When the of an is equal to another expression, either the expressions are equal or the opposite of the expressions are equal.
|ax+b|=cx+d
⇓
ax+b=cx+d or ax+b=- cx-d
To solve the given , we need to solve both of these cases for x.
|2x-19|=4x+1
lc 2x-19 ≥ 0:2x-19 = (4x+1) & (I) 2x-19 < 0:2x-19 = - (4x+1) & (II)
lc2x-19=4x+1 & (I) 2x-19=-4x-1 & (II)
l-19=2x+1 -19=-6x-1
l-20=2x -19=-6x-1
l-20=2x -18=-6x
l-10=x -18=-6x
l-10=x 3=x
(I), (II):Rearrange equation
lx= -10 x= 3
After solving an absolute value equation, it is necessary to check for . To do this, we substitute the found solutions into the given equation and determine if a true statement is made.
|2x-19|=4x+1
|2({\color{#0000FF}{\text{-}10}})-19|\stackrel{?}=4({\color{#0000FF}{\text{-}10}})+1
|\text{-}20-19|\stackrel{?}=\text{-}40+1
|\text{-}39|\stackrel{?}=\text{-}39
39=-39 *
Substituting - 10 for x in the equation does not result in a true statement, so x=-10 is extraneous solution. Now let's check whether or not x=3 is extraneous.
|2x-19|=4x+1
|2\cdot{\color{#FF0000}{3}}-19|\stackrel{?}=4\cdot{\color{#FF0000}{3}}+1
|6-19|\stackrel{?}=12+1
|\text{-}13|\stackrel{?}=13
13=13 ✓
Substituting 3 for x in the equation does result in a true statement, so x=3 is not an extraneous solution. Therefore, the equation has only one solution.
