# Equations: Linear and Systems

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Equations: Linear and Systems

Overview Three Topics: Solving Linear Equations for a given variable
In this module you will be learning 3 topics all involving linear equations. Linear Equations are equations with one or more variables whose power is of degree one. If the degree of any of the variables is not of degree one, then it is not a linear equation. Three Topics: Solving Linear Equations for a given variable Finding and using the Equation of a Line Solving and using Systems of Linear Equations *You will be able to navigate easily to the topics of interest by clicking on the links on each page. *The icon of a house in the upper right-hand corner of each page will return you to this page. *If you click on the title of any page, that will return you to that particular sub-topic of the module.

Solving Linear Equations
Writing Equations Using Application Problems Solving Equations by Using Addition and Subtraction Solving Equations by Using Multiplication and Division Solving Equations Using Algebra Tiles Solving Multi-Step Equations Solving Equations with the Variable on Each Side Application Problems

Writing Equations First, we need to be able to translate our words in mathematical problems. Here are basic terms which determine which mathematical operations to use. Addition Subtraction Multiplication Division Total Minus Times Divided by Plus Less than Of Per Added to Subtracted from Each Ratio More than Decreased by Factors Rate Sum(use parentheses to represent the answer.) Difference (use parentheses to represent the answer.) Product (use parentheses to represent the answer.) Quotient (use parentheses to represent the answer.)

Writing Equations - Example
You want to purchase an iPod and you know that it costs about \$160. You have \$60 dollars saved all ready. Your dad tells you that he will pay you \$20 each week for mowing the grass. How long will it take you to earn the additional money? 1st- You know that you have \$60 all ready. As a result, you need to earn \$160-\$60 = \$100.

Writing Equations-continued
2nd- You will earn \$20 each week and you need to earn \$100. Let x = the number of weeks you need to mow the lawn \$20 x = 100 Let’s find what number you have to multiply 20 by to get to 100? X = 5 It will take you 5 weeks to earn the \$100 which are left to purchase the iPod. Lets check your answer: \$160 - \$60 = \$100 left to earn \$20 ( 5) = \$100 \$100 + \$60 = \$160.

Writing Equations-continued
Click on the link below to practice writing equations. The beginning of the link will provide you with examples. If you continue to scroll down, you will be able to work with 5 problems by clicking on your answer choice and the computer will give you immediate feedback as to whether your choice is correct or not.

Solving Equations by Addition and Subtraction
When we balance equations, we have to remember to share the same amount with both sides. This is done by using the Addition and Subtraction Properties of Equality. x + 11= 15 x = 15 – 11 Subtract 11 from both sides so that the variable is the only term on the left side of the equal sign x + 0 = “x” plus 0 is equal to “ x” x = Simplify

Solving Equations by Addition and Subtraction-continued
To practice solving equations using addition and subtraction problems, open the below link. Once you click on the link, the first thing you will see is an example. If you scroll down you will see two choices; one choice is to review more examples and the second choice is for you to test yourself by working similar problems.

Solving Equations by Using Multiplication and Division
Review: 5x=20 therefore x has to be 4 because we know that 5(4) = 20. If we did not know that 5(4)=20, using 20 and 5 which operation would we have to perform to get 4? We would have to divide 20 by 5. As a result, division will undo multiplication and multiplication will undo division. These are called the Multlipication and the Division Properties of Equalities. Let’s apply this concept to solving equations.

Solving Equations by Using Multiplication and Division-continued
Solve: -13 x = 52 -13 x = Divide both sides by -13 because division will undo multiplication X = - 4 Let’s check to see if this is correct. -13(-4) = 52 therefore x = -4.

Solving Equations by Using Multiplication and Division-continued
Solve -5h = -2/3 -5h = -2/3 Divide both sides by -5 h= -2/3 (-1/5) Divide fractions by multiplying by its reciprocal h=2/15 Check: -5(2/15) =-10/15= -2/3 Let’s solve -5h = -2/3 by multiplying each side of this equation by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5 therefore: -1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s h= 2/ reciprocal is equal to 1.

Solving Equations by Using Multiplication and Division-continued
To practice solving equations using multiplication and division problems, open the below link. There are four steps once you click on the link. The first step will show you how to balance the equation. Step 2 will give you an in-depth explanation of how to solve equations. Step 3 will show you 5 additional examples. Step 4 will allow you to practice solving equations.

Solving Equations Using Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to solve equations.

