2∣≥0∣x+2∣>0{∣x+1∣−∣x+2∣≤0∣x+2∣<0
Solve the inequality
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Evaluate
∣x+1∣−∣x+2∣≥0
Separate the inequality into 4 possible cases
x+1−(x+2)≥0,x+1≥0,x+2≥0x+1−(−(x+2))≥0,x+1≥0,x+2<0−(x+1)−(x+2)≥0,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
The statement is false for any value of x
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Evaluate
x+1−(x+2)≥0
Remove the parentheses
x+1−x−2≥0
Simplify the expression
−1≥0
The statement is false for any value of x
x∈∅
x∈∅,x+1≥0,x+2≥0x+1−(−(x+2))≥0,x+1≥0,x+2<0−(x+1)−(x+2)≥0,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
x+1≥0
Move the constant to the right side
x≥0−1
Removing 0 doesn't change the value,so remove it from the expression
x≥−1
x∈∅,x≥−1,x+2≥0x+1−(−(x+2))≥0,x+1≥0,x+2<0−(x+1)−(x+2)≥0,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
x+2≥0
Move the constant to the right side
x≥0−2
Removing 0 doesn't change the value,so remove it from the expression
x≥−2
x∈∅,x≥−1,x≥−2x+1−(−(x+2))≥0,x+1≥0,x+2<0−(x+1)−(x+2)≥0,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
x+1−(−(x+2))≥0
Remove the parentheses
x+1+x+2≥0
Simplify the expression
2x+3≥0
Move the constant to the right side
2x≥0−3
Removing 0 doesn't change the value,so remove it from the expression
2x≥−3
Divide both sides
22x≥2−3
Divide the numbers
x≥2−3
Use b−a=−ba=−ba to rewrite the fraction
x≥−23
x∈∅,x≥−1,x≥−2x≥−23,x+1≥0,x+2<0−(x+1)−(x+2)≥0,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
x+1≥0
Move the constant to the right side
x≥0−1
Removing 0 doesn't change the value,so remove it from the expression
x≥−1
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x+2<0−(x+1)−(x+2)≥0,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
x+2<0
Move the constant to the right side
x<0−2
Removing 0 doesn't change the value,so remove it from the expression
x<−2
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x<−2−(x+1)−(x+2)≥0,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
−(x+1)−(x+2)≥0
Remove the parentheses
−x−1−x−2≥0
Simplify the expression
−2x−3≥0
Move the constant to the right side
−2x≥0+3
Removing 0 doesn't change the value,so remove it from the expression
−2x≥3
Change the signs on both sides of the inequality and flip the inequality sign
2x≤−3
Divide both sides
22x≤2−3
Divide the numbers
x≤2−3
Use b−a=−ba=−ba to rewrite the fraction
x≤−23
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x<−2x≤−23,x+1<0,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
x+1<0
Move the constant to the right side
x<0−1
Removing 0 doesn't change the value,so remove it from the expression
x<−1
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x<−2x≤−23,x<−1,x+2≥0−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
Evaluate
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Evaluate
x+2≥0
Move the constant to the right side
x≥0−2
Removing 0 doesn't change the value,so remove it from the expression
x≥−2
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x<−2x≤−23,x<−1,x≥−2−(x+1)−(−(x+2))≥0,x+1<0,x+2<0
The statement is true for any value of x
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Evaluate
−(x+1)−(−(x+2))≥0
Remove the parentheses
−x−1+x+2≥0
Simplify the expression
1≥0
The statement is true for any value of x
x∈R
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x<−2x≤−23,x<−1,x≥−2x∈R,x+1<0,x+2<0
Evaluate
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Evaluate
x+1<0
Move the constant to the right side
x<0−1
Removing 0 doesn't change the value,so remove it from the expression
x<−1
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x<−2x≤−23,x<−1,x≥−2x∈R,x<−1,x+2<0
Evaluate
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Evaluate
x+2<0
Move the constant to the right side
x<0−2
Removing 0 doesn't change the value,so remove it from the expression
x<−2
x∈∅,x≥−1,x≥−2x≥−23,x≥−1,x<−2x≤−23,x<−1,x≥−2x∈R,x<−1,x<−2
Find the intersection
x∈∅x≥−23,x≥−1,x<−2x≤−23,x<−1,x≥−2x∈R,x<−1,x<−2
Find the intersection
x∈∅x∈∅x≤−23,x<−1,x≥−2x∈R,x<−1,x<−2
Find the intersection
x∈∅x∈∅−2≤x≤−23x∈R,x<−1,x<−2
Find the intersection
x∈∅x∈∅−2≤x≤−23x<−2
Find the union
x≤−23
{x≤−23∣x+2∣>0{∣x+1∣−∣x+2∣≤0∣x+2∣<0
Solve the inequality
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Evaluate
∣x+2∣>0
Since the left-hand side is always positive or 0,and the right-hand side is always 0,the statement is true for any value of x,except when ∣x+2∣=0
∣x+2∣=0
Rewrite the expression
x+2=0
Move the constant to the right-hand side and change its sign
x=0−2
Removing 0 doesn't change the value,so remove it from the expression
x=−2
