Title: Lesson 1–1
Author: Teri Willard
Last saved by: MHE

Lesson 3–5

Example 1  One Solution
Solve the system of equations.
a + 2b – 4c = 8
2a – b + c = –8
–a – 3b + 2c = –9

Step 1	Use elimination to make a system of two equations in two variables.
a + 2b – 4c = 8 | Multiply by 2. | 2a + 4b – 8c = 16 | First equation
2a – b + c = –8 | | (–) 2a –    b +  c = –8 | Second equation
| | 5b – 9c = 24 | Subtract to eliminate a.

a + 2b – 4c =   8 | First equation
(+) –a – 3b + 2c = –9 | Third equation
–b – 2c = –1 | Add to eliminate a.

Notice that the a terms in each equation have been eliminated. The result is two equations with the same two variables b and c.

Step 2	Solve the system of two equations.

5b – 9c = 24 | | 5b –   9c = 24 |
–b – 2c = –1 | Multiply by 5. | (+) –5b – 10c = –5 |
| | – 19c = 19 | Add to eliminate b.
| | c = –1 | Divide each side by –19.

Use one of the equations with two variables to solve for b.
–b – 2c = –1 | Equation with two variables
–b – 2(–1) = –1 | Replace c with –1.
–b + 2 = –1 | Multiply.
b = 3 | Simplify.

The result is b = 3 and c = – 1.

Step 3	Solve for a using one of the original equations with three variables.
a + 2b – 4c = 8 | Original equation with three variables
a + 2(3) – 4(–1) = 8 | Replace b with 3 and c with –1.
a + 6 + 4 = 8 | Multiply.
a = –2 | Simplify.

The solution is (–2, 3, –1). Check this solution in the other two original equations.

Example 2  Infinite Solutions
Solve the system of equations.
2x + 4y – 2z = 10
–3x – 6y + 3z = –15
4x – 4y + 8z = 20

Eliminate x in the first two equations.
2x + 4y – 2z = 10 | Multiply by 3. | 6x + 12y – 6z =  30 |
–3x – 6y + 3z = –15 | Multiply by 2. | (+) –6x – 12y + 6z = –30 | Add the equations.
| | 0 =     0 |

The equation 0 = 0 is always true. This indicates that the first two equations represent the same plane. Check to see if this plane intersects the third plane.

2x + 4y – 2z = 10 | Multiply by 2. | 4x + 8y –   4z = 20 |
4x – 4y + 8z = 20 | | (–) 4x – 4y +   8z = 20 | Subtract the equations.
| | 12y – 12z =   0 |
| | y –     z =   0 | Divide each side by the GCF, 12

A solution does exist, and the planes intersect in the line y – z = 0. So, there are an infinite number of solutions.

Example 3  No Solution
Solve the system of equations.
x + y – z = 3
–2x – 2y + 2z = –14
3x – 4y + 5z = 8

Eliminate x in the first two equations.
x + y – z = 3 | Multiply by 2. | 2x + 2y – 2z =     6 |
–2x – 2y + 2z = –14 | | (+) –2x – 2y + 2z = –14 | Add the equations.
| | 0 =   –8 |

The equation 0 = –8 is never true. So, there is no solution of this system.

Example 4  Write and Solve a System of Equations
ENTERTAINMENT  Scott is in charge of purchasing a variety of tickets to a baseball game to be given to employees who met or exceeded their sales quotas for the past month. There are 3 types of tickets available for the game at a cost of $25 a piece, $35 a piece, or $50 a piece. He knows that he needs to buy 40 tickets in all and can spend $1670. He wants to buy twice as many $35 tickets as $25 tickets. How many of each type of ticket should he buy?

Explore	Read the problem and define the variables.
x = the number of $25 tickets
y = the number of $35 tickets
z = the number of $50 tickets

Plan	Scott has $1670 to spend on the tickets.
25x  + 35y + 50z = 1670

The number of tickets should be 40.
x + y + z = 40

He wants to buy twice as many $35 tickets as $25 tickets.
2x = y

Solve	Substitute 2x for y in each of the first two equations.
| 25x  + 35y + 50z = 1670 | First equation
| 25x  + 35(2x) + 50z = 1670 | Replace y with 2x.
| 25x + 70x + 50z = 1670 | Multiply.
| 95x + 50z = 1670 | Simplify.

| x + y + z = 40 | Second equation
| x + (2x) + z = 40 | Replace y with 2x.
| 3x + z = 40 | Simplify.

Now solve the system of two equations in two variables.
| 95x + 50z = 1670 | | 95x + 50z = 1670
| 3x + z = 40 | Multiply by 50. | (–) 150x + 50z = 2000
| | | –55x           = –330
| | | x = 6

Substitute 6 for x in the equation 2x = y.
| 2(6) = y | x = 6
| 12 = y | Multiply.

Substitute 6 for x and 12 for y in one of the original equations.
| x + y + z = 40 | Second equation
| 6 + 12 + z = 40 | x = 6, y = 12
| 18 + z = 40 | Add.
| z = 22 | Subtract 18 from each side.

So, Scott should buy 6 $25–tickets, 12 $35–tickets, and 22 $50 tickets.

Check	Check to see if all the criteria are met.
The total number of tickets is 40
6 + 12 + 22 = 40  ✓

The total amount of money spent is $1670.
6(25) + 12(35) + 22(50)  1670
           150 + 420 + 1100 = 1670  ✓

The number of $35 tickets is twice the number of $25 tickets.
12  2(6)  
12 = 12  ✓