1. Friday Warmup
Homework Check: Document Camera

2. SWBAT solve systems of equations by substitution.
Math 1B
Unit 0 Day 7
Objective:
SWBAT solve systems of equations by substitution.

3. EX 1: Solve using substitution. y = 2x + 2
Step 1: Write an equation containing only one variable and solve.
y = 2x + 2 Start with one equation.
–3x + 4 = 2x + 2 Substitute –3x + 4 for y in that equation.
4 = 5x + 2 Add 3x to each side.
2 = 5x Subtract 2 from each side.
0.4 = x Divide each side by 5.
Step 2: Solve for the other variable.
y = 2(0.4) + 2 Substitute 0.4 for x in either equation.
y = Simplify.
y = 2.8

4. Since x = 0.4 and y = 2.8, the solution is (0.4, 2.8).
EX1: (continued)
Since x = 0.4 and y = 2.8, the solution is (0.4, 2.8).
Check: See if (0.4, 2.8) satisfies y = –3x + 4 since y = 2x + 2 was used in Step 2.
–3(0.4) + 4 Substitute (0.4, 2.8) for (x, y) in the equation. –
2.8 = 2.8

5. REMEMBER>>>>>>3 TYPES OF SOLUTIONS
POINT x = number, y = number One Solution (x,y)
SAME LINE True Statement EX: 3 = 3 or 5 = Infinite pairs of Solutions  ∞
PARALLEL False Statement EX: 3 = 7 or 5 = 9 Zero Solutions  NO SOLUTION

6. Ex 2: Solve using substitution.
–2x + y = –1
4x + 2y = 12
Step 1: Solve the first equation for y because it has a coefficient of 1.
–2x + y = –1
y = 2x –1 Add 2x to each side.
Step 2: Write an equation containing only one variable and solve.
4x + 2y = 12 Start with the other equation.
4x + 2(2x –1) = 12 Substitute 2x –1 for y in that equation.
4x + 4x –2 = 12 Use the Distributive Property.
8x = 14 Combine like terms and add 2 to each side.
x = 1.75 Divide each side by 8.

7. Ex 2: (continued)
Step 3: Solve for y in the other equation.
–2(1.75) + y = -1 Substitute 1.75 for x.
–3.5 + y = –1 Simplify.
y = 2.5 Add 3.5 to each side.
Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5).

8. Ex 3: A youth group with 26 members is going to the beach
Ex 3: A youth group with 26 members is going to the beach. There will also be five chaperones that will each drive a van or a car. Each van seats 7 persons, including the driver. Each car seats 5 persons, including the driver. How many vans and cars will be needed?
Let v = number of vans and c = number of cars.
Drivers v c = 5
Persons v c = 31
Solve using substitution.
Step 1: Write an equation containing only one variable.
v + c = 5 Solve the first equation for c.
c = –v + 5

9. Ex 3: (continued)
Step 2: Write and solve an equation containing the variable v.
7v + 5c = 31
7v + 5(–v + 5) = 31 Substitute –v + 5 for c in the second equation.
7v – 5v + 25 = 31 Solve for v.
2v + 25 = 31
2v = 6
v = 3
Step 3: Solve for c in either equation.
3 + c = 5 Substitute 3 for v in the first equation.
c = 2

10. Ex 3: (continued)
Three vans and two cars are needed to transport 31 persons.
Check: Is the answer reasonable? Three vans each transporting 7 persons is 3(7), of 21 persons. Two cars each transporting 5 persons is 2(5), or 10 persons. The total number of persons transported by vans and cars is , or 31. The answer is correct.

11. Activity:
TBD
TBD

12. Systems of Equations by elimination or substitution
Homework:
Systems of Equations by elimination or substitution
WS must do 6 of 12…student choice using any of the three methods (must use at least 2 methods)