Bellringer

On piece of paper…

1. Put your name

2. Graph the system2x – 2y = 4

y – x = 6

3. State whether the solution is Independent, Dependent, or Inconsistent

3-2 Solving Systems Algebraically

M11.D.2.1.4: Write and/or solve systems of equations using graphing, substitution, and/or

elimination

Objectives

Solving Systems by Substitution

Solving Systems by Elimination

Solve the system by substitution. x + 3y = 12–2x + 4y = 9

Step 1:  Solve for one of the variables. Solving the first equationfor x is the easiest.

x + 3y = 12 x = –3y + 12

Step 2:  Substitute the expression for x into the other equation. Solve for y.

–2x + 4y = 9–2(–3y + 12) + 4y = 9 Substitute for x. 6y – 24 + 4y = 9 Distributive Property 6y + 4y = 33 y = 3.3

Step 3:  Substitute the value of y into either equation. Solve for x.x = –3(3.3) + 12x = 2.1

The solution is (2.1, 3.3).

Solving Systems by Substitution

At Renaldi’s Pizza, a soda and two slices of the pizza–of–the–

day costs $10.25. A soda and four slices of the pizza–of–the–day costs

$18.75. Find the cost of each item.

Relate:  2 • price of a slice of pizza + price of a soda = $10.25

4 • price of a slice of pizza + price of a soda = $18.75

Define: Let p = the price of a slice of pizza.    

Let s = the price of a soda.

Write: 2 p + s = 10.25

4 p + s = 18.75

2p + s = 10.25 Solve for one of the variables.s = 10.25 – 2p  

Real World Example

(continued)

4p + (10.25 – 2p) = 18.75Substitute the expression for s into the other equation. Solve for p.

p = 4.25

2(4.25) + s = 10.25Substitute the value of p into one of the equations. Solve for s.

s = 1.75

The price of a slice of pizza is $4.25, and the price of a soda is $1.75.

Continued

Use the elimination method to solve the system.3x + y = –9–3x – 2y = 12

y = –3

3x + (–3) = –9 Substitute y. Solve for x.

x = –2

The solution is (–2, –3).

3x + y = –9 Choose one of the original equations.

3x + y = –9 –3x – 2y = 12 Two terms are additive inverses, so add. –y = 3 Solve for y.

Solving by Elimination

Solve the system by elimination.2m + 4n = –43m + 5n = –3

To eliminate the n terms, make them additive inverses by multiplying.

m = 4 Solve for m.

2m + 4n = –4 Choose one of the original equations.

2(4) + 4n = –4 Substitute for m.8 + 4n = –4

4n = –12 Solve for n.n = –3

The solution is (4, –3).

–2m = –8 Add.

2m + 4n = –4 10m + 20n = –201 Multiply by 5.1

3m + 5n = –3 –12m – 20n = 122 Multiply by –4.2

Solving an Equivalent System

Solve each system by elimination.

Elimination gives an equation that is always false.

The two equations in the system represent parallel lines.

The system has no solution.

Multiply the first line by 2 to make the x terms additive inverses.

–3x + 5y = 66x – 10y = 0

–6x + 10y = 126x – 10y = 0

0 = 12

a. –3x + 5y = 66x – 10y = 0

Solving a System Without a Unique Solution

(continued)

Multiply the first line by 2 to make the x terms additive inverses.

–3x + 5y = 66x – 10y = –12

–6x + 10y = 12 6x + 10y = –12

0 = 0

b. –3x + 5y = 66x – 10y = –12

Elimination gives an equation that is always true.

The two equations in the system represent the same line.

The system has an infinite number of solutions: {(x, y)| y = x + }35

65

Continued

Homework

Pg 128 & 129 # 1,2,18,19,30,31