Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 2

Algebra 2 Solving Systems of Equations by Graphing

Name: Algebra 2: Solving Systems of Equations by Graphing
Brand: Educator.com
Price: 35 USD
Availability: InStock

Section 3: Systems of Equations and Inequalities: Lecture 1 | 17:13 min

Lecture Description

Next Lecture

Previous Lecture

Discussion
Answer Engine
Study Guides
Practice Questions
Download Lecture Slides
Table of Contents
Transcription
Related Books

Lecture Comments (3)

0 answers

Post by Alex Leung on August 25, 2020

At 4:02, I don't understand why the function is called a "dependent system." Is it because the points depend on the function?

1 answer

Last reply by: Ak Liu
Tue Jul 28, 2020 2:58 PM

Post by Ak Liu on July 28, 2020

On example 1, isn't the answer (1,2)? because I did the math, and the y-intercept is 1... But in your graph, it's -1.

Answer EngineGet answers to any question!Ask any question related to Algebra 2

Working on the solution...

Solving Systems of Equations by Graphing

You can solve a system of equations by graphing the equations and finding the point of intersection of the two graphs.
If the graphs are parallel lines, there is no point of intersection and no solution to the system.
If the graphs are the same lines, the system has an infinite number of solutions.
If the graphs intersect at one point, the system has a unique solution.

Solving Systems of Equations by Graphing

Solve by graphing
y = − 3x − 4
y = [1/2]x + 3

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept.
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:

ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - Y - Intercept + Slope
When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Given that the system is given in slope - intercept form, use Y - intercept + slope method.
Step 1: Find the y - intercep for equation 1 and plot it
b =
b = − 4, the point is (0, − 4)
Step 2: Find the slope for equation 1 and plot the next point starting from the y - intercept (0, − 4)
m = [rise/run] =
m = [rise/run] = [( − 3)/1] or [3/( − 1)]
Step 3: Draw the line. Make line extends your entire graph.
Step 4: Find the y - intercep for equation 2 and plot it
b =
b = 3, the point is (0,3)
Step 5: Find the slope for equation 2 and plot the next point starting from the y - intercept (0,3)
m = [rise/run] =
m = [rise/run] = [1/2] or [( − 1)/( − 2)]
Step 6: Draw the line. Make line extends your entire graph.
Step 7: Find the point of intersection.
Solution = ( , )
Solution = ( − 2,2)

Solve by graphing
y = − [1/3]x + 1
y = [2/3]x + 4

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form :
ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - Y - Intercept + Slope
When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Given that the system is given in slope - intercept form, use Y - intercept + slope method.
Step 1: Find the y - intercep for equation 1 and plot it
b =
b = 1, the point is (0,1)
Step 2: Find the slope for equation 1 and plot the next point starting from the y - intercept (0,1)
m = [rise/run] =
m = [rise/run] = [( − 1)/3] or [1/( − 3)]
Step 3: Draw the line. Make line extends your entire graph.
Step 4: Find the y - intercep for equation 2 and plot it
b =
b = 4, the point is (0,4)
Step 5: Find the slope for equation 2 and plot the next point starting from the y - intercept (0,3)
m = [rise/run] =
m = [rise/run] = [2/3] or [( − 2)/( − 3)]
Step 6: Draw the line. Make line extends your entire graph.
Step 7: Find the point of intersection.
Solution = ( , )
Solution = ( − 3,2)

Solve by graphing
y = − [3/2]x − 2
y = x + 3

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b
y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept
3. Draw the line with ruler, find the intersection

You would have to re-write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:
ax + by = c
a₂x + b₂y = c₂
Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - " Y - Intercept + Slope"
When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Given that the system is given in slope - intercept form, use Y - intercept + slope method.
Step 1: Find the y - intercep for equation 1 and plot it
b =
b = − 2, the point is (0, − 2)
Step 2: Find the slope for equation 1 and plot the next point starting from the y - intercept (0, − 2)
m = [rise/run] =
m = [rise/run] = [( − 3)/2] or [3/( − 2)]
Step 3: Draw the line. Make line extends your entire graph.
Step 4: Find the y - intercep for equation 2 and plot it
b =
b = 3, the point is (0,3)
Step 5: Find the slope for equation 2 and plot the next point starting from the y - intercept (0,3)
m = [rise/run] =
m = [rise/run] = [1/1] or [( − 1)/( − 1)]
Step 6: Draw the line. Make line extends your entire graph.
Step 7: Find the point of intersection.
Solution = ( , )
Solution = ( − 2,1)

