(6.6)Classic Applications of Systems

of Equations

2

Solving Word Problems Using Systems of Equations

• Define variables

• Make a table

• Write the linear system and solve using the most appropriate method

MUST practice to become proficient!!

Solving Word Problems Using Systems of Equations (Example 1)

Ex 1: Read pg. 292 Example 1

Define: x = the number of ounces of 15% acid solutiony = the number of ounces of 40% acid solution

How many ounces of a 15% acid solution should be mixed with a 40% acid solution to produce 60 ounces of a 25% acid solution?

% acid Amount of

solution

Amount of acid

First Solution 15% x 0.15x

Second Solution

40% y 0.40y

Final Solution 25% 60 0.25(60)

How many ounces of a 15% acid solution should be mixed with a 40% acid solution to produce 60 ounces of a 25% acid solution?

% acid Amount of solution

Amount of acid

First Solution 15% x 0.15x

Second Solution 40% y 0.40y

Final Solution 25% 60 0.25(60)

x + y = 60 0.15x + 0.40y = 0.25(60)

x + y = 600.15x + 0.40y = 0.25(60)

x + y = 600.15x + 0.40y = 0.25(60)

x = 36; y = 24

To get the required 60 ounces of the 25% acid solution, mix 36 ounces of the 15% acid solution with 24 ounces of the 40% acid solution.

Solving Word Problems Using Systems of Equations (Practice 1)

Practice 1:

Define: x = the number of ounces of 20% acid solutiony = the number of ounces of 70% acid solution

How many ounces of a 20% acid solution should be mixed with a 70% acid solution to produce 50 ounces of a 40% acid solution?

% acid Amount of

solution

Amount of acid

First Solution 20% x 0.20x

Second Solution

70% y 0.70y

Final Solution 40% 50 0.40(50)

How many ounces of a 20% acid solution should be mixed with a 70% acid solution to produce 50 ounces of a 40% acid solution?

% acid Amount of solution

Amount of acid

First Solution 20% x 0.20x

Second Solution 70% y 0.70y

Final Solution 40% 50 0.40(50)

x + y = 50 0.20x + 0.70y = 0.40(50)

x + y = 500.20x + 0.70y = 0.40(50)

x + y = 600.15x + 0.40y = 0.25(60)

x = 36; y = 24

To get the required 60 ounces of the 25% acid solution, mix 36 ounces of the 15% acid solution with 24 ounces of the 40% acid solution.

Solving Word Problems Using Systems of Equations (Example 2)

Ex 2: Read pg. 293 Example 2

Define: d = number of dimesq = number of quarters

In a coin bank, there are 250 dimes and quarters worth a total of $39.25. Find how many of each kind of coin are in the bank.

Number Coin Value

Value in cents

Quarters q .25 .25q

Dimes d .10 .10d

Total 250 39.25

In a coin bank, there are 250 dimes and quarters worth a total of $39.25. Find how many of each kind of coin are in the bank.

Number Coin Value

Value in cents

Quarters q .25 .25q

Dimes d .10 .10d

Total 250 39.25

q + d = 250 0.25q + 0.10d = 39.25

q + d = 2500.25q + 0.10d = 39.25

q + d = 2500.25q + 0.10d = 39.25

q = 95; d = 155

The coin bank has 95 quarters and 155 dimes.

Solving Word Problems Using Systems of Equations (Practice 2)

Practice 2:

Define: n = number of nickelsd = number of dimes

In a coin bank, there are 125 nickels and dimes worth a total of $10.50. Find how many of each kind of coin are in the bank.

Number Coin Value

Value in cents

Nickels n .05 .05n

Dimes d .10 .10d

Total 125 10.50

In a coin bank, there are 125 nickels and dimes worth a total of $10.50. Find how many of each kind of coin are in the bank.

