Exercise 26 Page 91

In order to rewrite this sentence into an absolute value inequality, we should first remember the standard form for an absolute value equation. |x-Midpoint|=Distance Let n be our variable. We can fill in the numerical information. & Twice a number... 10units from-1. & | 2n-( -1)|... 10Since we need the distance from -1 to be no less than 10 units, we will use ≥. & Twice a numberis no less than 10units from-1. & | 2n-( -1)|≥ 10 Now we can solve this inequality by splitting the absolute value into two cases. Because we want the distance to be greater than or equal to 10, we will need to write an "or" inequality. First case: &2n-(-1)≥10 Second case: &2n-(-1)≤-10 We can solve these cases separately and then combine the results. Let's begin with the first case.

2n-(-1)≥10

SubNeg

a-(- b)=a+b

2n+1≥10

SubIneq

LHS-1≥RHS-1

2x≥9

DivIneq

.LHS /2.≥.RHS /2.

x≥9/2

Now let's solve the second case.

2n-(-1)≤-10

SubNeg

a-(- b)=a+b

2n+1≤-10

SubIneq

LHS-1≤RHS-1

2n≤-11

DivIneq

.LHS /2.≤.RHS /2.

n≤-11/2

With this information we can write the solution set for this compound inequality. n≥9/2 or n≤-11/2