Solve by graphing
y = − [5/4]x − 2
y = − [5/4] + 2

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b
y₂ = m₂x + b
Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept.
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope. ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - "Y - Intercept + Slope"
When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Given that the system is given in slope - intercept form, use Y - intercept + slope method.
Step 1: Find the y - intercep for equation 1 and plot it
b =
b = − 2, the point is (0, − 2)
Step 2: Find the slope for equation 1 and plot the next point starting from the y - intercept (0, − 2)
m = [rise/run] =
m = [rise/run] = [( − 5)/4] or [5/( − 4)]
Step 3: Draw the line. Make line extends your entire graph.
Step 4: Find the y - intercep for equation 2 and plot it
b =
b = 2, the point is (0,2)
Step 5: Find the slope for equation 2 and plot the next point starting from the y - intercept (0,2)
m = [rise/run] =
m = [rise/run] = [( − 5)/4] or [5/( − 4)]
Step 6: Draw the line. Make line extends your entire graph.
Step 7: Find the point of intersection.
Solution = ( , )
Solution = Parallel Lines do not intersect, therefore, no solution.

Solve by graphing
y = [7/2]x + 3
y = [1/2]x − 3

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b
y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:
ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for ÿ". In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - "Y-Intercept + Slope"
When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Given that the system is given in slope - intercept form, use Y - intercept + slope method.
Step 1: Find the y - intercep for equation 1 and plot it
b =
b = 3, the point is (0,3)
Step 2: Find the slope for equation 1 and plot the next point starting from the y - intercept (0,3)
m = [rise/run] =
m = [rise/run] = [7/2] or [( − 7)/( − 2)]
Step 3: Draw the line. Make line extends your entire graph.
Step 4: Find the y - intercep for equation 2 and plot it
b =
b = − 3, the point is (0, − 3)
Step 5: Find the slope for equation 2 and plot the next point starting from the y - intercept (0, − 3)
m = [rise/run] =
m = [rise/run] = [1/2] or [( − 1)/( − 2)]
Step 6: Draw the line. Make line extends your entire graph.
Step 7: Find the point of intersection.
Solution = ( , )
Solution = ( − 2, − 4)

Solve by graphing
x − 2y = − 2
x + 2y = − 6

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b
y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept.
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:
ax + by = c
a₂x + b₂y = c₂

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - "Y-Intercept + Slope"

When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Given that the system is given in Standard Form, use X - intercept Y - Intercept method.
Step 1: Find the x - intercep for equation 1 by eliminating the y term, solve for x if necessary. Plot it.
x − 2y = − 2x = − 2
x - intercpet: ( − 2,0)
Step 2: Find the y - intercept for equation 1 by eliminating the x - term, solve for y if necessary. Plot it.
x − 2y = − 2 − 2y = − 2
y = 1
y - Intercept:(0,1)
Step 3: Draw the line through the x - and y - intercepts.
Step 4: Find the x - intercep for equation 2 by eliminating the y term, solve for x if necessary. Plot it.
x + 2y = − 6x = − 6
x - intercpet: ( − 6,0)
Step 5: Find the y - intercept for equation 2 by eliminating the x - term, solve for y if necessary. Plot it.
x + 2y = − 62y = − 6
y = − 3
y - Intercept:(0, − 3)
Step 6: Draw the line through the x - and y - intercepts.
Step 7: Find the point of intersection
Solution:( − 4, − 1)

Solve by graphing
x + y = 2
x + y = − 3

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form :
y = mx + b
y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept.
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:
ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - "Y-Intercept + Slope"

