Warm–up #2
1. Solve
3−2𝑥
𝑥+9
=0
2. Solve 𝑥 + 5 = 3𝑥 − 2
3. Solve for R:
A=𝜋 𝑟+𝑅 𝑆
Warm–up #2 Solutions
1. Solve
3−2𝑥
𝑥+9
=0
For a fraction to = 0, ONLY the numerator
0
needs to = 0 because = 0
#
3 − 2𝑥 = 0
3 – 2x = 0
– 2x = –3
x=
3
2
Warm–up #2 Solutions
2. Solve 𝑥 + 5 = 3𝑥 − 2
x + 5 = 3x – 2
or x + 5 = –(3x – 2)
–2x = –7
or x + 5 = –3x + 2
4x = –3
x=
7
2
or
x=
3
−
4
Check for extraneous solutions!
Nope!
Warm–up #2 Solutions
3. Solve for R:
A = (𝜋𝑟 + 𝜋𝑅)S
A = 𝜋rS + 𝜋RS
A – 𝜋rS = 𝜋RS
𝐴−𝜋𝑟𝑆
𝜋𝑆
𝐴
Or
𝜋𝑆
=𝑅
−𝑟 =𝑅
A=𝜋 𝑟+𝑅 𝑆
𝐴
=𝑟+𝑅
𝜋𝑆
𝐴
𝜋𝑆
−𝑟 =𝑅
Homework Log
Fri
10/2
Lesson
2–2
Learning Objective:
To solve number & investment
problems
Hw: #203 Pg. 110 # 1–6, 9–14
10/2/15 Lesson 2 – 2 Number &
Investment Problems Day 1
Advanced Math/Trig
Learning Objective
 To
solve number problems
 To
solve investment problems
Consecutive Integers
+1+1
Consecutive Integers:
Use: x, x + 1, x + 2, …
x x+1 x+2
+2
Consecutive Odd Integers:
Use: x, x + 2, x + 4, …
x
+2
x+2
x+4
+2
+2
Consecutive Even Integers:
Use: x, x + 2, x + 4, …
6
x
x+2
x+4
Consecutive Integers
1. Find 2 consecutive positive even integers
such that the difference of their squares is 52.
1st # = x
2nd # = x+ 2
(2nd #)2 – (1st #)2 = 52
(x + 2)2 – (x)2 = 52
x2 + 4x + 4 – x2 = 52
4x + 4 = 52
4x = 48
x = 12
12 & 14
Two-digit Numbers
Two Digit # 
23
tens digit
units digit
2
3
20
+
3
to get 23!!
Three-digit Numbers
Three Digit # 
215
hundreds digit tens digit
2
1
200 + 10 +
to get 215!
units digit
5
5
2. The sum of the digits in a 2-digit number is 5. If
the digits are reversed, the new number is 9 more than
the original number. What is the original number?
x 5 x
Original number:
10x + (5 – x)
5 x x
10x + (5 – x) + 9 = 10(5 – x) + x
New number: 10(5 – x) + x
10x + 5 – x + 9 = 50 – 10x + x
9x + 14 = –9x + 50
18x = 36
Orig # = 23
x=2
x=2
5–x=3
Simple Interest
I  prt
I = interest earned
p = principal
r = rate %  decimal
t = time (in years)
Simple Interest
3. Part of $9000 is to be invested at 12% and the
remainder at 8%. How much should be invested at
each rate to yield an annual interest income of $860.
Principal
Inv 1
Inv 2
Total
x
9000 – x
9000
•
rate
.12
.08
•
time
1
1
.12x + .08(9000 – x) = 860
=
Interest
.12x
.08(9000 – x)
860
equation!
Simple Interest #3 cont’d
.12x + .08(9000 – x) = 860
.12x + 720 – .08x = 860
.04x = 140
x = 3500
9000 – 3500 = 5500
$3500 at 12%
$5500 at 8%
Consecutive Integers
4. Find 3 consecutive odd integers whose sum is
111.
1st # = x
2nd # = x + 2
3rd # = x + 4
x + (x + 2) + (x + 4) = 111
3x + 6 = 111
3x = 105
x = 35
35, 37, 39
Digits of Numbers
5. In a three-digit number, the hundreds digit is
twice the tens digit and half of the units digits.
If the numbers are reversed, the new number is
32 less than twice the original number. Find the
original number.
Original #  100(2x) + 10(x) + 1(4x)
 214x
New # 
100(4x) + 10(x) + 1(2x)

412x
Digits of Numbers
5.
2(214x) – 32 = 412x
428x – 32 = 412x
– 32 = –16x
x=2
Original #  100(2x) + 10(x) + 1(4x)
 214x
 428
Simple Interest
6. Part of $10,000 is to be invested at 15% and the
remainder at 9%. How much should be invested at
each rate to yield an annual interest income of $1320.
Principal
Inv 1
Inv 2
Total
x
10000 – x
10000
•
rate
.15
.09
•
time
=
1
1
.15x + .09(10000 – x) = 1320
Interest
.15x
.09(10000 – x)
1320
equation!
Simple Interest #6 cont’d
.15x + .09(10000 – x) = 1320
.15x + 900 – .09x = 1320
.06x = 420
x = 7000
10000 – 7000 = 3000
$7000 at 15%
$3000 at 9%
Simple Interest
7. Jane has four times as much money invested in
bonds paying 12% as she does in stocks paying 10%. If
her total annual interest income from these
investments is $696, how much does she have invested
in each?
Principal
Bonds
Stocks
4x
x
•
rate
.12
.10
Total
.48x + .1x = 696
•
time
1
1
=
Interest
.48x
.1x
696
equation!
Simple Interest #7 cont’d
.48x + .1x = 696
.58x = 696
x = 1200
4(1200) = 4800
$1200 in stocks
$4800 in bonds
Ticket Out the Door

In a two-digit number, the
units digit is 3 less than twice
the tens digit. If the digits are
reversed, the new number is
18 more than the original
number. Find the original
number.
Homework
#203 Pg. 110 #1 – 6, 9 – 14