Exercise 17 Page 45

When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. |ax+b|=|cx+d| Although we can make 4 statements about this equation, there are actually only two possible cases to consider.

Statement	Result
Both absolute values are positive.	ax+b=cx+d
Both absolute values are negative.	-(ax+b)=-(cx+d)
Only the left-hand side is negative.	-(ax+b)=cx+d
Only the right-hand side is negative.	ax+b=-(cx+d)

Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given Equation:& |x-2|=|4+x| First Equation:& x-2 = 4+x Second Equation:& x-2=- (4+x) Let's solve for x in each of these equations simultaneously.

|x-2|=|4+x|

lc x-2 ≥ 0:x-2 = (4+x) & (I) x-2 < 0:x-2 = - (4+x) & (II)

lcx-2=4+x & (I) x-2=- (4+x) & (II)

▼

Solve for x

Distr

(II): Distribute -1

lx-2=4+x x-2=-4-x

SubEqn

(I):LHS-x=RHS-x

l-2≠4 x-2=-4-x

AddEqn

(II):LHS+x=RHS+x

l-2≠4 2x-2=-4

AddEqn

(II):LHS+2=RHS+2

l-2≠4 2x=-2

DivEqn

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

l-2≠4 x=-1

Notice that when solving Equation (I), we encountered a contradiction. This means that the first equation has no solution. The solution to Equation (II) is x=-1. Next we will substitute the found solution into the given equation to determine if a true statement is made and check for extraneous solutions.

|x-2|=|4+x|

Substitute

x= -1

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

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Simplify equation

AddSubTerms

Add and subtract terms

|\text{-}3|\stackrel{?}=|3|

AbsNeg

|-3|=3

3=3 ✓

Because the left-hand and right-hand sides are equal, we know that the solution is correct and x=-1 is not extraneous.