Solving Multi-Step Equations
There are several steps you will need to follow to solve multi-step equations. Think of using these steps as you would use the order of operations. Step 1: Use the Distributive Property to remove the grouping symbols. Step 2: Simplify the expressions on each side of the equal sign by combining like terms. Step 3: Combine like terms on different sides of the equal sign to get all the variables on the same side and all the numbers without variables on the other side together. Use what you learned in the first section for solving addition and subtraction equations. ( The addition and/or the subtraction properties of equality.) Step 4: Simplify each expression on each side of the equal sign. Step 5: Use what you learned in the second section for solving multiplication and division equalities. (The multiplication and /or the division property of equality.)

Solving Equations with the Variable on Each Side
The key is to combine like terms by using the previous strategies you learned from solving equations. Solve for f. -16f – 2 = -15f f – 2 = 17 Add 15f to both sides -f =  Add 2 to both sides f =  Multiply both sides by -1

Solving Equations with the Variable on Each Side-continued
Open the below link to practice solving multi-step equations and equations with variables on both sides of the equal sign. Once you open the link you will be able to practice solving 15 multi-step equations with there solutions at the bottom of the website.

Solving Multi-Step Equations Using Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to solve multi-step equations. To practice solving equations using algebra tiles click on the below link. The link allows you to represent the equations by clicking on the algebra tiles and move them around to solve the equations.

Finding the Equation of a Line
This unit will cover basic graphing of points and lines along with finding the slope of lines given two points and from the line graph. It will also cover how to find the equation of lines from the lines graph and given conditions. The unit concludes with finding the equations of parallel and perpendicular lines.

Finding the equation of a line
Review Graphing points Slope from a given line Slope between two points(Slope Formula) Graphing lines with Tables, Point & Slope and Slope Intercept Finding equation given slope and y-intercept Finding equation given point and slope(Slope Intercept and Point Slope) Finding equation of the line given graph Finding equation given two points Standard form ( Intercepts and putting into Slope Intercept) Parallel Lines Perpendicular Lines Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a position on a coordinate system. y-axis Quadrant II Quadrant I x-axis Quadrant III Quadrant IV Back to section Title page

To graph a point on a coordinate system You must travel left or right on the x-axis and the up or down on the y-axis. Right Left Up Down Back to section Title page

The Finding the equation of a line
(0,0) is called the origin. Back to section Title page

Points in Quadrant I have positive x values and positive y values
(3,4) Right 3 Up 4 x Back to section Title page

Points in Quadrant II have negative x values and positive y values
(-5,6) Left 5 Up 6 Back to section Title page

Points in Quadrant III have negative x values and negative y values
(-7,-3) Left 7 Down 3 Quadrant III Back to section Title page

Points in Quadrant IV have positive x values and negative y values
(8,-5) Right 8 Down 5 Quadrant IV Back to section Title page

Practice Graphing Lines
Go to this website if you would like to practice graphing points Back to section Title page

Slope of a given line In this section you will review how to find the slope of a given line from the graph of the line. Back to section Title page

𝑚= 𝑅𝑖𝑠𝑒 𝑅𝑢𝑛 Slope(m) is defined as the vertical
change of the line divided by the horizontal change of the line. Rise is the vertical change Run is the horizontal change 𝑚= 𝑅𝑖𝑠𝑒 𝑅𝑢𝑛 Back to section Title page

𝑚= 𝑅𝑖𝑠𝑒 𝑅𝑢𝑛 (1,1) and (5,4) 3 3 5 m = 5 Slope of a given line
Identify two points on the given line (1,1) and (5,4) 𝑚= 𝑅𝑖𝑠𝑒 𝑅𝑢𝑛 3 5 3 m = 5 Back to section Title page

(-4,7) and (5,2) -5 9 -5 m = 9 Slope of a given line Back to section
Title page

(-6,1) and (4,5) 5 10 5 m = 10 1 m = 2 Slope of a given line
(Simplify) 10 1 m = 2 Back to section Title page

No rise 11 m = 11 m = 0 Slope of a given line
m = 11 m = 0 The black line above is called a horizontal line. Horizontal lines always have a slope of 0. Back to section Title page

3 3 m = m = undefined Slope of a given line No Run
m = undefined No Run The black line above is a vertical line. A vertical line has an undefined slope. Back to section Title page

Click on the link below to practice finding the slope of a line
Back to section Title page

Finding Slope of Line (Slope Formula)
In this section you will review how to use the Slope Formula to find the slope of a line between two points. Back to section Title page

Given two points 1st Point 2nd Point Slope Formula Back to section
Title page

𝑚= 7 − 4 5 − 3 𝑚 = 3 2 Find the Slope between (3,4) and (5,7)
𝑚 = 3 2 Back to section Title page

m= 12 −8 m= 3 −2 Find the Slope between (6,-5) and (-2,7)
𝑚= 7 −(−5) −2 −6 m= 12 −8 m= 3 −2 Back to section Title page

