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Break down the given absolute value equation into two separate equations.
x=1 and x=1/2
When the absolute value of an expression is equal to an expression, either the expressions are equal or the opposite of the expressions are equal. Let's look at an example equation.
|ax+b|=cx
For this equation, there are two possible cases to consider.
lc 3x-2 ≥ 0:3x-2 = x & (I) 3x-2 < 0:3x-2 = - x & (II)
(I), (II):LHS+2=RHS+2
(I):LHS-x=RHS-x
(II):LHS+x=RHS+x
(I):.LHS /2.=.RHS /2.
(II):.LHS /4.=.RHS /4.
(II):a/b=.a /2./.b /2.
After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if it makes a true statement. Let's start with x=1.
x= 1
1* a=a
Subtract term
|a|=a
We ended with a true statement, so x=1 is not an extraneous solution. Now let's check x= 12.
x= 1/2
a* 1/b= a/b
a = 2* a/2
Subtract fractions
|-1/2|=1/2
We ended with another true statement, so x= 12 is also not an extraneous solution.