When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Given that the system is given in Standard Form, use X - intercept Y - Intercept method.
Step 1: Find the x - intercep for equation 1 by eliminating the y term, solve for x if necessary. Plot it.
x + y = 2x = 2
x - intercpet: (2,0)
Step 2: Find the y - intercept for equation 1 by eliminating the x - term, solve for y if necessary. Plot it.
x + y = 2y = 2
y = 2
y - Intercept:(0,2)
Step 3: Draw the line through the x - and y - intercepts.
Step 4: Find the x - intercep for equation 2 by eliminating the y term, solve for x if necessary. Plot it.
x + y = − 3x = − 3
x - intercpet: ( − 3,0)
Step 5: Find the y - intercept for equation 2 by eliminating the x - term, solve for y if necessary. Plot it.
x + y = − 3y = − 3
y = − 3
y - Intercept:(0, − 3)
Step 6: Draw the line through the x - and y - intercepts.
Step 7: Find the point of intersection
Since the lines are parallel, there is no solution.

Solve by graphing
x − 4y = 4
x + 2y = − 8

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b
y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept.
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:
ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left-"Y-Intercept + Slope"

When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x- and y-intercept and locate the point of intersection.
Given that the system is given in Standard Form, use X - intercept Y - Intercept method.
Step 1: Find the x - intercep for equation 1 by eliminating the y term, solve for x if necessary. Plot it.
x − 4y = 4x = 4
x - intercpet: (4,0)
Step 2: Find the y - intercept for equation 1 by eliminating the x - term, solve for y if necessary. Plot it.
x − 4y = 4 − 4y = 4
y = − 1
y - Intercept:(0, − 1)
Step 3: Draw the line through the x - and y - intercepts.
Step 4: Find the x - intercep for equation 2 by eliminating the y term, solve for x if necessary. Plot it.
x + 2y = − 8x = − 8
x - intercpet: ( − 8,0)
Step 5: Find the y - intercept for equation 2 by eliminating the x - term, solve for y if necessary. Plot it.
x + 2y = − 82y = − 8
y = − 4
y - Intercept:(0, − 4)
Step 6: Draw the line through the x - and y - intercepts.
Step 7: Find the point of intersection
Solution ( − 4, − 2)

Solve by graphing
x + 2y = 2
x + 2y = 4

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b
y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept.
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:
ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - "Y-Intercept + Slope"

When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x-and y-intercept and locate the point of intersection.
Given that the system is given in Standard Form, use X - intercept Y - Intercept method.
Step 1: Find the x - intercep for equation 1 by eliminating the y term, solve for x if necessary. Plot it.
x + 2y = 2x = 2
x - intercpet: (2,0)
Step 2: Find the y - intercept for equation 1 by eliminating the x - term, solve for y if necessary. Plot it.
x + 2y = 22y = 2
y = 1
y - Intercept:(0,1)
Step 3: Draw the line through the x - and y - intercepts.
Step 4: Find the x - intercep for equation 2 by eliminating the y term, solve for x if necessary. Plot it.
x + 2y = 4x = 4
x - intercpet: (4,0)
Step 5: Find the y - intercept for equation 2 by eliminating the x - term, solve for y if necessary. Plot it.
x + 2y = 42y = 4
y = 2
y - Intercept:(0,2)
Step 6: Draw the line through the x - and y - intercepts.
Step 7: Find the point of intersection
Since the lines are parallel, there is no solution.

Solve by graphing
− 2x − 2y = 6
6x + y = 2

There are three ways to solve system of equations by graphing. Choose the method that best fits the system of equation given.
Slope Intercept Form:
y = mx + b
y₂ = m₂x + b

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

This is the best method to use when the system of equations is in this format.
1. Graph the y - intercept b, @ (0,b)
2. Use the slope m = [rise/run] to graph additional points from the y - intercept.
3. Draw the line with ruler, find the intersection

You would have to re - write the equations into standard form. Use this method if you don't like graphing using y - intercept and slope.
Standard Form:
ax + by = c
a₂x + b₂y = c₂

Use the same values of x for both equations. Graph it. With a ruler, connect the points and find the solution if there's any.