𝑚= 10 −4 −1 −5 m = 6 −6 m=−1 Find the Slope between (5,4) and (-1,10)
Back to section Title page

Find the Slope between (8,4) and (-3,4) 𝑚= 4 − 4 −3 −8 𝑚 = 0 −11 𝑚=0
𝑚 = 0 −11 𝑚=0 This is a horizontal line Back to section Title page

Find the Slope between (-2,6) and (-2,9) 𝑚= 9 −6 −2 −(−2) 𝑚 = 3 0
𝑚 = 3 0 𝑚=𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 This is a vertical line. Back to section Title page

Go to this website if you would like to do a worksheet on finding
The slope between two points.

Graphing lines In this section you will review how to graph lines.
Using Tables Using a Point and the Slope Using the Slope and the y - intercept Back to section Title page

Using Tables Plot the points and connect them. x y -3 -10 -2 -8 -1 -6
-4 1 2 3 Every point on the line is a solution to the same line equation You need only two points to graph the line . Back to section Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point Point (1,2) 3 up m = 4 right or 3 down m = 4 left Back to section Title page

Using Slope and y-intercept
y-intercept is the point where the line goes through the y-axis Back to section Title page

Using Slope and y-intercept
-3 down m = 5 right 3 up m = -5 left Back to section Title page

Finding equation given slope and y-intercept
In this section you will review how to find the equation of a line given the slope of the line and the y-intercept. Back to section Title page

Slope Intercept Form y = mx + b where m is the slope and
b is the y-intercept Back to section Title page

b = 6 m = 5 y = mx + b y = 5x + 6 Find the equation of the line
Given a slope of 5 and y-intercept of 6 m = 5 b = 6 y = mx + b (substitute for m and b) y = 5x + 6 Back to section Title page

b = -3 m = 8 y = mx + b y = 8x - 3 Find the equation of the line
Given a slope of 8 and y-intercept of -3 m = 8 b = -3 y = mx + b (substitute for m and b) y = 8x - 3 Back to section Title page

y = mx + b Find the equation of the line Given:𝑚=− 3 4 𝑎𝑛𝑑 𝑏=6
(substitute for m and b) 𝑦=− 3 4 𝑥+6 Back to section Title page

Finding equation given point and slope
In this section you will review how to find the equation of a line given the slope and a point. Back to section Title page

Find the equation of the line
(Slope Intercept) Given: m = 6 and point (2,3) y = mx + b [substitute for m and (x,y)] 3 = 6(2) + b 3 = 12 + b -9 = b (substitute for m and b) y = 6x - 9 Back to section Title page

Find the equation of the line (Slope Intercept)
𝐺𝑖𝑣𝑒𝑛: 𝑚= 2 3 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7) y = mx + b [substitute for m and (x,y)] 7 = 𝑏 7=4+𝑏 3=𝑏 (substitute for m and b) y = 2 3 𝑥+3 Back to section Title page

Find the equation of the line (Slope Intercept)
𝐺𝑖𝑣𝑒𝑛: 𝑚=− 3 5 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2) y = mx + b [substitute for m and (x,y)] 2 = −4 +𝑏 2 = 𝑏 2 5 = 𝑏 (substitute for m and b) y =- 3 5 𝑥+ 2 5 Back to section Title page

Find the equation of the line (Slope Intercept)
Given: m = 0 and point (-2,-6) y = mx + b [substitute for m and (x,y)] -6 = 0(-2) + b -6 = b (substitute for m and b) y = 0x – 6 y = - 6 (This is the equation of a horizontal line) Back to section Title page

x = -2 Find the equation of the line (Slope Intercept) The equation is
Given: m = undefined and point (-2,-6) Since the slope is undefined, you know that the line is a vertical line. Vertical lines go through the x-axis at the x value of the ordered pair. The equation is x = -2 Back to section Title page

Point Slope Form 𝑦 − 𝑦 1 = m(x - 𝑥 1 ) 𝑥 1 𝑎𝑛𝑑 𝑦 1
You can use this form when you are given a point and a slope of the line. You will substitute for slope(m) and point for 𝑥 1 𝑎𝑛𝑑 𝑦 1 Back to section Title page

Point Slope Form 𝑦 − 𝑦 1 = m(x - 𝑥 1 ) Find the equation of the line
Given: m = 7 and (2,4) 𝑦 − 𝑦 1 = m(x - 𝑥 1 ) y – 4 = 7(x – 2) (This is an acceptable answer but you can also simplify it) y – 4 = 7x - 14 y = 7x - 10 Back to section Title page

