Exercise 19 Page 46

An absolute value measures an expression's distance from a midpoint on a number line. |x-1|= 5x+10 This equation means that the distance is 5x+10, either in the positive direction or the negative direction. |x-1|= 5x+10 ⇒ lx-1= 5x+10 x-1= - (5x+10) To find the solutions to the absolute value equation , we need to solve both of these cases for x.

|x-1|=5x+10

lc x-1 ≥ 0:x-1 = (5x+10) & (I) x-1 < 0:x-1 = - (5x+10) & (II)

lcx-1=5x+10 & (I) x-1= -(5x+10) & (II)

Distr

(II): Distribute -1

lx-1=5x+10 x-1= -5x-10

(I), (II): LHS+1=RHS+1

lx=5x+11 x= -5x-9

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Solve for x

SubEqn

(I): LHS-5x=RHS-5x

l- 4x=11 x=- 5x-8

AddEqn

(II): LHS+5x=RHS+5x

l-4x=11 6x=-9

DivEqn

(I): .LHS /(-4).=.RHS /(-4).

lx=- 114 6x=- 9

DivEqn

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

lx=- 114 x=- 96

ReduceFrac

(II): a/b=.a /3./.b /3.

lx= - 114 x= - 32

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answers by substituting them back into the original equation. Let's start with - 114.

|x-1|? =5x+10

Substitute

x= -11/4

| -11/4-1|? =5( -11/4)+10

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Simplify

MultPosNeg

a(- b)=- a * b

|-11/4-1|? =- 5*11/4+10

MoveLeftFacToNum

a*b/c= a* b/c

|-11/4-1|? =-55/4+10