This document is a lesson on systems of equations taught by Mrs. Cynthia, a math teacher. It introduces systems of equations as multiple linear equations dealt with simultaneously. Students learn to solve systems by graphing, substitution, and elimination. Examples show applying these methods to science problems involving forces and acceleration. Students are encouraged to try sample exercises and ask questions if confused.

2. HI there! My name is Mrs. Cynthia
and I am a math teacher here at
Math Wiz High School. Today, you
will join me and learn all about
Systems Of Equation.

3. This is will be our classroom for today. So go on and
pick a chair of your choice.

5. Let’s begin the class. Now, look at the board and click the
icon on what part of the lesson you want to study.

6. Definition
Solving by
Graphing
Solving by
Substitution
Solving by
Elimination
Exercises
Science Examples

8.  A "system" of equations is a set or
collection of equations that you deal
with all together at once.

9.  REMEMBER: A Linear Equation is an
equation for a line.

10.  THUS: A System of Equations is when we
have two or more equations working
together.
 Now consider the following two-variable
system of linear equations:
y = 3x – 2
y = –x – 6

11.  Since the two
equations above
are in a system,
we deal with
them together at
the same time. In
particular, we
can graph them
together on the
same axis system,
like this:

12.  A solution for a single equation is any
point that lies on the line for that
equation. A solution for a system of
equations is any point that lies on each
line in the system.

13. red point at right is
not a solution to the
system, because it is
not on either line

14. The blue point at right
is not a solution to the
system, because it lies
on only one of the
lines, not on both of
them:

15. The purple point at
right is a solution to
the system, because it
lies on both of the
lines:

16.  Thus, for systems of equations,
"solutions" are the "intersections".

17. There are different ways to solve systems of equations. The
three ways are by graphing, by substitution method and by
elimination. We will learn more of this as we proceed on the
lesson. Make sure you pay attention.

20.  When you are solving systems
graphically, finding
intersections of lines. This is a
cool method to start with
since it lets you see what's
going on.

22. X t y = 3 ; y=0
X=3
Y=3
(3,3)
X-2y =0
X = 0
-2y=0
Y=-2
(0,-2)

24. This is a cool method for getting you to see what's going
on. But, it has serious problems.

25. What if you are to get a graph such as this?

26. So, unless the problem is designed to cross at a nice, clean
point, the graphing method is pretty useless for solving
systems.

27. There are also some instances that something “freaky” may
happen when you’re graphing. Observe the following graph
on the next slides and read why they are like that.

31. Why use Algebra when graphs are so easy? Because: more
than 2 variables can't be solved by a simple graph. So
Algebra comes to the rescue with two popular methods:
Solving By Substitution and Solving By Elimination

32.  The method of solving "by substitution" works
by solving one of the equations (you
choose which one) for one of the variables
(you choose which one), and then
plugging this back into the other equation.

33. These are the steps:
1. Write one of the equations so it is in the
style "variable = ..."
2. Replace (i.e. substitute) that variable in
the other equation(s).
3. Solve the other equation(s)
4. (Repeat as necessary)

37. Well, were you able to follow the flow of the lesson? If yes
then job well done! By this time you may try out the sample
exercises in this lesson too! If you get confused, then just
come back here and try to answer the problems again.

40.  The addition method of solving
systems of equations is also called
the method of elimination. This
method is similar to the method you
probably learned for solving simple
equations.

47. Now it’s time for us to apply this methods on science
problems. Let’s look at the 2 examples and see how they
work.

48. Example 1:
A 10 kg body is placed
on a smooth inclined
plane at an angle of 30o
with the horizontal. A
string is attached to the
body and the string is
then passed over a small
pullet at the top of the
plane. At the other end of
the string an 8 kg body is
suspended. Find the
tension T in the string and
the acceleration a of the
system after it has been
released.

49. For the 8kg body, the equation F=ma is
 8g – T = 8a (eq. 1)
For the other block:
 T – log sin 300 = 10a (eq. 2)
 Using elimination method:
 8g – T = 8a
(+) -5g + T = 10a
3g = 18a or
a = 1/6
Answer: a = 1.63 meters/seconds2

50. Another way to obtain a is by solving for T
in equation:
 T = 10a – log sin 300
T = 10a – 5g (Substitute this to eq. 1)
For the other block:
 8g – (10a – 5g) = 8a
Simplifying:
3g = 18a
a = 1/6
Answer: a = 1.63 meters/seconds2

51. To find the value of T:
T = 10a – 5g (Substitute the value of a)
 T = 10(1/6) – 5g
 T = 6.67g or
 = 6.67 g x 9.8 m/s2
 = 65.2 Newtons
Did you get the same answer?

52. Let’s look at another example. Try to solve the problem as we
go on with the lesson.

53. Another way to obtain a is by solving for T
in equation:
 T = 10a – log sin 300
T = 10a – 5g (Substitute this to eq. 1)
For the other block:
 8g – (10a – 5g) = 8a
Simplifying:
3g = 18a
a = 1/6
Answer: a = 1.63 meters/seconds2