Warm–up #11. Solve

2.

3 32 2

7 9 22 2 2

3 34 4

911 44 4 4

5114 4

2 3( ) 1

2 3( ) 1

3

2

2 3 13 3

3 3 or (3 3)2 3 or 3 3

3 or 4 3

23

or 4

t tt t

t t t tt t t

t t

t

Warm–up #1 Solutions

Isolate Abs Val

Check in original!!

NOT a soln!

Abs Val either (+) or zero

But, NEVER (–)

NO SOLUTION

Warm–up #1 Solutions

3 7 4x

Homework LogThurs

10/1

Lesson 2 – 1

Learning Objective: To solve absolute values & solve for variables

Hw: #202 Pg.102 #33 – 59 odd, 67, 69

10/1/15 Lesson 2 – 1 Absolute Values & Formulas Day 2

Advanced Math/Trig

Learning Objective

To solve absolute value equations

To solve for variables

Solving for a Variable1. Solve for r: P = A(1 – rx)

P = A – Arx –A –A

P – A = –Arx –Ax –Ax

or

Solving for a Variable2. Solve for i:

(1+𝑛𝑖)(1+𝑛𝑖)

Absolute Value Equations

3.

x – 2 = 3x + 1 or x – 2 = –(3x + 1)

–2x = 3 or x – 2 = –3x –14x = 1

Check for extraneous solutions!

Absolute Value Equations

3. Check!

Amount of Loan

(1 )A P rt

A: amt owed on

loan

P: principal

r: interest rate

t: time

% decimal

Amount of Loan

interest = 11% r

= .11

t = 3

A = 4522

P = ?

4. A sum of $ is borrowed at 11% per year & amt required to pay off loan after 3 years is $4522. How much $ is borrowed?

principal

A = P(1 + rt)

4522 = P(1+ (.11)(3))

4522 = P(1.33)

P = $ 3400

Discount Rate

(1 )P A rn A: maturity value

P: proceeds

r: discount rate

n: years discounted

% decimal

Discount Rate

n = 2

P = 889.20

rate = 11% r = .11

A = ?

5. Find the maturity value of a note discounted for 2 years if the proceeds are $889.20 and discount rate is 11%.

P = A(1 – rn)

889.20 = A(1 – (.11)

(2))

889.20 = A(.78)

A = $ 1140

Discount Rate

n = 5

P = 225

A = 1500

r = ?

6. The maturity value of a note discounted for 5 years is $1500 with the proceeds of $225. Find the discount rateP = A(1 – rn)

225 = 1500(1 – (r)(5))

225 = 1500 – 7500r

–1275 = –7500r

0.17 = r r =

17%

7.

8x – 12 = 8x – 8 Or 8x – 12 = – (8x – 8)

– 12 = – 8 8x – 12 = – 8x+8

No solution 16x = 20x =

Absolute Value Equations

7. x =

Check for Extraneous Solutions

8.

2x – 1 = 5 or 2x – 1 = – 5 +1 +1 +1 + 12x = 6 2x = – 4 2 2 2 2

x = 3 or x = – 2 {–2, 3}

Solving an Absolute Value Equation

Ticket Out the Door

Solve for a

Explain your process.

Homework

#202 Pg. 102 #33 – 59 odd, 67, 69