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Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

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Consistent Systems  The lines will intersect.  The point where the lines intersect is your solution.  The solution of this graph is (1, 2) (1,2)

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Page 1: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Module 1 Lesson 5

SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Page 2: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

What is a system of equations? A system of equations is when you have two

or more equations using the same variables. The solution to the system is the point that

satisfies ALL of the equations. This point will be an ordered pair.

When graphing, you will encounter three possibilities. Consistent Systems (one solution) Inconsistent Systems (no solutions) Dependent Systems (Infinite number of

solutions)

Page 3: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Consistent Systems

The lines will intersect. The point where the lines

intersect is your solution. The solution of this graph

is (1, 2)

(1,2)

Page 4: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Inconsistent Systems These lines never

intersect as the lines are parallel!

Since the lines never cross, there is NO SOLUTION!

Parallel lines have the same slope with different y-intercepts.

2Slope = = 21

y-intercept = 2y-intercept = -1

Page 5: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Dependent Systems These lines are the

same! Since the lines are on

top of each other, there are INFINITELY MANY SOLUTIONS!

Coinciding lines have the same slope and y-intercepts.

2Slope = = 21

y-intercept = -1

Page 6: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

What is the solution of the system graphed below?

1. (2, -2)2. (-2, 2)3. No solution4. Infinitely many solutions

Page 7: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

1) Find the solution to the following system by graphing:

2x + y = 4x - y = 2

Graph both equations. I will graph using x- and y-intercepts (plug in

zeros), but you can also rewrite in to y =mx + b form.

Graph the ordered pairs.

2x + y = 4(0, 4) and (2, 0)

x – y = 2(0, -2) and (2, 0)

Page 8: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Graph the equations.

2x + y = 4(0, 4) and (2, 0)

x - y = 2(0, -2) and (2, 0)

Where do the lines intersect?(2, 0)

2x + y = 4

x – y = 2

Page 9: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Check your answer!

To check your answer, plug the point back into both equations.

2x + y = 4 2(2) + (0) = 4

x - y = 2(2) – (0) = 2 Nice job…let’s try another!

Page 10: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

2) Find the solution to the following system:

y = 2x – 3-2x + y = 1

Graph both equations. Put both equations in slope-intercept or standard form. I’ll do slope-intercept form on this one!

y = 2x – 3y = 2x + 1

Graph using slope and y-intercept

Page 11: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Graph the equations.y = 2x – 3m = 2 and b = -3y = 2x + 1m = 2 and b = 1Where do the lines intersect?

No solution!

Notice that the slopes are the same with different y-intercepts. If you recognize this early, you don’t

have to graph them!

Page 12: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Check your answer!

Not a lot to check…Just make sure you set up your equations correctly.

I double-checked it and I did it right…

Page 13: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

What is the solution of this system?

1. (3, 1)2. (4, 4)3. No solution4. Infinitely many

solutions

3x – y = 82y = 6x -16

Page 14: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Solving a system of equations by GraphingLet's summarize! There are 3 steps to

solving a system using a graph.

Step 1: Graph both equations.

Step 2: Do the graphs intersect?

Step 3: Check your solution.

Graph using slope and y – intercept or x- and y-intercepts. Be sure to use a ruler and graph paper!

This is the solution! LABEL the solution!

Substitute the x and y values into both equations to verify the point is a solution to both equations.

Page 15: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Solving Systems of Equations using Substitution

Steps:1.Solve one equation for one variable (y= ; x= ; a=)2.Substitute the expression from step one into the other equation.3.Simplify and solve the equation.4.Substitute back into either original equation to find the value of the other variable.5. Check the solution in both equations of the system.

Page 16: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Example #1: y = 4x3x + y = -21

Step 1: Solve one equation for one variable.

y = 4x (This equation is already solved for y.)Step 2: Substitute the expression from step one into the other equation.

3x + y = -213x + 4x = -21

Step 3: Simplify and solve the equation. 7x = -21

x = -3

Page 17: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

y = 4x3x + y = -21

Step 4: Substitute back into either original equation to find the value of the other variable.

3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12

Solution to the system is (-3, -12).

Example #1 cont:

Page 18: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

y = 4x3x + y = -21

Step 5: Check the solution in both equations.

y = 4x

-12 = 4(-3)

-12 = -12

3x + y = -21

3(-3) + (-12) = -21

-9 + (-12) = -21

-21= -21

Solution to the system is (-3,-12).

