Exercise 5 Page 47

When the absolute value of an expression 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 absolute value equation, 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)

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Solve for x

(I), (II):LHS-2x=RHS-2x

l-19=2x+1 -19=-6x-1

SubEqn

(I):LHS-1=RHS-1

l-20=2x -19=-6x-1

AddEqn

(II):LHS+1=RHS+1

l-20=2x -18=-6x

DivEqn

(I):.LHS /2.=.RHS /2.

l-10=x -18=-6x

DivEqn

(II):.LHS /(-6).=.RHS /(-6).

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 extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made.

|2x-19|=4x+1

Substitute

x= -10

|2({\color{#0000FF}{\text{-}10}})-19|\stackrel{?}=4({\color{#0000FF}{\text{-}10}})+1

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Simplify equation

MultPosNeg

a(- b)=- a * b

|\text{-}20-19|\stackrel{?}=\text{-}40+1

AddSubTerms

Add and subtract terms

|\text{-}39|\stackrel{?}=\text{-}39

AbsNeg

|-39|=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

Substitute

x= 3

|2\cdot{\color{#FF0000}{3}}-19|\stackrel{?}=4\cdot{\color{#FF0000}{3}}+1

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Simplify equation

Multiply

Multiply

|6-19|\stackrel{?}=12+1

AddSubTerms

Add and subtract terms

|\text{-}13|\stackrel{?}=13

AbsNeg

|-13|=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.