Calculus Examples

2+|3x|=2+3
Step 1
Add 2 and 3.
2+|3x|=5
Step 2
Move all terms not containing |3x| to the right side of the equation.
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Step 2.1
Subtract 2 from both sides of the equation.
|3x|=5-2
Step 2.2
Subtract 2 from 5.
|3x|=3
|3x|=3
Step 3
Remove the absolute value term. This creates a ± on the right side of the equation because |x|=±x.
3x=±3
Step 4
The complete solution is the result of both the positive and negative portions of the solution.
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Step 4.1
First, use the positive value of the ± to find the first solution.
3x=3
Step 4.2
Divide each term in 3x=3 by 3 and simplify.
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Step 4.2.1
Divide each term in 3x=3 by 3.
3x3=33
Step 4.2.2
Simplify the left side.
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Step 4.2.2.1
Cancel the common factor of 3.
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Step 4.2.2.1.1
Cancel the common factor.
3x3=33
Step 4.2.2.1.2
Divide x by 1.
x=33
x=33
x=33
Step 4.2.3
Simplify the right side.
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Step 4.2.3.1
Divide 3 by 3.
x=1
x=1
x=1
Step 4.3
Next, use the negative value of the ± to find the second solution.
3x=-3
Step 4.4
Divide each term in 3x=-3 by 3 and simplify.
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Step 4.4.1
Divide each term in 3x=-3 by 3.
3x3=-33
Step 4.4.2
Simplify the left side.
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Step 4.4.2.1
Cancel the common factor of 3.
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Step 4.4.2.1.1
Cancel the common factor.
3x3=-33
Step 4.4.2.1.2
Divide x by 1.
x=-33
x=-33
x=-33
Step 4.4.3
Simplify the right side.
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Step 4.4.3.1
Divide -3 by 3.
x=-1
x=-1
x=-1
Step 4.5
The complete solution is the result of both the positive and negative portions of the solution.
x=1,-1
x=1,-1
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 [x2  12  π  xdx ] 
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