In order to use this method when the system of equations is in Standard Form, you must solve for y. In other words, re - write the system into slope - intercept form.

This is the best method to use if and only if you do not get fractions when finding the x - and y - intercepts. If you get fractions, the best way is to solve for y and use the method on the left - "Y-Intercept + Slope"

When using this method, you will get four points (x,0) and (0,y) from equation 1 and (x,0), (0,y) from equation 2. Carefully draw the lines through the x - and y - intercept and locate the point of intersection.
Eventhough this system of equations is in standard form, you should notice right away that there will be a fraction on equation two when solving for the x - intercept.
To avoid fractions, solve this system of equations using a table of values.
Step 1:Create two tables of values. Choose 4 points to solve for y.
x 2x-2y=6
-1
0
1
2
x 2x-2y=6
-1 -2
0 -3
1 -4
2 -5
x 6x+y=2
-1
0
1
2
x 6x+y=2
-1 8
0 2
1 -4
2 -10
Step 2: Plot the points. Draw the lines through the points.
Step 3: Locate the point of intersection
Solution (1, − 4)

« Previous Question | Next Question »

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Solving Systems of Equations by Graphing

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Systems of Equations

Example: Two Equations

Solving by Graphing

Point of Intersection

Types of Systems

Example 1: Solve by Graphing

Example 2: Solve by Graphing

Example 3: Solve by Graphing

Example 4: Solve by Graphing

Intro 0:00
Systems of Equations 0:09

Example: Two Equations

Solving by Graphing 0:53

Point of Intersection

Types of Systems 2:29

Independent (Single Solution)
Dependent (Infinite Solutions)
Inconsistent (No Solution)

Example 1: Solve by Graphing 5:20
Example 2: Solve by Graphing 9:10
Example 3: Solve by Graphing 12:27
Example 4: Solve by Graphing 14:54

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Solving Systems of Equations by Graphing

Welcome to Educator.com.0000

In today's Algebra II lesson, we will be discussing solving systems of equations by graphing.0002

Recall that a system of equations, for our purposes, is defined as two equations and two variables.0009

Later on, we will be looking at systems of equations involving more than two variables.0016

But right now, we are just going to stick to this definition.0021

For example, a system of equations could be something such as 2x - y = 7, when it is considered along with another equation, such as x + 3y = -4.0025

And a solution to a system of equations would be a value for x and a value for y--a set of values that satisfies both of the equations.0040

The first technique we are going to discuss in solving systems of equations is solving by graphing.0053

In addition, later on, we will be talking about how these systems can be solved algebraically.0059

So, one way to solve is to graph each equation: the solution for the system is the point of intersection of the two graphs.0065

And we will do some examples in a minute; but for now, imagine that you are given an equation,0075

and you figure out some points, or one of the points in the slope, and it turns out that this is the graph of the line described by the equation.0081

And you are given another equation as part of that system; and you go ahead and graph that one out.0098

And it comes out to a line that looks like this.0105

The solution for the system is the point of intersection, so this is the solution.0110

The x- and y-coordinate for this point are the solution.0117

Now, here you can see a weakness in this method: and that is that, if the solution doesn't land right on an integer, it is difficult to get an exact value.0121

And that is one advantage to the algebraic methods.0136

But if you graph carefully, and the solution does land on, say, (3,4), then you could get an exact solution.0138

There are several types of systems of equations; we call a system independent if it has exactly one solution.0150

For example, just as I showed you in the previous slide, sometimes you will graph out a system, and usually,0159

in the problems you will be doing, you will see a single point of intersection, which is the solution--the single solution.0167

And this set of lines describes an independent system.0175

Other times, you will go along, and you will graph out the system of equations.0185

You have graphed the first equation, and perhaps the graph looks like this.0191

Then, you go to graph the second equation, and it turns out that it describes the same line.0196

So, if the system of equations--both equations within that system describe the same line, then there would be an infinite number of solutions.0202

