Questions over hw? Elimination Practice.

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Presentation transcript:

Questions over hw? Elimination Practice

CCGPS Coordinate Algebra Day 19 (9-7-12) UNIT QUESTION: How do I justify and solve the solution to a system of equations or inequalities? Standard: MCC9-12.A.REI.1, 3, 5, 6, and 12 Today’s Question: When is it better to use substitution than elimination for solving systems? Standard: MCC9-12.A.REI.6

Solve Systems of Equations by Graphing

Steps Make sure each equation is in slope-intercept form: y = mx + b. Graph each equation on the same graph paper. The point where the lines intersect is the solution. If they don’t intersect then there’s no solution. Check your solution algebraically.

1. Graph to find the solution. y = 3x – 12 y = -2x + 3 Solution: (3, -3)

2. Graph to find the solution.

3. Graph to find the solution. No Solution

4. Graph to find the solution.

5. Graph to find the solution.

Types of Systems There are 3 different types of systems of linear equations 3 Different Systems: Consistent-independent Inconsistent Consistent-dependent

Type 1: Consistent-independent A system of linear equations having exactly one solution is described as being consistent-independent. y  The system has exactly one solution at the point of intersection x

Type 2: Inconsistent A system of linear equations having no solutions is described as being inconsistent. y  The system has no solution, the lines are parallel x Remember, parallel lines have the same slope

Type 3: Consistent-dependent A system of linear equations having an infinite number of solutions is described as being consistent-dependent. y  The system has infinite solutions, the lines are identical x

So basically…. If the lines have the same y-intercept b, and the same slope m, then the system is consistent-dependent If the lines have the same slope m, but different y-intercepts b, the system is inconsistent If the lines have different slopes m, the system is consistent-independent

Solve Systems of Equations by Substitution

Steps One equation will have either x or y by itself, or can be solved for x or y easily. Substitute the expression from Step 1 into the other equation and solve for the other variable. Substitute the value from Step 2 into the equation from Step 1 and solve. Your solution is the ordered pair formed by x & y. Check the solution in each of the original equations.

Solve by Substitution 1. x = -4 3x + 2y = 20 1. (-4, 16)

Solve by Substitution 2. y = x - 1 x + y = 3 2. (2, 1)

Solve by Substitution 3. 3x + 2y = -12 y = x - 1 3. (-2, -3)

Solve by Substitution 4. x = 1/2 y - 3 4x - y = 10 4. (8, 22)

Solve by Substitution 5. x = -5y + 4 3x + 15y = -1 5. No solution

Solve by Substitution 6. 2x - 5y = 29 x = -4y + 8 6. (12, -1)

Solve by Substitution 7. x = 5y + 10 2x - 10y = 20 7. Many solutions

Solve by Substitution 8. 2x - 3y = -24 x + 6y = 18 9. (-6, 4)

CW/HW

Graphing and Substitution WS HW Graphing and Substitution WS