Download notes 6.2 day 2

Substitution Method
Day 2
Today’s Objective:
I can solve a system using
substitution.
1. Solve for 1 variable
3x + 2y = 15
2. Substitute the variable you just solved for
into the other equation AND solve.
x=y–5
3. Solve for the other variable (substitute)
4. Write the ordered pair
Step 1
Step 2
x=y–5
3x + 2y = 15
3( y – 5 ) + 2y = 15
3y – 15 + 2y = 15
5y – 15 = 15
5y = 30
y =6
Step 3
x=y–5
x=6–5
x=1
Step 4
(1, 6)
1. Solve for 1 variable
2x – 5y = 18
2. Substitute the variable you just solved for
into the other equation AND solve.
2x + y = 6
3. Solve for the other variable (substitute)
4. Write the ordered pair
Step 1
Step 2
2x – 5y = 18
2x – 5( -2x + 6 ) = 18
-2x
-2x
2x + 10x – 30 = 18
y = -2x + 6 12x – 30 = 18
12x = 48
x =4
2x + y = 6
Step 3
y = -2x + 6
y = -2(4) + 6
y = -2
Step 4
(4, -2)
How many solutions can a system have?
x = -2y + 4
3.5x + 7y = 14
3.5x + 7y = 14
3.5 ( -2y + 4 ) + 7y = 14
-7y + 14 + 7y = 14
-7y + 7y + 14 = 14
14 = 14
The system has infinitely many
solutions
y = 3x – 11
y – 3x = -13
y - 3x = -13
( 3x - 11) - 3x = -13
3x – 11 – 3x = -13
3x – 3x – 11 = - 13
-11 = -13
-11 ≠ -13
The system has NO solutions
A snack bar sells two sizes of snack packs. A large snack pack is $5 and a
small snack pack is $3. In one day, the snack bar sold 60 snack packs for a
total of $220. How many small snack packs did the snack bar sell?
Identify variables:
Let L = Large packs
Total # of snack packs:
L
Money earned from
snack packs:
5L + 3S = 220
1. Solve for 1 variable
L = 60 - S
+
Let S= small snack packs
S = 60
2. Substitute the variable you just solved for
into the other equation AND solve.
5L + 3S = 220
5( 60 - S ) + 3S = 220
300 – 5S + 3S = 220
300 – 2S= 220
– 2S = -80
S = 40
They sold
40 small
snack packs
I can solve a system using substitution.
Assignment:
Book pg 375: 18-24, 26, 28,30, 38
18.(4,6)
19.(-12,-5)
20.(5,2)
21.(0, -.5)
22.(1,3)