Transcript x –y - howesmath

Applications of Linear
Equations in Two Variables
When solving an application that involves two unknowns,
sometimes it is convenient to use a system of linear equations in
two variables.
Solution:
In this application we have two unknowns,
which we can represent by x and y.
Let x represent the cost of one large popcorn.
Let y represent the cost of one drink.
We must now write two equations. Each of the first
two sentences in the problem gives a relationship
between x and y
(
(
Cost of 1 ) + (cost of 2) = (total)
large popcorn
drinks
cost
x + 2y = 5.75
Cost of 2
) + (cost of 2) = (total)
large popcorns
drinks
cost
2x + 5y = 13.00
x + 2y = 5.75
2x + 5y = 13.00
x + 2y = 5.75
x = -2y + 5.75
2 x + 5 y = 13.00
2(-2y + 5.75) + 5y = 13.00
-4y + 11.50 + 5y = 13.00
y + 11.50 = 13.00
y = 1.50
Isolate x in the
first equation.
Substitute x = -2y + 5.75
into the second equation.
Solve for y.
x = -2y + 5.75 First equation after solving for x
Substitute y = 1.50 into this equation.
x = -2(1.50) + 5.75
x = -3.00 + 5.75
x = 2.75
The cost of one large popcorn is $2.75 and the cost of one drink is
$1.50
Check by verifying that the solutions meet the specified conditions
1 popcorn + 2 drinks = 1($2.75) + 2 ($1.50) = $5.75
2 popcorn + 5 drinks = 2($2.75) + 5($1.50) = $13.00
TRUE
TRUE
I = Prt
where P is the principal.
r is the annual interest rate, and
t is the time in years
Example 2
Joanne has a total of $6000 to deposit in two accounts.
One account earns 3.5% simple interest and the other
earns 2.5% simple interest. If the total amount of interest
at the end of 1 year is $195, find the amount she
deposited in each account.
Solution:
Let x represent the principal deposited in the 2.5%
Let y represent the principal deposited in the 3.5%
Principal
Interest (I=Pr)
2.5%Account
3.5% account
Total
x
y
6000
0.025x
0.03y
195
(Principal) + (principal) = (total
)
invested
invested
principal
at 2.5%
at 3.5%
( Interest) + (interest) = (total )
earned
earned
interest
at 2.5%
at 3.5%
x + y = 6000
0.025x + 0.035y = 195
We will choose the addition method to solve the system of
equations. First multiply the second equation by 1000 to
clear decimals.
Multiply by -25
x + y = 6000
0.025x + 0.035y = 195
Multiply by 1000
10y = 45,000
-25x
x +- y25y
= 6000
= -150,000
25x++35y
35y==195,000
195,000
25x
10y = 45,000
After eliminating the x-variable, solve for y.
10y = 45,000
10
10
y = 4500 The amount invested in the 3.5% account is $4500
x + y = 6000
Substitute y = 4500 into the equation x + y =6000
x + 4500 = 6000
x = 1500
The amount invested in the 2.5% account is $1500.
To check the solution, verify that the conditions of the problem
have been met.
1. The sum of $1500 and $4500 is $6000 as desired.
TRUE
2. The interest earned on $1500 at 2.5% is: 0.025($1500) = $37.50
The interest earned on $4500 at 3.5% is: 0.035($4500) = $157.50
$195.00
TRUE
Example 3
Using a System of linear Equations in a Mixture Application
A 10% Alcohol solution is mixed with a 40% alcohol solution
to produce 30L of a 20% alcohol solution. Find the number
of liters of 10% solution and the number of liters of 40%
solution required for this mixture.
Solution:
Each solution contains a percentage of alcohol plus some
other mixing agent such as water. Before we set up a
system of equations to model this situation, it is helpful
to have background understanding of the problem.
10%
x liters of
solution
y liters of
+ solution
0.10x L
of pure
alcohol
20%
40%
30 liters
= of
solution
0.40y L of
pure alcohol
.20(30)L of
pure alcohol
Number of liters
of solution
Number of liters
of pure alcohol
10% Alcohol
40% Alcohol
20% Alcohol
x
y
30
0.10x
0.40y
0.20(30) = 6
From the first row, we have
( Amount of ) + ( amount of ) = ( total amount )
10% solution
40% solution
of 20% solution
x + y = 30
From the second row, we have
( Amount of ) + ( amount of ) = (total amount of)
alcohol in
alcohol in
alcohol in
10% solution
40% solution
20% solution
0.10x + 0.40y = 6
We will solve the system with the addition method by first clearing decimals
x + y = 30
0.10xx + 0.40y
4y = 60
=6
Multiply by -1
Multiply by 10
-x –y = -30
x +4 y = 60
3y = 30
3y = 30
After eliminating the x – variable, solve for y
y = 10
10 L of 40% solution is needed
Substitute y = 10 into either of the original equations
x +x(10)
+ y ==30
30
x = 20
20 L of 10% solution is needed
10 L of 40% solution must be mixed with 20 L of 10% solution
The following formula relates the distance traveled to the rate and time of travel.
d = rt
distance = rate x time
Example 4
A plane travels with a tail wind from Kansas City, Missouri, to
Denver, Colorado, a distance of 600 miles in 2 hours. The return trip
against a head wind takes 3 hours. Find the speed of the plane in
still air, and find the speed of the wind.
Solution
Let p represent the speed of the plane in still air.
Let w represent the speed of the wind.
Distance
Rate
Time
With a tail
wind
600
p+w
2
Against a
head wind
600
p-w
3
To set up two equation in p and w, recall that d = rt
From the first row, we have
(Distance
) = (rate with )(time traveled )
with the wind
the wind with the wind
600 = (p + w)2
From the second row, we have
(Distance
) = (rate against)(time traveled
)
against the wind
the wind
against the wind
600 = (p – w)3
Using the distributive property to clear parentheses, produces
the following system
Multiply by 3
2p + 2w = 600
6p + 6w = 1800
3p - 3w = 600
6p -6w = 1200
Multiply by 2
12p
= 3000
12p = 3000
12p = 3000
The speed of the
12
12
plane in still air
p = 250
is 250 mph.
Substitute p = 250 into the
first equation
2(250) + 2w = 600
500 + 2w = 600
2w = 100
w = 50
The speed of the plane in still air is 250 mph. The speed of the wind is 50 mph.