Solving Linear Systems by graphing 3.1 Solving Linear Systems by graphing What you should learn: Goal 1 Graph and solve systems of linear equations in two variables. Goal 2 Use linear systems to solve real-life problems. 3.1 Solving Linear Systems by graphing

{ Decide whether the ordered pairs are solution (2, -1), (3, 0) Example 1) Decide whether the ordered pairs are solution { (2, -1), (3, 0) 3.1 Solving Linear Systems by graphing

{ Solve the linear system by graphing Example 2) 1.Graph each line. 2. Use any method to graph the lines. 3.Where they intersect is the solution 3.1 Solving Linear Systems by graphing

{ Solve the linear system by graphing x-int = 3 x-int = 5 y-int = 6 3.1 Solving Linear Systems by graphing

{ Solve the linear system by graphing 1.Graph each line. example 3) Solve the linear system by graphing { 1.Graph each line. 2. Use any method to graph the lines. 3.Where they intersect is the solution 3.1 Solving Linear Systems by graphing

Where do they intersect? Graph the line y- intercept = 3 or (0,3) Slope = Graph the line x-int = 2 y-int = -4 Where do they intersect? 3.1 Solving Linear Systems by graphing

3.1 Solving Linear Systems by graphing

1. Set-up 2 linear equations. (1) for days (1) for money Example 1) Your friends are planning a 7 day trip to Florida. You estimate that it will cost $275 per day in Tampa and $400 per day in Orlando. Your total budget for the 7 days is $2300. How many day should you spend in each location? 1. Set-up 2 linear equations. (1) for days (1) for money

Reflection on the Section Explain how to use a graph to determine how many solutions there are for a system of linear equations. assignment 3.1 solving linear systems by graphing

Solving Linear Systems Algebraically 3.2 Solving Linear Systems Algebraically What you should learn: Goal 1 Use algebraic method to solve systems of linear equations . Substitution Method and Combination Method Use linear systems to solve real-life problems. Goal 2 3.2 Solving Linear Systems Algebraically

{ Solve the linear system by substitution example 1) 1. Solve either equation for either one of its variables. 2. Substitute this expression into the other equation and solve for the other variable. 3. Substitute this value into the revised first equation and solve. 4. Check the solution pair in each of the original equations. 3.2 Solving Linear Systems Algebraically

{ Solve the linear system by substitution equation 1 Begin by solving Equation 1 for y + x + x Substitute this expression for y in Equation 2 and solve for x. 3.2 Solving Linear Systems Algebraically

equation 2 -1 -1 Substitute this value into the revised first equation and solve for y 3.2 Solving Linear Systems Algebraically

The solution is (-1, 0). Check to see that it satisfies each of the original equations. 3.2 Solving Linear Systems Algebraically

{ Solve the linear system by substitution example 2) { Begin by solving Equation 2 for x + 4y + 4y Substitute this expression for x in Equation 1 and solve for y. 3.2 Solving Linear Systems Algebraically

+2 +2 Substitute this value into the revised first equation and solve for x 3.2 Solving Linear Systems Algebraically

The solution is (1, 1/2). Check to see that it satisfies each of the original equations. 3.2 Solving Linear Systems Algebraically

Modeling Real – Life situation example 3) Modeling Real – Life situation You are selling tickets at a high school football game. Student tickets cost $2 and general admission tickets cost $3. You sell 1957 tickets and collect $5035. How many of each type of ticket did you sell? Students 2x General 3y Total $ collected 5035 3.2 Solving Linear Systems Algebraically

The other equation is the # of student plus the # of general tickets is 1957. { Let’s solve Equation 1 for y 3.2 Solving Linear Systems Algebraically

Substitute in Equation 2 for y and solve for x # of student tickets # of general tickets 3.2 Solving Linear Systems Algebraically

Solve the linear system by Linear Combination 1. Arrange the equations with like terms in columns. 2. Study the coefficients of x (or y). 3. Multiply one or both equations by an appropriate number to obtain new coefficients for x (or y) that are opposites. Add the equations and solve for the remaining variable. substitute the value obtained in Step 4 into either of the original equations and solve for the other variable. 3.2 Solving Linear Systems Algebraically

{ Solve the linear system by Linear Combination Example 1) In this linear system, the coefficients for y are opposites. By adding the two equations, you obtain and equation that has only one variable, x. 3.2 Solving Linear Systems Algebraically