Point Slope Form Here is a website to visit for more practice on Point Slope form Back to section Title page

Finding equation of the line given graph
In this section you will review how to find the equation of the line given the graph of the line Back to section Title page

Finding equation of the line given graph
y = mx + b Find the y-intercept (b) b = -3 Find the slope(m) 3 7 m = 3 7 (substitute for m and b) y = 3 7 𝑥−3 Back to section Title page

Finding equation of the line given graph
y = mx + b Find the y-intercept (b) b = 4 -2 Find the slope(m) 7 -2 m = 7 (substitute for m and b) y = −2 7 𝑥+4 Back to section Title page

The equation is y = 2 Finding equation of the line given graph y-axis
This is a horizontal line. It goes through the y-axis at 2. The equation is y = 2 Back to section Title page

The equation is x = 5 Finding equation of the line given graph x-axis
This is a vertical line. It goes through the x-axis at 5. x-axis The equation is x = 5 Back to section Title page

Finding the equation of given two points
In this section you will review how to find the equation of a line when you are given two points on the line. Back to section Title page

Finding the equation of given two points
Given: (2,3) and (5,8) 1st Find the slope between the two points 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 𝑚= 8 −3 5 −2 𝑚= 5 3 2nd Use the slope and one of the points 𝑚= 5 3 and (2,3) 𝑏=− 1 3 y = mx + b (substitute for m and b) 3= 𝑏 𝑦= 5 3 𝑥 − 1 3 3= 𝑏 Back to section Title page

Standard form of equations of lines
In this section you will review the standard form of the equation of a line. You will also learn how to find the x and y intercepts. You will also change from slope intercept to standard form and from standard form to slope intercept. Back to section Title page

Ax + By = C The standard form of the equation of
a line has both x and y on the same side and the constant on the other with no fractions present. Ax + By = C A, B, and C must be integers The coefficient of x (A) must be positive (and must be the first term in equation) C is the constant term Back to section Title page

Plot the x and y intercepts and connect with the line.
Finding the intercepts from standard form 3x + 4y = 12 To find the x intercept Let y = 0 To find the y intercept Let x = 0 3x + 4(0) = 12 3(0) + 4y = 12 3x = 12 4y = 12 x = 4 y = 3 Plot the x and y intercepts and connect with the line. Back to section Title page

Changing from standard to slope intercept
Get y on a side by itself 3x + 4y = 16 -3x -3x 4y = 16 – 3x 4 4 4 Slope Intercept form 𝑦=− 3 4 𝑥+4 Back to section Title page

Multiply each term by 3 to remove the fraction
Changing from slope intercept to standard Multiply each term by 3 to remove the fraction 𝑦= 2 3 𝑥−8 (3)𝑦= 𝑥−(3)8 3𝑦=2𝑥−24 -3y y +24 2x – 3y = 24 24 = 2x – 3y Back to section Title page

Parallel Lines In this section you will review how to find the equations of parallel lines. You will review how to find parallel lines to a given line and when you are given a slope. Back to section Title page

Parallel lines do not intersect.
Parallel lines have the same slope Both lines have a slope of 𝑚= −7 9 The arrows on the lines indicate that the lines are parallel Back to section Title page

Line parallel to given line through the given point
You will now use the given point and the slope you found to find the next point (5,8) -3 4 -3 4 Continue to the next page to find the new line’s equation You must first find the slope of (red)given line -3 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = 4 Back to section Title page

You must use the given point(5,8) the slope of (red)given line
Finding Parallel Line Equation (5,8) You must use the given point(5,8) the slope of (red)given line -3 4 -3 4 5= −15 4 +𝑏 35 4 = 𝑏 𝑎𝑛𝑑 𝑚= −3 4 y = mx + b (Replace for m and b) 𝑚= −3 4 and point (5,8) 𝑦= −3 4 𝑥 5= − 𝑏 Back to section Title page

Line parallel to given line through the given point
Find the equation of the line that is parallel to the line y = 2x + 7 going through the point (3,4) Parallel lines have the same slope so you must find the slope of the given line Given m = 2 so the parallel m = 2 Parallel m = 2 Point is (3,4) use y = mx + b 4 = 2(3) + b 4 = 6 + b -2 = b (substitute for m and b) So parallel line is y = 2x - 2 Back to section Title page

Line parallel to given line through the given point
Parallel 𝑚= −4 5 through (-7,8) Use y = mx + b 8= −4 5 −7 +𝑏 8= 𝑏 12 5 =𝑏 (substitute for m and b) 𝑦= −4 5 𝑥+ 12 5 Back to section Title page