Page 19: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Example #2: x + y = 10 5x – y = 2

Step 1: Solve one equation for one variable. x + y = 10 y = -x +10Step 2: Substitute the expression from step one into the other equation. 5x - y = 2

5x -(-x +10) = 2

Page 20: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

x + y = 10 5x – y = 2

5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2

6x = 12 x = 2

Step 3: Simplify and solve the equation.

Example #2 cont:

Page 21: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

x + y = 10 5x – y = 2

Step 4: Substitute back into either original equation to find the value of the other variable.

x + y = 102 + y = 10 y = 8Solution to the system is

(2,8).

Example #2 cont:

Page 22: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

x + y = 10 5x – y = 2

Step 5: Check the solution in both equations.

x + y =102 + 8 =10

10 =10

5x – y = 25(2) - (8) = 2

10 – 8 = 22 = 2

Solution to the system is (2, 8).

Example #2 cont:

Page 23: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Solving Systems of Equations using Elimination (also called Solving by using Addition)

Steps:1.Place both equations in Standard Form, Ax + By = C.2.Determine which variable to eliminate with Addition or Subtraction.3.Solve for the variable left.4.Go back and use the found variable in step 3 to find second variable.5. Check the solution in both equations of the system.

Page 24: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

EXAMPLE #1:

STEP 2: Multiply the 2nd equation by -1. 5x + 3y =11

-5x + 2y =1

5x + 3y = 115x = 2y + 1

STEP 3: Add like terms and solve. 5x + 3y =11

-5x + 2y = -1 5y =10 y = 2

STEP 1: Write both equations in Ax + By = C form.

5x + 3y =1 5x - 2y =1

Page 25: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

STEP 4: Solve for the other variable by substituting

into either equation.5x + 3y =11

5x + 3(2) =11 5x + 6 =11 5x = 5 x = 1

5x + 3y = 115x = 2y + 1

The solution to the system is (1,2).

Page 26: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

5x + 3y= 115x = 2y + 1

Step 5: Check the solution in both equations.

5x + 3y = 11 5(1) + 3(2) =11 5 + 6 =11 11=11

5x = 2y + 15(1) = 2(2) + 1 5 = 4 + 1 5=5

The solution to the system is (1,2).

Page 27: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Solving Systems of Equations using EliminationSteps:1. Place both equations in Standard Form, Ax + By = C.2. Determine which variable to eliminate with Addition or Subtraction.3. Solve for the remaining variable.4. Go back and use the variable found in step 3 to find the second variable.5. Check the solution in both equations of the system.

Page 28: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Example #2: x + y = 10 5x – y = 2

Step 1: The equations are already in standard form: x + y = 10

5x – y = 2 Step 2: Adding the equations will eliminate y.

x + y = 10 x + y = 10 +(5x – y = 2) +5x – y = +2Step 3: Solve for the variable.

x + y = 10 +5x – y = +2

6x = 12 x = 2

Page 29: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

x + y = 10 5x – y = 2

Step 4: Solve for the other variable by

substituting into either equation.

x + y = 102 + y = 10 y = 8Solution to the system is (2,8).

Page 30: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

x + y = 10 5x – y = 2

x + y =10

2 + 8 =10

10=10

5x – y =2 5(2) - (8) =2 10 – 8 =2 2=2

Step 5: Check the solution in both equations.Solution to the system is (2,8).

Page 31: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word Problem:

Two angles are supplementary. The measure of one angle is 10 degrees more than three times the other. Find the measure of each angle.

Page 32: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word Problem:

Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle.

x = degree measure of angle #1 y = degree measure of angle #2Therefore x + y = 180

Page 33: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word Problem:Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle.

x + y = 180x =10 + 3y

Page 34: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word Problem, cont:

x + y = 180x =10 + 3y

x + y = 180

-x + 3y = -10

4y =170 y = 42.5

x + y = 180 x - 3y = 10

I will multiply the second equation by -1 then add like terms.

Page 35: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word Problem:

Substitute the 42.5 to find the other angle.

x + 42.5 = 180 x = 180 - 42.5

x = 137.5

(137.5, 42.5)

Page 36: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word problem:

The sum of two numbers is 70 and their difference is 24. Find the two numbers.x = first number

y = second numberTherefore, x + y = 70

Page 37: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word Problem:

The sum of two numbers is 70 and their difference is 24. Find the two numbers.

x + y = 70x – y = 24

Page 38: Module 1 Lesson 5 SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES

Using Elimination to Solve a Word Problem:

x + y =70

x - y = 24 2x = 94

x = 47

47 + y = 70

y = 70 – 47

y = 23(47, 23)