{ + Solve the linear system by Linear Combination 6 6 Example 1) + 6 6 Now, we take this value and sub it in for either equation… 3.2 Solving Linear Systems Algebraically

So, our solution is I used the 1st equation… 3.2 Solving Linear Systems Algebraically

{ (4) (3) Solve the linear system by Linear Combination Example 2) (4) (3) Now, take this and sub it in… 3.2 Solving Linear Systems Algebraically

So, our solution is 3.2 Solving Linear Systems Algebraically

Solve the linear system with Many Solutions { example 1) You can use any of the three methods to show that the system has many solutions. -2 ( ) 0 = 0 This means all numbers work. 3.2 Solving Linear Systems Algebraically

{ If we graphed these two lines, they will be the same line. Solve the linear system with Many Solutions { If we graphed these two lines, they will be the same line. 3.2 Solving Linear Systems Algebraically

Solve the linear system with NO Solutions { example 2) You can use any of the three methods to show that the system has no solutions. 0 = 4 This means no numbers work. 3.2 Solving Linear Systems Algebraically

{ If we graphed these two lines, they will be parallel lines. Solve the linear system with NO Solutions { If we graphed these two lines, they will be parallel lines. 3.2 Solving Linear Systems Algebraically

Parallel Lines – the slopes are the same the y-intercepts are different Same lines – the slopes are the same the y-intercepts are the same. Lines are either Parallel, the same, or they just intersect somewhere. Perpendicular Lines – *slopes are opposite inverses y-intercepts are different. 3.2 Solving Linear Systems Algebraically

Example 1) In one week a music store sold 7 violins for a total of $1600. Two different types of violins were sold. One type cost $200 and the other type cost $300. How many of each type of violin did the store sell? What to do? Set-up two equations. # of type A # of type B equation 1 + = total # sold . . equation 2 Price of type A # of type A Price of type B # of type B + = Total sales

. equation 1 equation 2 + = Total sales total # sold Price of type A Price of type B # of type B = Total sales total # sold

Which method would you like to use to solve the linear system? equation 1 equation 2 Price of type A # of type A + . Price of type B # of type B = Total sales total # sold Which method would you like to use to solve the linear system?

Example 2) You have $3.55 of change in a bag. There are dimes and quarters only. If there is a total of 22 coins. How many of each are there? Set-up two equations. # of dimes # of Quarters equation 1 = total # coins + . . equation 2 0.10 # of dimes 0.25 # of quarters + = Total change

. . D + Q = 22 0.10D + 0.25Q = 3.55 total # coins + = Total change + = # of dimes # of Quarters total # coins + = . . Total change 0.10 # of dimes 0.25 # of quarters + = D + Q = 22 0.10D + 0.25Q = 3.55

(-0.10) D + Q = 22 0.10D + 0.25Q = 3.55 -0.10D - 0.10Q = -2.2 0.10D + 0.25Q = 3.55 0.15Q = 1.35 0.15 0.15 Q = 9

Example 3) A music Club A has a $30 membership fee and charges $8 per CD. Music club B has a $45 membership fee and charges $7 per CD. How many CD’s would a person need to buy for Club A to more expensive than Club B? What to do? Set-up two equations. Membership fee Per CD expression A + Membership fee Per CD expression B +

Set these equal to eachother. Membership fee Per CD + Membership fee Per CD +

= Per CD Membership fee Per CD Membership fee + + 30 + 8x = 45 + 7x -7x -7x 30 + x = 45 So, they are equally as good at 15 CD’s. -30 -30 x = 15 How many CD’s would a person need to buy for Club A to more expensive than Club B? 16 or more

Example 2) You combine 2 solutions to form a mixture that is 40% acid. One solution is 20% acid and the other is 50% acid. If you have 90 milliliters of the mixture, how much of each solution was used to create the mixture? Let’s look at we have… Try setting-up 2 equations. One equation that has the amount of 1st solution plus amount of 2nd solution.

Setting-up equations…let’s say that x is the amount of 1st solution y is the amount of 2nd solution 90 is total amount Next equation should be the % of acid times the amount each solution.

Solve using any method you like. Probably substitution.

