Exercise 8 Page 30

When the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Let's look at an example equation. |ax+b|=|cx+d| For this equation, there are two possible cases to consider. ax+b=cx+d or ax+b=- (cx+d) To solve the given absolute value equation, we will write two equations similar to those above when we remove the absolute value.

|x+8|=|2x+1|

lc x+8 ≥ 0:x+8 = (2x+1) & (I) x+8 < 0:x+8 = - (2x+1) & (II)

lcx+8=2x+1 & (I) x+8=- (2x+1) & (II)

Distr

(II):Distribute -1

lx+8=2x+1 x+8=- 2x -1

(I), (II):LHS-8=RHS-8

lx=2x-7 x=- 2x -9

SubEqn

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

l- x=- 7 x=- 2x -9

DivEqn

(I):.LHS /- 1.=.RHS /- 1.

lx=7 x=- 2x -9

AddEqn

(II):LHS+2x=RHS+2x

lx=7 3x=- 9

DivEqn

(II):.LHS /3.=.RHS /3.

lx_1=7 x_2=- 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.

|x+8|=|2x+1|

Substitute

x= 7

| 7+8|? =|2( 7)+1|

Multiply

Multiply

|7+8|? =|14+1|

AddTerms

Add terms

|15|? =|15|

AbsPos

|15|=15

15=15 ✓

We found that x=7 is not extraneous.

|x+8|=|2x+1|

Substitute

x= - 3

|{\color{#0000FF}{\text{-} 3}}+8|\stackrel{?}=|2({\color{#0000FF}{\text{-} 3}})+1|

Multiply

Multiply

|\text{-} 3+8|\stackrel{?}=|\text{-} 6+1|

AddTerms

Add terms

|5|\stackrel{?}=|\text{-} 5|

AbsPos

|5|=5

5\stackrel{?}=|\text{-} 5|

AbsNeg

|-5|=5

5=5 ✓

Since we obtained a true statement, we know that x=- 3 is not extraneous.