Exercise 11 Page 31

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. ax+b=cx or ax+b=- cx To solve the given absolute value equation we will write two equations — similar to those above — when we remove the absolute value.

|3x-2|=x

lc 3x-2 ≥ 0:3x-2 = x & (I) 3x-2 < 0:3x-2 = - x & (II)

lc3x-2=x & (I) 3x-2=- x & (II)

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Solve for x

(I), (II):LHS+2=RHS+2

l3x=x+2 3x=- x+2

SubEqn

(I):LHS-x=RHS-x

l2x=2 3x=- x+2

AddEqn

(II):LHS+x=RHS+x

l2x=2 4x=2

DivEqn

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

lx=1 4x=2

DivEqn

(II):.LHS /4.=.RHS /4.

lx=1 x= 24

ReduceFrac

(II):a/b=.a /2./.b /2.

lx_1= 1 x_2= 12

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.

|3x-2|=x

Substitute

x= 1

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

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Simplify

OneMult

1* a=a

|3-2|\stackrel{?}=1

SubTerm

Subtract term

|1|\stackrel{?}=1

AbsPos

|a|=a

1=1

We end with a true statement, so x=1 is not an extraneous solution.

|3x-2|=x

Substitute

x= 1/2

\left|3\left({\color{#009600}{\dfrac{1}{2}}}\right)-2\right|\stackrel{?}={\color{#009600}{\dfrac{1}{2}}}

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Simplify

MoveLeftFacToNumOne

a* 1/b= a/b

\left|\dfrac{3}{2}-2\right|\stackrel{?}=\dfrac{1}{2}

NumberToFrac

a = 2* a/2

\left|\dfrac{3}{2}-\dfrac{4}{2}\right|\stackrel{?}=\dfrac{1}{2}

SubFrac

Subtract fractions

\left|\text{-}\dfrac{1}{2}\right|\stackrel{?}=\dfrac{1}{2}

AbsNeg

|-1/2|=1/2

1/2=1/2

Again we end with a true statement, so x= 12 is also not an extraneous solution.