Line parallel to given line through the given point
Find the equation of the line parallel to 7x + 2y = 10 and going through point (3,-5) 1st Find the slope from given equation 𝑚= −7 2 point (3,-5) y = mx + b −5= − 𝑏 7x + 2y = 10 −5= − 𝑏 2y = 10 – 7x 𝑦=5 − 7 2 x 11 2 =𝑏 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚= −7 2 (substitute for m and b) 𝑦= −7 2 𝑥 Back to section Title page

All horizontal lines are parallel
y = 6 y = 3 y = -4 Back to section Title page

y = -2 Find the equation of the line parallel to the
given line(red) through the given point(blue) y = 7 y = -2 Given (0,-2) Back to section Title page

All vertical lines are parallel
x = -8 x = -1 x = 3 Back to section Title page

x = 4 Find the equation of the line parallel to the
given line(red) through the given point(blue) Given (4,3) x = 4 x = -5 Back to section Title page

Perpendicular Lines In this section you will review how to find the equations of perpendicular lines. You will review how to find perpendicular lines to a given line and when you are given a slope. Back to section Title page

Perpendicular lines have slopes that are negative reciprocals.
3 5 Blue Line Red Line 𝑚= 5 3 -3 𝑚=− 3 5 5 Back to section Title page

If you multiply the slopes of two lines together and the result is -1, then the two lines are perpendicular. 1st Line 2nd Line 𝑚= 5 3 𝑚=− 3 5 5 3 − = -1 Back to section Title page

Finding Perpendicular Line through given point
Black Line 𝑚= 5 8 Perpendicular Line 𝑚=− 8 5 Back to section Title page

Finding equation of line perpendicular to given line through the given point
𝐺𝑖𝑣𝑒𝑛: 𝑦=4𝑥+8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5) Slope of given line m = 4 so the perpendicular slope is 𝑚=− and point (2,5) y = mx + b 5=− 𝑏 5=− 1 2 +𝑏 (substitute for m and b) 11 2 =𝑏 y=− 1 4 𝑥+ 11 2 Back to section Title page

𝐺𝑖𝑣𝑒𝑛: 𝑦= 3 4 𝑥−5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
Finding equation of line perpendicular to given line through the given point 𝐺𝑖𝑣𝑒𝑛: 𝑦= 3 4 𝑥−5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6) 𝐺𝑖𝑣𝑒𝑛 𝑚= 𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚=− 4 3 𝑚=− and point (-2,6) y = mx + b 6=− 4 3 −2 +𝑏 (substitute for m and b) 6= 𝑏 y=− 4 3 𝑥+ 10 3 10 3 =𝑏 Back to section Title page

1st Find the slope from given equation
Finding equation of line perpendicular to given line through the given point Given : 3x + 4y = 7 and going through (4,-6) 1st Find the slope from given equation Use 𝑚= 𝑎𝑛𝑑 (4.−6) y = mx + b −6= 𝑏 4y = 7 – 3x −6= 𝑏 𝑦= 𝑥 −6= 𝑏 −34 3 = 𝑏 𝐺𝑖𝑣𝑒𝑛: 𝑚=− 3 4 (substitute for m and b) 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 :𝑚= 4 3 𝑦= 4 3 𝑥 − 34 3 Back to section Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue) y = 6 (5,-3) Perpendicular Line x = 5 Back to section Title page

y = 3 (-4,3) x = 5 Find the equation of the line perpendicular to the
given line(red) through the given point(blue) y = 3 (-4,3) x = 5 Back to section Title page

Applications In this section you will find sample problems that are applications of equations of lines. Back to section Title page

Tennis Lessons Click here for answer The graph above charts the cost of tennis lessons at a local tennis club. The cost includes a one time membership fee and price per hour of the lessons 1. Find the linear equation that fits given chart. Back to section Title page

Tennis Lessons-continued
Click here for answer 2. How much will the total cost for 52 lessons be? Back to section Title page

Tennis Lessons-continued

Linear Equation that fits the chart. Two points that are on the line are (0,150) and (15,300) so you can find the slope of the line. Return to problem m = 300 − −0 = =10 (This is the cost per lesson.) The y intercept(b) is 150. (This is the one time membership fee.) The Linear Equation is y = 10x + 150 2. How much will the total cost for 52 lessons be? Return to problem Substitute 52 in for x in y = 10x + 150 y = 10 (52) + 150 y = 670 so 52 lessons will cost \$670 3. If John has \$850, how many lesson can he buy? Substitute 850 for y in y = 10x + 150 850 = 10x + 150 700 = 10x x = 70 so \$850 will pay for 70 lessons Return to problem

Solving Systems of Linear Equations
Topics Solving systems graphically Solving systems by using tables Solving systems algebraically Substitution method Elimination method Systems with more than 2 variables Using matrix equations Using augmented matrices Applications of systems

Graphical Representation of Linear Systems of Equations

Graphical Representation of Systems of Linear Equations

Graphical Representation of Systems of Linear Equations

Graphical Representation of Systems of Linear Equations

Graphical Representation of Systems of Linear Equations

Solving Systems Graphically
Click this link for a video of solving systems of linear equations graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear equations graphically.