Exclude the impossible values of x
x=−2
{x≤−23x=−2{∣x+1∣−∣x+2∣≤0∣x+2∣<0
Solve the inequality
More Steps
Evaluate
∣x+1∣−∣x+2∣≤0
Separate the inequality into 4 possible cases
x+1−(x+2)≤0,x+1≥0,x+2≥0x+1−(−(x+2))≤0,x+1≥0,x+2<0−(x+1)−(x+2)≤0,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
The statement is true for any value of x
More Steps
Evaluate
x+1−(x+2)≤0
Remove the parentheses
x+1−x−2≤0
Simplify the expression
−1≤0
The statement is true for any value of x
x∈R
x∈R,x+1≥0,x+2≥0x+1−(−(x+2))≤0,x+1≥0,x+2<0−(x+1)−(x+2)≤0,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
More Steps
Evaluate
x+1≥0
Move the constant to the right side
x≥0−1
Removing 0 doesn't change the value,so remove it from the expression
x≥−1
x∈R,x≥−1,x+2≥0x+1−(−(x+2))≤0,x+1≥0,x+2<0−(x+1)−(x+2)≤0,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
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Evaluate
x+2≥0
Move the constant to the right side
x≥0−2
Removing 0 doesn't change the value,so remove it from the expression
x≥−2
x∈R,x≥−1,x≥−2x+1−(−(x+2))≤0,x+1≥0,x+2<0−(x+1)−(x+2)≤0,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
More Steps
Evaluate
x+1−(−(x+2))≤0
Remove the parentheses
x+1+x+2≤0
Simplify the expression
2x+3≤0
Move the constant to the right side
2x≤0−3
Removing 0 doesn't change the value,so remove it from the expression
2x≤−3
Divide both sides
22x≤2−3
Divide the numbers
x≤2−3
Use b−a=−ba=−ba to rewrite the fraction
x≤−23
x∈R,x≥−1,x≥−2x≤−23,x+1≥0,x+2<0−(x+1)−(x+2)≤0,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
More Steps
Evaluate
x+1≥0
Move the constant to the right side
x≥0−1
Removing 0 doesn't change the value,so remove it from the expression
x≥−1
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x+2<0−(x+1)−(x+2)≤0,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
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Evaluate
x+2<0
Move the constant to the right side
x<0−2
Removing 0 doesn't change the value,so remove it from the expression
x<−2
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x<−2−(x+1)−(x+2)≤0,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
More Steps
Evaluate
−(x+1)−(x+2)≤0
Remove the parentheses
−x−1−x−2≤0
Simplify the expression
−2x−3≤0
Move the constant to the right side
−2x≤0+3
Removing 0 doesn't change the value,so remove it from the expression
−2x≤3
Change the signs on both sides of the inequality and flip the inequality sign
2x≥−3
Divide both sides
22x≥2−3
Divide the numbers
x≥2−3
Use b−a=−ba=−ba to rewrite the fraction
x≥−23
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x<−2x≥−23,x+1<0,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
More Steps
Evaluate
x+1<0
Move the constant to the right side
x<0−1
Removing 0 doesn't change the value,so remove it from the expression
x<−1
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x<−2x≥−23,x<−1,x+2≥0−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
Evaluate
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Evaluate
x+2≥0
Move the constant to the right side
x≥0−2
Removing 0 doesn't change the value,so remove it from the expression
x≥−2
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x<−2x≥−23,x<−1,x≥−2−(x+1)−(−(x+2))≤0,x+1<0,x+2<0
The statement is false for any value of x
More Steps
Evaluate
−(x+1)−(−(x+2))≤0
Remove the parentheses
−x−1+x+2≤0
Simplify the expression
1≤0
The statement is false for any value of x
x∈∅
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x<−2x≥−23,x<−1,x≥−2x∈∅,x+1<0,x+2<0
Evaluate
More Steps
Evaluate
x+1<0
Move the constant to the right side
x<0−1
Removing 0 doesn't change the value,so remove it from the expression
x<−1
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x<−2x≥−23,x<−1,x≥−2x∈∅,x<−1,x+2<0
Evaluate
More Steps
Evaluate
x+2<0
Move the constant to the right side
x<0−2
Removing 0 doesn't change the value,so remove it from the expression
x<−2
x∈R,x≥−1,x≥−2x≤−23,x≥−1,x<−2x≥−23,x<−1,x≥−2x∈∅,x<−1,x<−2
Find the intersection
x≥−1x≤−23,x≥−1,x<−2x≥−23,x<−1,x≥−2x∈∅,x<−1,x<−2
Find the intersection
x≥−1x∈∅x≥−23,x<−1,x≥−2x∈∅,x<−1,x<−2
Find the intersection
x≥−1x∈∅−23≤x<−1x∈∅,x<−1,x<−2
Find the intersection
x≥−1x∈∅−23≤x<−1x∈∅
Find the union
x≥−23
{x≤−23x=−2{x≥−23∣x+2∣<0
Since the left-hand side is always positive or 0,and the right-hand side is always 0,the statement is false for any value of x
{x≤−23x=−2{x≥−23x∈/R
Find the intersection
x∈(−∞,−2)∪(−2,−23]{x≥−23x∈/R
Find the intersection
x∈(−∞,−2)∪(−2,−23]x∈/R
Find the union
x∈(−∞,−2)∪(−2,−23]
Check if the solution is in the defined range
x∈(−∞,−2)∪(−2,−23],x=−2
Solution
x∈(−∞,−2)∪(−2,−23]
Show Solution