And the reason is that every single point along this line, every single set of (x,y) coordinates is an intersection of the two systems.0216

The line...every single point along this line (and there is an infinite number of points along the line)0226

is the set of solutions, the set of (x,y) values where the two lines intersect, since they are the same line.0233

Here, we call this a dependent system; here we have one solution; here an infinite number of solutions; and this is a dependent system.0239

The third possibility is that there are no solutions; and we call this an inconsistent system.0254

So, thinking about a situation where there would not be any solution, where there is no point of intersection: it would be a set of parallel lines.0263

Drawing this over here for clarity: here is my x and y axis; and if I graphed the first line, the first linear equation,0273

and then I went ahead and graphed the second one, and those two turned out to have the same slope, those would be parallel lines.0284

They are never going to intersect; therefore, we can see on the graph: there is no solution.0292

So, no solution--that is an inconsistent system.0298

So, there are three possibilities: one solution--intersection at one point; an infinite number of solutions--0306

all the points along the line; or no solution, because it is a set of parallel lines that do not intersect.0312

The first example gives us a system of equations with two variables: x + y = 3 and x - y = 1.0322

So, graphing this first line, I am going to start with x + y = 3; and I find a few x and y values, some ordered pairs, so I can do the graphing.0331

First, I am going to let x equal 0; well, if x is 0, I can see easily that y is 3.0343

So, that gives me the y-intercept; next, I am going to let y equal 0 to find the x-intercept.0351

So, this would be 0; x would be 3; and just one more point to make it a better graph0361

(because it is especially important when you are not just graphing one line--you are actually looking0367

for a solution to a system of equations--to have a really good graph, so you can find a point of intersection accurately):0372

so, when x is 1, x plus y equals 3; so when x is 1, y equals 2.0378

So, I am going to graph this line: first, I have the y-intercept at 3; I have the x-intercept at 3; and another point in between those--x is 1; y is 2.0387

And this gives me a better sense of the slope of the line than if I had just done two points.0399

OK, the second equation--graphing that: x - y =1: again, I am finding a few values for x and y--a few sets of values.0406

When x is 0, that would give me 0 - y = 1, so - y equals 1; therefore, y is -1; the y-intercept is -1.0422

Now, finding the x-intercept, let y equal 0; if y is 0, x is 1.0434

One final point: I am going to just let x equal 2; so that would be 2 - y = 1; and that is going to give me -y = -1, so y = 1.0442

Again, I have three sets of points to graph.0457

So, the y-intercept is at -1; the x-intercept is at 1; and one more point: when x is 2, y is 1.0460

OK, drawing this line, I can immediately see that I have a single point of intersection, right here.0476

This is the solution; and this occurs at (2,1).0487

So, the solution for this is x = 2, y = 1.0493

And therefore, this is an independent system, since it has one solution.0501

OK, and you can always check your work by substituting x and y values back into these equations.0515

I have x + y = 3, so that is 2 + 1 = 3; and that checks out.0528

I have x - y = 1, so 2 - 1 = 1, and that checks out; so you can easily check and make sure that both of these are valid solutions for both of the equations.0534

The second example is a little bit more complicated, but using the same system.0551

And this time, instead of just finding (x,y) values, I am going to graph these out by putting both equations into slope-intercept form.0557

So, the first one is 2x - 4y = 12, so I am going to subtract 2x from both sides, and then I am going to divide both sides by -4.0565

And that is going to give me y = 1/2x - 3.0577

Slope-intercept form is very helpful when you are graphing: this is -2, -4, -6, -8, -10, 2, 4, 6, 8, 10.0582

OK, here I have a y-intercept at -3; and I know the slope--the slope is 1/2.0597

That tells me, when I increase y by 1, I am going to increase x by 2; if I increase y by another 1, I am going to increase x by 2.0606

So, you can see the slope of this line right here; that gives me enough to work with.0620

I am going to do the same with the second equation, 4x - y = 10, putting this into slope-intercept form, y = mx + b,0638

subtracting 4x from both sides, and then dividing both sides by -1 to give me y = 4x - 10.0646