Solving Systems of Linear Equations using tables
Another method you can use to find the solution to a system of linear equations is to make a table for each of the equations and then compare to see if any of the ordered pairs of solutions is in both tables of values. Example: y= 3x and y= x+6 x 3x+2 2 1 5 8 3 11 4 14 17 6 20 X x+6 6 1 7 2 8 3 9 4 10 5 11 12 Solution

Solving Systems of Linear Equations using tables
Using tables of values to find a solution to a system of equations is very inefficient. About the only time that you would use this method is if the tables have already been generated for you. Then you would just inspect the ordered pairs to find the same values in each table, if they happen to be listed. Example: x y -3 -10 -2 -8 -1 -4 1 2 3 x y -3 5 -2 16/3 -1 14/3 4 1 10/3 2 8/3 3 The solution is:(3,2)

Solving Systems of Linear Equations using tables
However, with today’s technology. Your calculator can generate tables of values very quickly. From the graph above, you can tell that the x-coordinate of the intersection is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear Equations using tables
Using your calculator you can refine the x-coordinates in you table to begin to get closer to the actual point of intersection. Notice in the table after several refinements to the x-coordinate, we have the point of intersection rounded to 4 decimal places ( , ).

Solving Systems of Linear Equations algebraically
Using Substitution Using Elimination Systems with more than 2 variables Using Matrix Equations- If you have never worked with matrices, then skip this section Using reduced row echelon form of augmented matrices- If you have never worked with matrices, then skip this section.

Solving Systems of Linear Equations algebraically
Substitution Method The concept of solving systems of linear equations using substitution is this: Take one of the equations and isolate one of the variables on one side of the equation (Get a variable by itself on one side of the equals sign). Now use the other equation and wherever the isolated variable is, substitute the expression that equals that variable from the first equation for that variable. Now you have an equation with only one variable. Solve this equation using methods demonstrated in the “Solving Linear Equations” section of this powerpoint. Last, take the value solved-for from the previous step, substitute this value into either equation and then solve for the other variable. Your answer will typically be written as an ordered pair of numbers

Substitution Method: Example
Solving Systems of Linear Equations algebraically Substitution Method: Example To solve a system of linear equations using substitution: Isolate one of the variables on one side of one of the two equations. 2x + 5y = x + 4y = 2 x = 2 - 4y Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable. 2(2 - 4y) + 5y = 7 Now simplify and solve the equation for the remaining variable. 4 – 8y + 5y = 7 4 – 3y = 7 -3y = 3 Y = -1 Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that equation. X + 4y = 2 X + 4(-1) = 2 X – 4 = 2 X = 6 Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order. (x,y) = (6, -1)

Click the link below to see a video using the substitution method
Solving Systems of Linear Equations algebraically Click the link below to see a video using the substitution method Note: this video begins with a review of solving systems graphically. If you drag the slider at the bottom of the video to minute 1:37, that is when substitution begins.

Practice problems using substitution
Solving Systems of Linear Equations algebraically Practice problems using substitution Note: I find this site to be better for just checking your answers to a system of equations, rather than trying to follow their solution. It is correct, but it may be difficult to follow all of their steps.

Solving Systems of Linear Equations algebraically
Elimination Method Simplify each equation so that each equation is in standard form (ax + by = c). Place the two equations so that all the like terms are in the same position and align them vertically. Example: x + 3y = 7 4x – 3y = 11 If you are lucky you will be able to either add or subtract the two equations with one of the variables disappearing and what remains is an equation with only one variable. Example: x + 3y = 7 Add vertically: 6x = 18 Solve for x: x = 3 Now substitute “3” for x into either of the two original equations and solve for y. 2 (3) + 3y = 7 (the first equation from above) 6 + 3y = 7 3y = 1 Y=1/3 Solution: ( 3, 1/3)

