4.3

Systems of Linear

Equations and

Problem Solving

Problem-Solving Steps

1. UNDERSTAND the problem. During this step, become

comfortable with the problem. Some ways of doing this

are to

Read and reread the problem.

Choose two variables to represent the two unknowns.

Construct a drawing.

Propose a solution and check Pay careful attention to

check your proposed solution. This will help when writing

equations to model the problem.

Problem Solving Steps

Continued

2. TRANSLATE the problem into two equations.

3. SOLVE the system of equations.

4. INTERPRET the results. Check the proposed

solution in the stated problem and state your

conclusion.

Problem Solving Steps

Example

Continued

One number is 4 more than twice the second number.

Their total is 25. Find the numbers.

Read and reread the problem. Suppose that the second

number is 5. Then the first number, which is 4 more than

twice the second number, would have to be 14 (4 + 2•5).

Is their total 25? No: 14 + 5 = 19. Our proposed solution

is incorrect, but we now have a better understanding of

the problem.

Since we are looking for two numbers, we let

x = first number

y = second number

1. UNDERSTAND

Continued

2. TRANSLATE

Example (cont)

One number is 4 more than twice the second number.

x = 4 + 2y

Their total is 25.

x + y = 25

3. SOLVE

Continued

Example (cont)

Using substitution method, we substitute the solution for x from the first equation into the second equation.

x + y = 25

(4 + 2y) + y = 25 (replace x with result from first equation)

4 + 3y = 25 (simplify left side)

3y = 25 – 4 = 21 (subtract 4 from both sides and simplify)

y = 7 (divide both sides by 3)

Now we substitute the value for y into the first equation.

x = 4 + 2y = 4 + 2(7) = 4 + 14 = 18

We are solving the system x = 4 + 2y and x + y = 25

4. INTERPRET

Example (cont)

Check: Substitute x = 18 and y = 7 into both of the

equations.

First equation,

x = 4 + 2y

18 = 4 + 2(7) true

Second equation,

x + y = 25

18 + 7 = 25 true

State: The two numbers are 18 and 7.

Example

Continued

Hilton University Drama club sold 311 tickets for a play.

Student tickets cost 50 cents each; non student tickets

cost $1.50. If total receipts were $385.50, find how

many tickets of each type were sold.

1. UNDERSTAND

Read and reread the problem. Suppose the number of

students tickets was 200. Since the total number of

tickets sold was 311, the number of non student tickets

would have to be 111 (311 – 200).

Example (cont)

Continued

Are the total receipts $385.50? Admission for the 200

students will be 200($0.50), or $100. Admission for the

111 non students will be 111($1.50) = $166.50. This gives

total receipts of $100 + $166.50 = $266.50. Our proposed

solution is incorrect, but we now have a better

understanding of the problem.

Since we are looking for two numbers, we let

s = the number of student tickets

n = the number of non-student tickets

1. UNDERSTAND (cont)

Continued

2. TRANSLATE

Example (cont)

Hilton University Drama club sold 311 tickets for a play.

s + n = 311

total receipts were $385.50

0.50s

Total

receipts

= 385.50

Admission for

students

1.50n

Admission for

non students

+

3. SOLVE

Continued

Example (cont)

We are solving the system s + n = 311 and 0.50s + 1.50n =

385.50

Since the equations are written in standard form (and we might like to get rid of the decimals anyway), we’ll solve by the addition method. Multiply the second equation by –2.

s + n = 311

2(0.50s + 1.50n) = 2(385.50)

s + n = 311

s – 3n = 771

2n = 460

n = 230

simplifies to

Now we substitute the value for n into the first equation.

s + n = 311 s + 230 = 311 s = 81

4. INTERPRET

Example (cont)

Check: Substitute s = 81 and n = 230 into both of the

equations.

First equation,

s + n = 311

81 + 230 = 311 true

Second equation,

0.50s + 1.50n = 385.50

40.50 + 345 = 385.50 true

0.50(81) + 1.50(230) = 385.50

State: There were 81 student tickets and 230 non

student tickets sold.

Example

Continued

Terry Watkins can row about 10.6 kilometers in 1 hour

downstream and 6.8 kilometers upstream in 1 hour. Find

how fast he can row in still water, and find the speed of the

current.

1. UNDERSTAND

Read and reread the problem. We are going to propose a

solution, but first we need to understand the formulas we will

be using. Although the basic formula is d = r • t (or r • t = d),

we have the effect of the water current in this problem. The

rate when traveling downstream would actually be r + w and

the rate upstream would be r – w, where r is the speed of the

rower in still water, and w is the speed of the water current.

Example

Continued

1. UNDERSTAND (cont)

Suppose Terry can row 9 km/hr in still water, and the water

current is 2 km/hr. Since he rows for 1 hour in each direction,

downstream would be (r + w)t = d or (9 + 2)1 = 11 km

Upstream would be (r – w)t = d or (9 – 2)1 = 7 km

Our proposed solution is incorrect (hey, we were pretty close

for a guess out of the blue), but we now have a better

understanding of the problem.