So here, I have a y-intercept at -10 and a slope of 4; I am going to increase y by 4 (that is 2, 4), and increase x by 1.0656

Increase y by 2, 4; increase x by 1; increase y by 4 (2, 4); increase x by 1; OK.0669

So, this is a much steeper line; the slope is much steeper--the slope of 4; so I am going to go ahead and draw it like that.0687

And the point of intersection is right here, and again, we have an independent system with a single solution.0695

And it is at (2,-2), so that is my point of intersection: x is 2; y is -2.0708

Again, you could always check your answers by substituting either or both equations with these x and y values, and ensuring that the solutions are valid.0723

So again, I have an independent system; it has one solution--one set of valid solutions.0733

OK, this next example: again, solve by graphing; and my approach is going to be to put these into y-intercept form and use that to graph.0748

First, 2x - 3y = 6, so -3y = -2x + 6; divide both sides by -3; that is going to give me...-2/-3 is 2/3x, and then 6/-3 is -2.0758

So, this line has a y-intercept of -2 and a slope of 2/3; so increase y by 2; increase x by 3.0781

OK, that is my first equation; the second equation--again, putting it into the slope-intercept form, which is an easy way of graphing:0799

This gives me 6y = 4x - 12; dividing both sides by 6 gives me 4/6x - 2.0808

And I am going to go ahead and simplify this to 2/3x - 2.0823

So, you may already see that these are the same equation--these are going to describe the same line.0828

Even if you didn't notice it right away, when you start graphing, you are going to see: the y-intercept is -2; the slope is 2/3.0835

You are going to end up with the same line.0842

Because this is the same line, they intersect at every single point; these two equations--their graphs intersect at every single point,0844

which is an infinite number of points, along the line; they don't intersect everywhere, but everywhere on this line.0852

Therefore, this is a dependent system, and there is an infinite number of solutions; all points along this line are solutions.0857

And this is known as a dependent system.0871

So initially, you looked at this; you might not have recognized that these are actually describing the same line.0877

But once you started to plot it, either by putting it in the y-intercept form or by finding points along these lines,0882

you would have quickly realized that this is the same line.0890

In this last example, again, there is a system of equations that we need to solve by graphing.0896

Again, using the method of slope-intercept form for graphing: 4x - 2y = 8; subtract 4x from both sides, and then divide both sides by -2 to give me y = 2x - 4.0905

OK, the y-intercept is -4; the slope is 2; increase y by 2; increase x by 1; increase y by 2; increase x by 1; and so on.0925

I have three points; that is plenty to graph this line...so that is my first line.0942

The second line: again, slope-intercept form: add 6x to both sides--it gives you 3y = 6x + 12.0947

Now, divide both sides by 3 to get y = 2x + 4.0961

So here, I have a y-intercept of 4 and a slope of 2; so increase x by 2; increase y by 1.0968

So, I am going to get a line looking like this; and what you may quickly realize is that these are parallel lines.0978

One thing you could have noted is that the slope, m, equals 2 for this line, and it equals 2 for this line.0991

Since these are parallel lines, they are never going to intersect; and this is an inconsistent system, and there is no solution.0998

And what we say here is: you could say the solution is just the empty set; there is no solution.1008

This is described as an inconsistent system.1016

OK, that concludes this lesson of Educator.com, describing graphing to solve systems of equations.1024

And I will see you next time.1032

Related Books

Algebra and Trigonometry: Structure and Method, Book 2

Author: McDougal Littell

ISBN: 395977258

Publisher: McDougal Littell

Year: 1999

This classic textbook does a great job of explaining concepts and includes lots of in-depth, worked out examples in every section as well as practice problems.

Related Books

Name	Description	Link
BookRenter.com	BookRenter.com is simply the most reliable online textbook rental service.	Visit BookRenter.com
PhysicsForums.com Homework Help	Physics Forums is a scientific community for students looking for math & science help.	Visit PhysicsForums.com Homework Help

Dr. Carleen Eaton

Solving Systems of Equations by Graphing

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro