Using a venn diagram, compare and contrast Graphing,Elimination and Substitution Methods in solving system of linear equation in two variables.
Answers
your final answer, is this system consistent or inconsistent?
If you said inconsistent, you are right!
If you get no solution for your final answer, would the equations be dependent or independent?
If you said independent, you are correct!
The graph below illustrates a system of two equations and two unknowns that has no solution:
no solution
Infinite Solutions
If the two lines end up lying on top of each other, then there is an infinite number of solutions. In this situation, they would end up being the same line, so any solution that would work in one equation is going to work in the other.
If you get an infinite number of solutions for your final answer, is this system consistent or inconsistent?
If you said consistent, you are right!
If you get an infinite number of solutions for your final answer, would the equations be dependent or independent?
If you said dependent, you are correct!
The graph below illustrates a system of two equations and two unknowns that has an infinite number of solutions:
all
notebook Example 1: Determine whether each ordered pair is a solution of the system.
example 1a (3, -1) and (0, 2)
Let’s check the ordered pair (3, -1) in the first equation:
example 1b
*Plug in 3 for x and -1 for y
*True statement
So far so good, (3, -1) is a solution to the first equation x + y = 2.
Now, let’s check (3, -1) in the second equation:
example 1c
*Plug in 3 for x and -1 for y
*True statement
Hey, we ended up with another true statement, which means (3, -1) is also a solution to the second equation x - y = 4.
Here is the big question, is (3, -1) a solution to the given system?????
Since it was a solution to BOTH equations in the system, then it is a solution to the overall system.
Now let’s put (0, 2) into the first equation:
example 1d
*Plug in 0 for x and 2 for y
*True statement
This is a true statement, so (0, 2) is a solution to the first equation x + y = 2.
Finally, let’s put (0,2) into the second equation:
example 1e
*Plug in 0 for x and 2 for y
*False statement
This time we got a false statement, you know what that means. (0, 2) is NOT a solution to the second equation x - y = 4.
Here is the big question, is (0, 2) a solution to the given system?????
Since it was not a solution to BOTH equations in the system, then it is not a solution to the overall system.
Three Ways to Solve Systems of Linear
Equations in Two Variables
There are three ways to solve systems of linear equations in two variables:
graphing
substitution method
elimination method
Solve by Graphing
Step 1: Graph the first equation.
Unless the directions tell you differently, you can use any "legitimate" way to graph the line. Our tutorials show three different ways. Feel free to review back over them if you need to: Tutorial 12: Graphing Equations shows how to graph by plotting points, Tutorial 14: Graphing Linear Equations shows how to graph using intercepts, and Tutorial 16: Equations of Lines shows how to graph using the y-intercept and slope.
Step 2: Graph the second equation on the same coordinate system as the first.
You graph the second equation the same as any other equation. Refer to the first step if you need to review the different ways to graph a line.
The difference here is you will put it on the same coordinate system as the first. It is like having two graphing problems in one.
Step 3: Find the solution.
If the two lines intersect at one place, then the point of intersection is the solution to the system.
If the two lines are parallel, then they never intersect, so there is no solution.
If the two lines lie on top of each other, then they are the same line and you have an infinite number of solutions.. In this case you can write down either equation as the solution to indicate they are the same line.
Step 4: Check the proposed ordered pair solution in BOTH equations.
You can plug in the proposed solution into BOTH equations. If it makes BOTH equations true then you have your solution to the system.
If it makes at least one of them false, you need to go back and redo the problem.