I decided to do the blog today since I happened to check on the blog and didn't see an update. I hope Dr.Eviatar doesn't mind if I scribe here with "no permission" :)

Well, today we had to hand in #6 of Exercise 24 from the book. There were some volunteers who put the answer for it on the board. I didn't quite know how to do it, so I didn't volunteer.

Here are the answers for #6 a.)&b.)

**a.)**

*3x + 2y = 4 -->equation [1]*

x - y = 3 -->equation [2]

x - y = 3 -->equation [2]

Multiply the equation [2] by 3 like this:

*3(x - y - 3) = 0*

=> 3x - 3y - 9 = 0

=> 3x - 3y - 9 = 0

Then you subtract the new equation from the first [1] equation:

*3x + 2y - 4 = 0*

5y + 5

__- 3x - 3y - 9 = 0__5y + 5

Find the value of y:

*5y = -5*

=> y = -1

=> y = -1

Substitute the value of y in the equation [1] to find the value of x:

*3x + 2(-1) = 4*

3x - 2 = 4

3x = 6

x = 2

3x - 2 = 4

3x = 6

x = 2

Answer is: (2,-1)

**b.)**

*2x + 3y = 48*

3x + 2y = 42

3x + 2y = 42

Same process, but this time you multiply both equations to get rid of y:

*2(2x + 3y =48)*

=> 4x + 6y = 96

3(3x + 2y = 42)

=> 9x + 6y = 126

=> 4x + 6y = 96

3(3x + 2y = 42)

=> 9x + 6y = 126

Addition-Subtraction Method:

*4x + 6y = 96*

-5x = -30

__- 9x + 6y = 126__-5x = -30

Value of x:

*-5x = -30*

=> x = 6

=> x = 6

Substitute the value of x to get the value of y:

*2(6) + 3y = 48*

12 + 3y =48

3y =36

y = 12

12 + 3y =48

3y =36

y = 12

Answer is: (6,12)

After that, we started a new topic. It was sort of the same as our previous topic (Systems of Linear Equations in Two Variables), but this time it is for THREE variables. I'm still in the process of understanding it, so pardon my explanations. Well, all I have right now are some notes from the board (-_-"). It was so much better when we can just look at the slides again in the blog for information.

But anyway, here are the notes:

__Independent__- 1 Solution (all 3 equations intersect)

__Dependent__- Infinite solutions -> concide

-> common line

__Inconsistent__- No solutions

The graph for the "Ordered Triple" example (3,3,3)

The graph was sort of hard to draw, and to understand. As a matter of fact, I don't really get how you draw the points in it. That's why we're not really asked to draw the graph.

Example of how to solve equations with THREE variables:

*x + y - z = 2 --> [Equation 1]*

*x - 2y + 2 = -1 --> [Equation 2]*

*3x + y - 2z = 4 --> [Equation 3]*

Steps:

*1st*:

[Equation 1] + [ Equation 2]

*=> 2x - y = 1 --> [Equation 4]*

*2nd*:

2 * [Equation 2]

*=> 2x - 4y + 2z = -2 --> [Equation 5]*

*3rd*:

Equation 3 + Equation 5

*=> 5x - 3y = 2 --> [Equation 6]*

*4th*:

3 * [Equation 4]

*=> 6x - 3y = 3*

*--> [Equation 7]*

*5th*:

[Equation 6] - [Equation 7]

*=> x=1*

*Substitute x=1 into [Equation 4]*:

*=> 2 - y = 1*

*=> y = 1*

*Substitute x=1 and y=1 into [Equation 1]*:

*=> 1 + 1 - z = 2*

*=> z = 0*

Answer to the equations: (1,1,0)

Well, that's all the notes that was for today. Things for homework are Exercise 25 #1, and the problem she wrote on the board.

In case you weren't in class or didn't get to write it down, here is the problem (Parts of it are already summarized/put into an equation by Dr.Eviatar):

*w + l + h = 80cm*

Length is 10cm less than twice the sum of width and height and twice the width exceeds the height by 6cm. Find the width, length, and height of the box.

*l = 2 (w + h) - 10*

*2w = h + 6*

That's practically what we did for class today. I hope I helped in the new topic in some way. I'm not going to pick who's going to be the next scribe since I sort of just barged in the blog to put what we did today [kind of saying that I scribed with "no permission", haha.] I have to do the homework myself. I'll at least "try" to do it (-_-") I'm still trying to understand the given problem.

Till next time my fellow classmates,

*Charizze*^-^

## No comments:

Post a Comment