Precalculus Examples

3+|2x|=3+33+|2x|=3+3
Step 1
Add 33 and 33.
3+|2x|=63+|2x|=6
Step 2
Move all terms not containing |2x||2x| to the right side of the equation.
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Step 2.1
Subtract 33 from both sides of the equation.
|2x|=6-3|2x|=63
Step 2.2
Subtract 33 from 66.
|2x|=3|2x|=3
|2x|=3|2x|=3
Step 3
Remove the absolute value term. This creates a ±± on the right side of the equation because |x|=±x|x|=±x.
2x=±32x=±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.
2x=32x=3
Step 4.2
Divide each term in 2x=32x=3 by 22 and simplify.
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Step 4.2.1
Divide each term in 2x=32x=3 by 22.
2x2=322x2=32
Step 4.2.2
Simplify the left side.
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Step 4.2.2.1
Cancel the common factor of 22.
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Step 4.2.2.1.1
Cancel the common factor.
2x2=32
Step 4.2.2.1.2
Divide x by 1.
x=32
x=32
x=32
x=32
Step 4.3
Next, use the negative value of the ± to find the second solution.
2x=-3
Step 4.4
Divide each term in 2x=-3 by 2 and simplify.
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Step 4.4.1
Divide each term in 2x=-3 by 2.
2x2=-32
Step 4.4.2
Simplify the left side.
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Step 4.4.2.1
Cancel the common factor of 2.
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Step 4.4.2.1.1
Cancel the common factor.
2x2=-32
Step 4.4.2.1.2
Divide x by 1.
x=-32
x=-32
x=-32
Step 4.4.3
Simplify the right side.
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Step 4.4.3.1
Move the negative in front of the fraction.
x=-32
x=-32
x=-32
Step 4.5
The complete solution is the result of both the positive and negative portions of the solution.
x=32,-32
x=32,-32
Step 5
The result can be shown in multiple forms.
Exact Form:
x=32,-32
Decimal Form:
x=1.5,-1.5
Mixed Number Form:
x=112,-112
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 [x2  12  π  xdx ] 
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