agcaoili, mikaela systems of linear equation

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.

This is will be our classroom for today. So go on and

pick a chair of your choice.

Let’s begin the class. Now, look at the board and click the

icon on what part of the lesson you want to study.

Definition

Solving by

Graphing

Solving by Substitution

Solving by

Elimination

Exercises

Science Examples

A "system" of equations is a set or

collection of equations that you deal

with all together at once.

REMEMBER: A Linear Equation is an

equation for a line.

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

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:

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.

red point at right is

not a solution to the

system, because it is

not on either line

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:

The purple point at

right is a solution to

the system, because it

lies on both of the

lines:

Thus, for systems of equations,

"solutions" are the "intersections".

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.

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.

X t y = 3 ; y=0

X=3

Y=3

(3,3)

X-2y =0

X = 0

-2y=0

Y=-2

(0,-2)

This is a cool method for getting you to see what's going

on. But, it has serious problems.

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

So, unless the problem is designed to cross at a nice, clean

point, the graphing method is pretty useless for solving

systems.

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.

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

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.

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)

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.

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.

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.

Example 1:

A 10 kg body is placedon a smooth inclinedplane at an angle of 30o

with the horizontal. Astring is attached to thebody and the string isthen passed over a smallpullet at the top of theplane. At the other end ofthe string an 8 kg body issuspended. Find thetension T in the string andthe acceleration a of thesystem after it has beenreleased.

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

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

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?

Let’s look at another example. Try to solve the problem as we

go on with the lesson.

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