Since we are looking for two rates, we let

r = the rate of the rower in still water

w = the rate of the water current

Continued

2. TRANSLATE

Example (cont)

rate

downstream

(r + w)

time

downstream

• 1

distance

downstream

= 10.6

rate

upstream

(r – w)

time

upstream

• 1

distance

upstream

= 6.8

3. SOLVE

Continued

Example (cont)

We are solving the system r + w = 10.6 and r – w = 6.8

Since the equations are written in standard form, we’ll solve by the addition method. Simply combine the two equations together.

r + w = 10.6

r – w = 6.8

2r = 17.4

r = 8.7

Now we substitute the value for r into the first equation.

r + w = 10.6 8.7 + w = 10.6 w = 1.9

4. INTERPRET

Example (cont)

Check: Substitute r = 8.7 and w = 1.9 into both of the

equations.

First equation, (r + w)1 = 10.6

(8.7 + 1.9)1 = 10.6 true

Second equation,

(r – w)1 = 1.9

(8.7 – 1.9)1 = 6.8 true

State: Terry’s rate in still water is 8.7 km/hr and the rate

of the water current is 1.9 km/hr.

Example

Continued

A Candy Barrel shop manager mixes M&M’s worth $2.00 per

pound with trail mix worth $1.50 per pound. Find how many

pounds of each she should use to get 50 pounds of a party

mix worth $1.80 per pound.

1. UNDERSTAND

Read and reread the problem. We are going to propose a

solution, but first we need to understand the formulas we

will be using. To find out the cost of any quantity of items

we use the formula

price per unit • number of units = price of all units

Example (cont)

Continued

1. UNDERSTAND (cont)

Suppose the manage decides to mix 20 pounds of M&M’s.

Since the total mixture will be 50 pounds, we need 50 – 20 = 30

pounds of the trail mix. Substituting each portion of the mix into

the formula,

M&M’s $2.00 per lb • 20 lbs = $40.00

Trail mix $1.50 per lb • 30 lbs = $45.00

Mixture $1.80 per lb • 50 lbs = $90.00

Example (cont)

Continued

1. UNDERSTAND (cont)

Since $40.00 + $45.00 $90.00, our proposed solution is

incorrect (hey, we were pretty close again), but we now

have a better understanding of the problem.

Since we are looking for two quantities, we let

x = the amount of M&M’s

y = the amount of trail mix

Continued

2. TRANSLATE

Example (cont)

Fifty pounds of party mix

x + y = 50

price per unit • number of units = price of all units Using

Price of

M&M’s

2x

Price of

trail mix

+ 1.5y

Price of

mixture

= 1.8(50) = 90

3. SOLVE

Continued

Example (cont)

We are solving the system x + y = 50 and 2x + 1.50y = 90

Since the equations are written in standard form (and we might like to get rid of the decimals anyway), we’ll solve by the addition method. Multiply the first equation by 3 and the second equation by –2.

3(x + y) = 3(50)

–2(2x + 1.50y) = –2(90)

3x + 3y = 150

–4x – 3y = – 180

– x = – 30 x = 30

simplifies to

Now we substitute the value for x into the first equation.

x + y = 50 30 + y = 50 y = 20

4. INTERPRET

Example (cont)

Check: Substitute x = 30 and y = 20 into both of the equations.

First equation,

x + y = 50

30 + 20 = 50 true

Second equation,

2x + 1.50y = 90

60 + 30 = 90 true

2(30) + 1.50(20) = 90

State: The store manager needs to mix 30 pounds of M&M’s and

20 pounds of trail mix to get the mixture at $1.80 a pound.

Example

Continued

The measure of the largest angle of a triangle is 90° more

than the measure of the smallest angle, and the measure of

the remaining angle is 30° more than the measure of the

smallest angle. Find the measure of each angle.

1. UNDERSTAND

Read and reread the problem. We are going to propose a solution.

Suppose the measure of the smallest angle is 30º.

Then the largest angle is 30º + 90º = 120º (90º more than the

smallest), and the remaining angle is 30º + 30º = 60º (30º more

than the smallest).

In a triangle, the sum of the measures of the 3 angles is 180º.

Since 30º + 120º + 60º 180º, our proposed solution is incorrect,

but we now have a better understanding of the problem.

Continued

2. TRANSLATE

Example (cont)

If we let

s = the measure of the smallest angle, then

m = the measure of the middle angle, and

l = the measure of the largest angle, then

we are solving the system

s + m + l = 180

l = 90 + s

m = 30 + s

3. SOLVE

Continued

Example (cont)

Substitute the last two equations into the first equation.

s + 30 + s + 90 + s = 180

3s + 120 = 180

3s = 60

s = 20

Substitute this value for s into the 2nd and 3rd equations

from the original set.

l = 90 + s = 90 + 20 = 110

m = 30 + s = 30 + 20 = 50

4. INTERPRET

Example (cont)

Check: Substitute s = 20, m = 50, and l = 110 into all

three equations.

20 + 50 + 110 = 180 true

110 = 90 + 20 true

50 = 30 + 20 true

State: The 3 angles of the triangle are 20º, 50º and 110º.