Reflection on the Section When using the linear combination method for solving a linear system, why would you want to have the coefficients of one of the variables be opposites? assignment 3.2 solving linear systems by algebraically

Solving Linear Systems of Linear Inequalities 3.3 Solving Linear Systems of Linear Inequalities What you should learn: Goal 1 Graph a system of linear inequalities to find the solutions of the system. Goal 2 Use linear systems to solve real-life problems. 3.3 Solving Linear Systems of Linear Inequalities

{ Graph the system of linear inequalities example 1) Graph the system of linear inequalities { What we are going to do is graph and shade each inequality separately, but both on the same coordinated plane. 33 Graphing and Solving Systems of Linear Inequalities

False Sketch the graph of … Graph the line either solve for y, (slope-intercept form) or find x- and y-int Test an easy to deal with point…like (0,0) So, shade in the other side. False 3.3 Graphing and Solving Systems of Linear Inequalities

False sketch the graph of … Graph the line either use the slope and y-intercept, (slope-intercept form) or find x- and y-int Test an easy to deal with point…like (0,0) So, shade in the other side. False 33 Graphing and Solving Systems of Linear Inequalities

When both are graphed and shaded, the solution is the doubled shaded area. 33 Graphing and Solving Systems of Linear Inequalities

{ example 2 ) sketch the graph of … 3.3 Graphing and Solving Systems of Linear Inequalities

{ example ) sketch the graph of … 3.3 Graphing and Solving Systems of Linear Inequalities

3.3 Graphing and Solving Systems of Linear Inequalities

33 Graphing and Solving Systems of Linear Inequalities

assignment Reflection on the Section What is the procedure used to graph a system of linear inequalities? assignment 3.3 solving linear systems of Inequalities

Linear Programming 3.4 What you should learn: Goal 1 Solve linear programming problems. Goal 2 Use linear systems to solve real-life problems. 3.4 Linear Programming

Reflection on the Section Why do you need to find the vertices of the feasible region when using linear programming? assignment 3.4 linear Programming

Graphing Linear Equations in Three Variables. 3.5 Graphing Linear Equations in Three Variables. What you should learn: Goal 1 Graph linear equations in three variables and evaluate linear functions of two variables. Goal 2 Use functions of two variables to solve real-life problems. 3.5 Graphing Linear Equations in Three Variables

Graphing in Three Dimensions Solutions of equations in three variables can be pictured with a three-dimensional coordinate system. To construct such a system, begin with the xy-coordinate plane in a horizontal position. Then draw the z-axis as a vertical line through the origin. In much the same way that points in a two dimensional coordinate system are represented by ordered pairs, each point in space can be represented by an ordered triple (x, y, z). Drawing the point represented by an ordered triple is called plotting the point. The three axes, taken two at a time, determine three coordinate planes that divide space into eight octants. The first octant is one for which all three coordinates are positive.

Plotting Points in Three Dimensions Plot the ordered triple in a three-dimensional coordinate system. (–5, 3, 4) SOLUTION To plot (–5, 3, 4), it helps to first find the point (–5, 3) in the xy-plane. The point (–5, 3, 4), lies four units above.

Plotting Points in Three Dimensions Plot the ordered triple in a three-dimensional coordinate system. (–5, 3, 4) (3, – 4, –2) SOLUTION SOLUTION To plot (–5, 3, 4), it helps to first find the point (–5, 3) in the xy-plane. To plot (3, – 4, –2) it helps to first find the point (3, – 4) in the xy-plane. The point (–5, 3, 4), lies four units above. The point (3, – 4, –2) lies two units below.

Graphing in Three Dimensions A linear equation in three variables x, y, z is an equation in the form ax + by + cz = d where a, b, and c are not all 0. An ordered triple (x, y, z) is a solution of this equation if the equation is true when the values of x, y, and z are substituted into the equation. The graph of an equation in three variables is the graph of all its solutions. The graph of a linear equation in three variables is a plane. A linear equation in x, y, and z can be written as a function of two variables. To do this, solve the equation for z. Then replace z with f(x, y).

Graphing a Linear Equation in Three Variables Sketch the graph of 3x + 2y + 4z = 12. SOLUTION Begin by finding the points at which the graph intersects the axes. Let x = 0, and y = 0, and solve for z to get z = 3. This tells you that the z-intercept is 3, so plot the point (0, 0, 3). In a similar way, you can find the x-intercept is 4 and the y-intercept is 6. After plotting (0, 0, 3), (4, 0, 0) and (0, 6, 0), you can connect these points with lines to form the triangular region of the plane that lies in the first octant.