Solving Systems of Linear Equations algebraically
Elimination Method Often once the equations have been simplified into standard form and placed vertically aligning all the like terms, you will not be able to just and or subtract the two equations to make one of the variables disappear. Notice below, if you add or subtract the two equations you still have two variables. This is not good. Example: x + 3y = x + 3y = 9 4x – 5y = x – 5y = 9 add: x - 2y = subtract: 8x +8y = 0 When this happens, you need to multiply (or divide) every term of one or both of the equations by something to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add the two equations to eliminate that variable). I have found that my students tend to make many less careless errors when adding than when subtracting. This is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other. Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom equation). So we want to multiply the second equation by “-3”. Example: x + 3y = x + 3y = 9 -3(4x – 5y) = -3(9) -12x + 15y = -27 Add like terms vertically: y = -18 Solve for x: y = -1 Now substitute “-1” for y into either of the two original equations and solve for x. 4x – 5(-1) = 9 (I chose the second equation from above) -4x = 9 -4x = 4 x= -1 Solution: ( -1, -1)

Solving Systems of Linear Equations algebraically
Elimination Method Here is an example where you need to do even more work to prepare the equations before adding (or subtracting) them. 2x + 5y = 14 3x – 2y = First decide which variable to eliminate, x or y. You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y to be 10 and Your choice, either way works. I’m going to chose to force the coefficient on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then force the coefficient on y in the second equation to be -10 by multiplying both sides of the second equation by 5. 2(2x) + 2(5y) = 2(14) x + 10y = 5(3x) – 5(2y) =5(-36) 15x - 10y = -180 Now add: x = Solve for x: x = -8 Now solve for Y: 2(-8) +5y =14 y = 14 5y = 30 y = 6 Solution: (-8, 6)

Elimination Method: Click the link below to watch a video.
Solving Systems of Linear Equations algebraically Elimination Method: Click the link below to watch a video.

Solving Systems of Linear Equations algebraically
Elimination Method: Click the link below for practice problems and solutions for solving linear systems using elimination.

System of linear equations with 3 variables
Solving Systems of Linear Equations algebraically System of linear equations with 3 variables If you have an equation with three variables, this is still called a linear equation, but the graph of it is not a line. Since there are three variables and each variable represents a dimension, this means our graph will be a plane in 3-D. If you have a system of three equations representing three planes, you could have the intersection at one point such as the intersection of 2 corner walls and the ceiling in a room. This is the ideal situation because we will get an ordered triple (x,y,z) for the solution. Other times we could get a line for the intersection of three planes. Think of the pages of a book with the pages representing planes, and the intersection of these planes is a line representing the binding of the book. We could have three parallel planes that never intersect and have no points of intersection. We are not going to consider all of the possibilities in this document.

System of linear equations with 3 variables
Solving Systems of Linear Equations algebraically System of linear equations with 3 variables *We have one-point of intersection(consistent & independent) a point, *We have an infinite set of points of intersection (consistent and dependent) a line or a plane(not shown) *We have no points of intersection (inconsistent). *Below are three possible solutions for three planes.

Elimination Method: system with 3 variables
Solving Systems of Linear Equations algebraically Elimination Method: system with 3 variables This is a good video showing how to solve a system of three equations with three variables.

Elimination Method: system with 3 variables
Solving Systems of Linear Equations algebraically Elimination Method: system with 3 variables

Elimination Method: system with 3 variables
Solving Systems of Linear Equations algebraically Elimination Method: system with 3 variables

Elimination Method: system with 3 variables
Solving Systems of Linear Equations algebraically Elimination Method: system with 3 variables

Solving Systems of Linear Equations algebraically
Matrix Equations Write this system using matrix equations: 2x + y + 3z = 2 x + y + 8z = 2 x + y + z = 3 This system is already in standard form. Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix. Notice that the coefficient on a variable is understood to be one if no other number is multiplied by the variable.

Elimination Method: system with 3 variables
Solving Systems of Linear Equations algebraically Elimination Method: system with 3 variables

Solving Systems of Linear Equations algebraically
Matrix Equations: This section assumes that you have had some experience working with matrices. If you have not, then skip this part. Write this system using matrix equations: 2x + y + 3z = 2 x + y + 8z = 2 x + y + z = 3 First the equations need to be in standard form: AX + BY = C (If there are two variables X & Y.) AX + BY + CZ = D (if there are three variables, X, Y and Z.) AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.) and so on.

Solving Systems of Linear Equations algebraically
Matrix Equations Next multiply this matrix by the 3 by 1 variable matrix and set this matrix equation equal to the 3 by 1 matrix created from the constants in the original equations.

