Systems of Linear

Equations and

Problem Solving

Steps in Solving Problems1) Understand the problem.

• Read and reread the problem• Choose a variable to represent the

unknown• Construct a drawing, whenever possible• Propose a solution and check

2) Translate the problem into two equations.3) Solve the system of equations.4) Interpret the results.

• Check proposed solution in the problem• State your conclusion

Problem Solving Steps

Finding an Unknown NumberExampl

eOne 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 numbery = second number

1.) Understand

2.) Translate

Example continued

Finding an Unknown Number

One number is 4 more than twice the second number.

x = 4 + 2y

Their total is 25.

x + y = 25

Example continued3.) Solve

Finding an Unknown Number

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

Example continued4.)

Interpret

Finding an Unknown Number

Check: Substitute x = 18 and y = 7 into both of the equations.First equation, x = 4 + 2y

18 = 4 + 2(7) trueSecond equation,

x + y = 2518 + 7 = 25 true

State: The two numbers are 18 and 7.

Solving a Problem

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).

Solving a Problem

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 ticketsn = the number of non-student tickets

Example continued1.) Understand (continued)

2.) TranslateExample

continued

Solving a Problem

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

Solving a Problem

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 = 3112(0.50s + 1.50n) =

2(385.50)

s + n = 311s – 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

Solving a Problem

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.

Solving a Rate Problem

ExampleTerry 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.) UnderstandRead 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.

Solving a Rate Problem

Example

Continued

1.) Understand (continued)

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 kmUpstream would be (r – w)t = d or (9 – 2)1 = 7 kmOur 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 waterw = the rate of the water current

2.) Translate

Solving a Rate Problem

rate downstream

(r + w)

time downstream

• 1

distance downstream

= 10.6

rate upstream

(r – w)

time upstream

• 1

distance upstream

= 6.8

3.) Solve

Solving a Rate Problem

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 = 8.7Now we substitute the value for r into the first equation.

r + w = 10.6

8.7 + w = 10.6

w = 1.9

4.) Interpret

Solving a Rate Problem

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.

Solving a Mixture Problem

ExampleA 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.) UnderstandRead 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

Solving a Mixture Problem

Continued

1.) Understand (continued)

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

Solving a Mixture Problem

1.) Understand (continued)

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’sy = the amount of trail mix

2.) Translate

Solving a Mixture Problem

Fifty pounds of party mix

x + y = 50

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

Price of M&M’s

2x

Price of trail mix

+ 1.5y

Price of mixture

= 1.8(50) = 90

3.) Solve

Continued

Solving a Mixture Problem

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

Solving a Mixture Problem

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.