1. 3-2: Solving Linear Systems

2. Solving Linear Systems
There are two methods of solving a system of equations algebraically:
Elimination
Substitution

3. Elimination
The key to solving a system by elimination is getting rid of one variable.
Let’s review the Additive Inverse Property.
What is the Additive Inverse of: 3x? -5y? 8p? q?
-3x y -8p -q
What happens if we add two additive inverses?
We get zero. The terms cancel.
We will try to eliminate one variable by adding, subtracting, or
multiplying the variable(s) until the two terms are additive inverses.
We will then add the two equations, giving us one equation with one variable.
Solve for that variable.
Then insert the value into one of the original equations to find the other variable.

4. Elimination Solve the system: m + n = 6 m - n = 5
Notice that the n terms in both equations are additive inverses. So if we add the equations the n terms will cancel.
So let’s add & solve: m + n = m - n = 5
2m + 0 = 11
2m = 11
m = 11/2 or 5.5
Insert the value of m to find n: n = 6
n = .5
The solution is (5.5, .5).

5. Elimination Solve the system: 3s - 2t = 10 4s + t = 6
We could multiply the second equation by 2 and the
t terms would be inverses. OR
We could multiply the first equation by 4 and the second equation
by -3 to make the s terms inverses.
Let’s multiply the second equation by 2 to eliminate t. (It’s easier.)
3s - 2t = s – 2t = 10
2(4s + t = 6) s + 2t = 12
Add and solve: s + 0t = 22
11s = 22
s = 2
Insert the value of s to find the value of t (2) - 2t = t = -2
The solution is (2, -2).

6. Answer these questions.
When you solve systems using elimination, plan a strategy. The flowchart like the one below can help you decide how to eliminate a variable When you solve systems using elimination, plan a strategy. Here are a few hints to help you decide how to eliminate a variable.
Answer these questions.
Can I eliminate a variable by
adding or subtracting the
given equations?
Do It.
YES Or NO
Can I multiply one of
the equations by a
number, and then
add or subtract the
equations?
YES or NO
Do It.
Multiply both equations
by different numbers.
then add or subtract the
equations

8. Elimination Solve the system by elimination: 1. -4x + y = -12
4. 5m + 2n = -8
4m +3n = 2

9. The student will be able to:
Objective
The student will be able to:
Solve Systems of Equations by Graphing.
Designed by Skip Tyler, Varina High School

10. Intersecting Lines
The point where the lines intersect is your solution.
The solution of this graph is (1, 2)
(1,2)

11. Parallel Lines These lines never intersect!
Since the lines never cross, there is NO SOLUTION!
Parallel lines have the same slope with different y-intercepts.

12. Coinciding Lines These lines are the same!
Since the lines are on top of each other, there are INFINITELY MANY SOLUTIONS!
Coinciding lines have the same slope and y-intercepts.

13. What is the solution of the system graphed below?
(2, -2)
(-2, 2)
No solution
Infinitely many solutions

14. 1) Find the solution to the following system:
2x + y = 4
x - y = 2
Graph both equations. I will graph using x- and y-intercepts (plug in zeros).
Graph the ordered pairs.
2x + y = 4
(0, 4) and (2, 0)
x – y = 2
(0, -2) and (2, 0)

15. Graph the equations. 2x + y = 4 (0, 4) and (2, 0) x - y = 2
Where do the lines intersect?
(2, 0)
2x + y = 4
x – y = 2

16. Check your answer!
To check your answer, plug the point back into both equations.
2x + y = 4
2(2) + (0) = 4
x - y = 2
(2) – (0) = 2
Nice job…let’s try another!

17. 2) Find the solution to the following system:
y = 2x – 3
-2x + y = 1
Graph both equations. Put both equations in slope-intercept or standard form. I’ll do slope-intercept form on this one!
y = 2x + 1
Graph using slope and y-intercept

18. Graph the equations. y = 2x – 3 m = 2 and b = -3 y = 2x + 1
Where do the lines intersect?
No solution!
Notice that the slopes are the same with different y-intercepts. If you recognize this early, you don’t have to graph them!

19. Check your answer!
Not a lot to check…Just make sure you set up your equations correctly.
I double-checked it and I did it right…

20. What is the solution of this system?
3x – y = 8
2y = 6x -16
(3, 1)
(4, 4)
No solution
Infinitely many solutions

21. Solving a system of equations by graphing.
Let's summarize! There are 3 steps to solving a system using a graph.
Graph using slope and y – intercept or x- and y-intercepts. Be sure to use a ruler and graph paper!
Step 1: Graph both equations.
This is the solution! LABEL the solution!
Step 2: Do the graphs intersect?
Substitute the x and y values into both equations to verify the point is a solution to both equations.
Step 3: Check your solution.

22. Solving systems of equations in real-world problems.
April sold 75 tickets to a school play and collected a total of $495. If the adult tickets cost $8 each and child tickets cost $5 each, how many adult tickets and how many child tickets did she sell?
Solution: Let a represent the adult tickets and c represent the child tickets.
Individual tickets sold equaled 75, so a + c = 75
All total April sold $495 in tickets, since adult tickets are $8 and child tickets are $5, so 8a + 5c = 495.
System of Equations a + c = 75
8a + 5c == 495

23. Solution a + c = 75 8a + 5c = 495 a = 75 – c 8(75 – c) + 5c = – 8c + 5c = c = = 3c 35 = c a + c = 75 a + 35 = 75 a = 40 There were 40 adult tickets and 35 child tickets sold = 75 8(40) + 5(35) = = 495

24. Write a system of equations and solve
Write a system of equations and solve. At a baseball game, Jose bought five hot dogs and three sodas for $17. At the same time, Allison bought two hot dogs and four sodas for $11. Find the cost of one hot dog and one soda.

25. Suppose your community center sells a total of 292 tickets for a basketball game. An adult ticket cost $3. A student ticket cost $1. The sponsors collected $470 in ticket sales. Write and solve a system to find the number of each type of ticket sold.

26. Let a = number of adult tickets Let s = number of student tickets total number of ticket total number of sales a + s = a + 1s = 470 Solve by elimination (get rid of s) because the difference of the coefficients of s is zero. a + s = 292 3a + s = a + 0 = -178 a = 89
This is the number of adult tickets sold.
That means you must subtract the two equations so,
-3a – a = -470 is what you must subtract.
Next Step

27. This is the number of student tickets sold.
Solve for the eliminated variable using either of the original equations. a + s = s = 292 s = 203 There were 89 adult tickets sold and 203 student tickets sold. Is the solution reasonable? The total number of tickets is = The total sales is $3(89) + $1(203) = $470. The solution is correct.
This is the number of student tickets sold.

28. 7) The sum of two numbers is 20. Their difference is 4
7) The sum of two numbers is 20. Their difference is 4. Write and solve a system of equations. 8) Your school sold 456 tickets for a school play. An adult ticket cost $ A student ticket cost $1. Total ticket sales equaled $ Let a = adult tickets sold, and s = student tickets sold. How many tickets of each were sold? 9) Suppose the band sells cans of popcorn for $5 each and mixed nuts for $8 each. The band sells a total of 240 cans and makes a total of $ Find the number of cans of each sold.
7. (12 and 8) 8. (270 adult, 186 student) 9. (102 cans of popcorn, 138 cans of nuts)
One more problem,
my favorite.

29. 10) A farmer raises chicken and cows
10) A farmer raises chicken and cows. He has a total of 34 animals in his barnyard. His six-year son came in one day all excited saying, “Daddy, daddy, did you know all your animals have a total of 110 legs.” Write a system of equations to represent this situation. How many chickens and how many cows does the farmer have?