Number Coin Value

Value in cents

Quarters n .05 .05n

Dimes d .10 .10d

Total 125 10.50

n + d = 125 0.05n + 0.10d = 10.50

n + d = 1250.05n + 0.10d = 10.50

n = 40; d = 85

The coin bank has 40 nickels and 85 dimes.

n + d = 1250.05n + 0.10d = 10.50

Solving Word Problems Using Systems of Equations (Example 3)

Ex 3: Read pg. 294 Example 3

Define: u = units digit of the original numbert = tens digit of the original number

The sum of the digits of a two-digit number is 7. The original two-digit number is 3 less than 4 times the number with its digits reversed. Find the original two-digit number.

The sum of the digits of a two-digit number is 7. The original two-digit number is 3 less than 4 times the number with its digits reversed. Find the original two-digit number.

The two-digit number : 10t + u

The number with its digits reversed: 10u + t

t + u = 7sum of digits is 7

10t + u = 4(10u + t) - 3

original two-digit number is 3 less than 4 times the number with its digits reversed

t + u = 7

10t + u = 4(10u + t) - 3

t = 6; u = 1

The number 61 is a two-digit number which is 3 less than 4 times itself with its digits reversed.

Solving Word Problems Using Systems of Equations (Practice 3)

Practice 3:

Define: u = units digit of the original numbert = tens digit of the original number

The sum of the digits of a two-digit number is 6. The original two-digit number is 6 less than twice the number with its digits reversed. Find the original two-digit number.

The sum of the digits of a two-digit number is 6. The original two-digit number is 6 less than twice the number with its digits reversed. Find the original two-digit number.

The two-digit number : 10t + u

The number with its digits reversed: 10u + t

t + u = 6sum of digits is 6

10t + u = 2(10u + t) - 6

original two-digit number is 6 less than twice the number with its digits reversed

t + u = 6

10t + u = 2(10u + t) - 6

t = 4; u = 2

The number 42 is a two-digit number which is 6 less than twice itself with its digits reversed.

Solving Word Problems Using Systems of Equations (Example 4)

Example 4: Read pg. 294 Example 4

Define: x = rate of plane without windy = rate of wind

A plane leaves New York City and heads for Chicago, which is 750 miles away. The plane, flying against the wind, takes 2.5 hours to reach Chicago. After refueling the plane returns to New York City, traveling with the wind, in 2 hours. Find the rate of the wind and the rate of the plane with no wind.

Rate Time Distance

Rate against wind

x – y 2.5 750

Rate with wind x + y 2 750

A plane leaves New York City and heads for Chicago, which is 750 miles away. The plane, flying against the wind, takes 2.5 hours to reach Chicago. After refueling the plane returns to New York City, traveling with the wind, in 2 hours. Find the rate of the wind and the rate of the plane with no wind.

(x – y)(2.5) = 750

(x + y)(2) = 750

(x – y)(2.5) = 750

(x + y)(2) = 750

Rate Time Distance

Rate against wind

x – y 2.5 750

Rate with wind x + y 2 750

x = 337.5; y = 37.5

The rate of the plane is 337.5 miles per hour and the rate of the wind is 37.5 miles per hour.

Solving Word Problems Using Systems of Equations (Practice 4)

Practice 4:

Define: x = rate of boat without currenty = rate of current

A boat went upstream (against the current) a distance of 90 miles in 4.5 hours. The boat went the same distance downstream (with the current) in 3 hours. Find the rate of the current and the rate of the boat with no current.

Rate Time Distance

Rate against current

x – y 4.5 90

Rate with current

x + y 3 90

A boat went upstream (against the current) a distance of 90 miles in 4.5 hours. The boat went the same distance downstream (with the current) in 3 hours. Find the rate of the current and the rate of the boat with no current.

(x – y)(4.5) = 90

(x + y)(3) = 90

(x – y)(4.5) = 90

(x + y)(3) = 90x = 25; y = 5

The rate of the boat is 25 miles per hour and the rate of the current is 5 miles per hour.

Rate Time Distance

Rate against current

x – y 4.5 90

Rate with current

x + y 3 90