( ) Evaluating Functions of Two Variables Write the linear equation 3x + 2y + 4z = 12 as a function of x and y. Evaluate the function when x = 1 and y = 3. Interpret the result geometrically. SOLUTION 3x + 2y + 4z = 12 Write original function. 4z = 12 –3x –2y Isolate z-term. 1 4 (12 – 3x – 2y) z = Solve for z. 1 4 (12 – 3x – 2y) f(x, y) = Replace z with f(x, y). 1 4 (12 – 3(1) – 2(3)) f(1, 3) = 3 4 = Evaluate when x = 1 and y = 3. This tells you that the graph of f contains the point 1, 3, . 3 4 ( )

Using Functions of Two Variables in Real Life Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35. Write a model for the total amount you will spend as a function of the number of pounds of bluegrass and rye. SOLUTION Your total cost involves two variable costs (for the two types of seed) and one fixed cost (for the spreader). Verbal Model Total cost Bluegrass cost Bluegrass amount Rye cost Rye amount Spreader cost = • + • + …

Modeling a Real-Life Situation Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35. Write a model for the total amount you will spend as a function of the number of pounds of bluegrass and rye. … Labels Total cost = C (dollars) Bluegrass cost = 2 (dollars per pound) Bluegrass amount = x (pounds) Rye cost = 1.5 (dollars per pound) Rye amount = y (pounds) Spreader cost = 35 (dollars) …

Modeling a Real-Life Situation Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35. Write a model for the total amount you will spend as a function of the number of pounds of bluegrass and rye. … Total cost Bluegrass cost Rye cost Rye amount Spreader cost Bluegrass amount = • + Algebraic Model C = 2 x + 1.5y + 35 Evaluate the model for several different amounts of bluegrass and rye, and organize your results in a table. To evaluate the function of two variables, substitute values of x and y into the function.

Modeling a Real-Life Situation Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35. Evaluate the model for several different amounts of bluegrass and rye, and organize your results in a table. For instance, when x = 10 and y = 20, then the total cost is: C = 2 x + 1.5 y + 35 Write original function. C = 2 (10) + 1.5(20) + 35 Substitute for x and y. = 85 Simplify.

Modeling a Real-Life Situation Landscaping You are planting a lawn and decide to use a mixture of two types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound and the rye costs $1.50 per pound. To spread the seed you buy a spreader that costs $35. Evaluate the model for several different amounts of bluegrass and rye, and organize your results in a table. The table shows the total cost for several different values of x and y. $175 $160 $145 $130 40 $155 $140 $125 $110 30 $135 $120 $105 $90 20 $115 $100 $85 $70 10 x Bluegrass (lb) y Rye (lb)

assignment Reflection on the Section Name a point on the graph of f if f(-2,3) = 7 assignment 3.5 Graphing Linear Equations in Three Variables

Solve systems of linear equations in three variables. 3.6 Solving Systems of Linear Equations in Three Variables. What you should learn: Goal 1 Solve systems of linear equations in three variables. Goal 2 Use linear systems in 3 variables to solve real-life problems. 3.6 Solving Systems of Linear Equations in 3 variables

They are written as an ordered triple (x, y, z) System of three linear equations looks like this: We can solve by using: Linear Combination Method Solutions can be one, none, or many. They are written as an ordered triple (x, y, z)

Solve the system. Ex) Eliminate one of the variables in two of the original equations. 1 (2) Add 2 times the second equation to the first.

Solve the system. Ex) Eliminate one of the variables in two of the original equations. 1 Add 2 times the second equation to the first. Eliminate the same variable by adding two other equations.

Solve the system. Ex) Eliminate one of the variables in two of the original equations. 1 (-3) Add -3 times the first equation to the second.

Solve the system. Ex) Eliminate one of the variables in two of the original equations. 1 Add -3 times the first equation to the second. Now take those 2 equations …

Solve the new system. 2 Add 7 times the second equation to the first. Solve for z. -18 -18 z = 4 Substitute into new 1 or 2.

z = 4 x = -3 Substitute x = -3 and z = 4 into an original equation and solve for y 3 Solve for y. The solution is x = -3, y = 2, and z = 4, or the triple ordered pair (-3, 2, 4). y = 2

Reflection on the Section How do you decide if a linear system in 3 variables has one, infinitely many, or no solutions using the equations? Using the graphs? assignment 3.6 Solving Systems of Linear Equations in 3 variables