Transcript Verbal Systems

Aim: How do we solve verbal problems
using two variables?
Do Now:
Jonathan left his home by car,
traveling on a certain road at the rate of 45 mph.
Three hours later, his brother Jessie left the home
and started after him on the same road, traveling
at a rate of 60 mph. In how many hours did Jessie
overtake Jonathan?
Aim: Verbal Systems
Course: Math Literacy
Do Now
D = rt
Jonathan traveled 3 hours longer
45(h + 3)
Jonathan
60h
Jessie
h = the number of hours Jessie traveled.
h + 3 = hours traveled by Jonathan
60h = the distance traveled by Jessie
45(h + 3) = the distance traveled by Jonathan
60h = 45(h + 3)
60h = 45h + 135
15h = 135
h=9
Aim: Verbal Systems
60h = 45(h + 3)
60(9) = 45(9 + 3)
540 = 540
Course: Math Literacy
Using Two Variables
The larger of two numbers is 4 times the
smaller. If the larger number exceeds the
smaller by 15, find the number.
Let x = smaller #
4x = x + 15
3x = 15
Let 4x = larger #
x=5
4x = 20
Use a system of equations to solve the
same problem
Let x = smaller #
Let y = larger #
y = 4x
x = y - 15
Substitution
x = 4x - 15
-3x = -15
y = 4x = 4(5) = 20
x=5
Aim: Verbal Systems
Course: Math Literacy
Model Problem
Use a system of equations to solve
The sum of two numbers is 8.6. Three times
the larger number decreased by twice the
smaller is 6.3. What are the numbers?
Let x = smaller #
Let y = larger #
x + y = 8.6
3y – 2x = 6.3
x + y = 8.6
3y – 2x = 6.3
2(x + y = 8.6)
x + y = 8.6
x + 4.7 = 8.6
x = 3.9
Additive
inverse
2x + 2y = 17.2
-2x + 3y = 6.3
5y = 23.5
y = 4.7
Verbalnumbers
Systems
Course: 3.9
Math Literacy
TheAim:
two
are 4.7 and
Do Now – Two Variables
Jonathan left his home by car, traveling on a
certain road at the rate of 45 mph. Three hours
later, his brother Jessie left the home and started
after him on the same road, traveling at a rate of
60 mph. In how many hours did Jessie overtake
Jonathan?
h = the number of hours Jessie traveled.
t = hours traveled by Jonathan
60h = the distance traveled by Jessie
45t = the distance traveled by Jonathan
60h = 45t
h+ 3=t
60h = 45(h + 3)
15h = 135
h=9
Aim: Verbal Systems
Course: Math Literacy
Aim: How do we solve verbal problems
using two variables?
Do Now:
Mario had $6.50, consisting of dimes
and quarters, in a coin bank. The number of
quarters was 10 less than twice the number of
dimes. How many coins of each kind did he have?
Aim: Verbal Systems
Course: Math Literacy
Use a system of equations to solve the
same problem.
Let d = # of dimes
.10d = value of
dimes
.10d + .25q = 6.50
q = 2d - 10
q = # of quarters
.25q = value of
quarters
10d + 25q = 650
Substitution
10d + 25(2d – 10) = 650
10d + 50d – 250 = 650
60d – 250 = 650
60d = 900
d = 15
q = 2d – 10 = 2(15) – 10 = 20
15 dimes = $1.50 20 quarters = $5.00
1.50 + 5.00 = $6.50Course: Math Literacy
Aim: Verbal Systems
Model Problem
The owner of a men’s clothing store bought six belts
and eight hats for $140. A week later, at the same
prices, he bought nine belts and six hats for $132.
Find the price of a belt and the price of a hat.
Let b = belt
Let h = hat
6b + 8h = 140
Additive inverse
- eliminate h
9b + 6h = 132
6b + 8h = 140
9b + 6h = 132
6(6) + 8h = 140
36 + 8h = 140
8h = 104
h = 13
3(6b + 8h = 140)
-4(9b + 6h = 132)
18b + 24h = 420
-36b - 24h = -528
= -108
b=6
The belts costs $6 ea. And the hats cost $13 ea.
Aim: Verbal Systems
-18b
Course: Math Literacy
Use a system of equations to solve
A dealer wishes to obtain 50 pounds of mixed cookies to
sell for $3.00 per pound. If he mixes cookies worth $3.60
per pound with cookies worth $2.10 per pound, find the
number of pounds of each kind he should use.
Let x = #lb. of
Let y = #lb. of
cookie 1
cookie 2
Value of cookie #1
Value of cookie #2
3.60x
2.10y
x + y = 50
Substitution
3.60x + 2.10y = 150
x = 50 - y
360x + 210y = 15000
360(50 – y) + 210y = 15000
18000 – 360y + 210y = 15000
– 150y = -3000
y = 20 lb. of cookie 1 - $2.10
x + y = 50  xAim:+ Verbal
20 Systems
= 50  x = 30 lb.Course:
of Math Literacy
cookie 2 - $3.60
Use a system of equations to solve
A motor boat can travel 60 miles downstream in 3
hours. It requires 5 hours to make the return trip
against the current. Find the rate of the boat in still
water and the rate of the current.
Let r = boat’s rate in
still water
Let c = current’s
rate
r+c=
boat’s rate going downstream
20mph
r-c=
boat’s rate going upstream 12mph
r + c = 20
r - c = 12
Additive
inverse eliminate c
2r = 32
r = 16mph rate of boat in still
water
Course: Math Literacy
r + c = 20  Aim:
16Verbal
+ cSystems
= 20  c = 4mph
rate of current