Solving Systems of Linear Equations algebraically
Matrix Equations Note: How to find inverses of matrices is not included in this tutorial. However, if you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear Equations algebraically
Matrix Equations

Solving Systems of Linear Equations algebraically
Matrix Equations

Solving Systems of Linear Equations algebraically
Matrix Equations

Matrix Equations: Practice problems
Solving Systems of Linear Equations algebraically Matrix Equations: Practice problems This is the same link from earlier. The only difference is that you now may want to solve these problems using matrix equations. Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear Equations algebraically
Augmented Matrices: You may want to skip this part of the unit. This may be beyond the scope of this unit. This is a quick way to solve systems, but in no way does it cover all of the concepts needed to understand and use matrices. Note: The second equation has no “Y”, so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear Equations algebraically
Augmented Matrices

Solving Systems of Linear Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices: Click below to watch a video using augmented matrices to solve a 3 by 3 system of linear equations. This video is worth watching even if you don’t want to know how to solve systems by hand using augmented matrices. The person in the video clearly loves what he is doing.

Augmented Matrices using your graphics calculator
Solving Systems of Linear Equations algebraically Augmented Matrices using your graphics calculator

Solving Systems of Linear Equations algebraically
Augmented Matrices

Solving Systems of Linear Equations algebraically
Augmented Matrices

Solving Systems of Linear Equations algebraically
Augmented Matrices

Solving Systems of Linear Equations algebraically
Augmented Matrices

Why study systems of linear equations?
Applications of Systems of Linear Equations Why study systems of linear equations? Systems of linear equations are everywhere. When studying mathematics, economics, sciences, business, etc. you will encounter situations where needing to answer a problem will involve solving a system of equations. Examples: Rate of work problems, Percent mixture problems, Rate-time-distance problems, Problems finding when the cost of multiple scenarios is the least, money and age problems, etc. Solving systems is a nice skill to have, but it even more valuable when you have the ability to use that skill to set-up and solve real world problems.

Application of Linear Sytems
Applications of Systems of Linear Equations Application of Linear Sytems Suppose the local market place sells a fruit smoothie for \$2.50 and a cup of hot chocolate for \$ On Thursday, Malinda sold 30 more fruit smoothies than hot chocolates for a total of \$ worth of fruit smoothies and hot chocolate. How many cups of each did Malinda sell? First, we need to underline the important information. Suppose the local market place sells a fruit smoothie for \$2.50 and a cup of hot chocolate for \$ On Thursday, Malinda sold 30 more fruit smoothies than hot chocolates for a total of \$ worth of fruit smoothies and hot chocolate. How many cups of each did Malinda sell?

Application of Linear Systems- continued
Applications of Systems of Linear Equations Application of Linear Systems- continued Second, we must define our variables. Let “c” represent the number of cups of hot chocolate and “s” represent the number of fruit smoothies. Next, write equations to represent the underlined information from step 1. c = # of cups of hot chocolate for \$2.00 s = # of fruit smoothie for \$2.50 s = 30 + c (30 more than hot chocolates) c + s = \$ (for a total of \$ worth of fruit smoothies and hot chocolate) c s = total number of cups sold # of cups of # of smoothies hot chocolate

Application of Linear Systems- continued
Applications of Systems of Linear Equations Application of Linear Systems- continued c = # of cups of hot chocolate for \$2.00 s = # of fruit smoothies for \$2.50 s = 30 + c c + s = total number of cups sold \$2c + \$2.50s = \$ (for a total of \$ worth of fruit smoothies and hot chocolate) 2 c (30 + c ) = Solve for c 2c (30) c = Use the distributive property 2c c + 75 = \$ Simplify 4.50c + 75= Combine like terms 4.50c = Subtract 75 4.50c = Divide by 4.50 c = Number of cups of hot chocolate

Application of Linear Systems- continued
Applications of Systems of Linear Equations Application of Linear Systems- continued How many cups of each did Malinda sell? c = 23 cups of hot chocolate s = 30 + c (30 more smoothies than hot chocolates) s = s = 53 fruit smoothies Malinda sold 23 cups of hot chocolate and 53 fruit smoothies Check your answer: \$2 (23) + \$2.50 (53) = \$178.50 \$ \$ = \$178.50

Rate-Time Distance Application
Applications of Systems of Linear Equations Rate-Time Distance Application

Rate-Time Distance Application
Applications of Systems of Linear Equations Rate-Time Distance Application

Rate-Time Distance Application
Applications of Systems of Linear Equations Rate-Time Distance Application

Rate-Time Distance Application
Applications of Systems of Linear Equations Rate-Time Distance Application

Rate-Time Distance Application
Applications of Systems of Linear Equations Rate-Time Distance Application

Click the links below for practice problems
Applications of Systems of Linear Equations Click the links below for practice problems This site gives 6 problems to solve followed by a second page of the solutions. It is good for practice. This site shows 3 examples of applications. The first example involves money and tickets, the second example beginning at minute 4:00 is a percent mixture problem, and the third example beginning at minute 8:00 is rate-time-distance problem.

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