XYVER MATH

It’s all about MATH

Xyvermath is back…….. September 9, 2008

Filed under: Uncategorized — jhano @ 3:29 pm

This site will be posted of examples about the MATH II topics….

wait nyo lang okie….

always remember…..MATH IS EASY!!!!!!!

 

Periodical Test…is coming your way… August 5, 2008

Filed under: MATH — jhano @ 4:01 am
Tags:

IIPiety, Peace, Joy, and Forbearance…..

Good luck to all of you!!!!

Kaya nyo yan….

sorry if i can’t give more examples here in my website….

i’m so busy kasi eh….

siguro 2nd grading period nalang ulit…

iperfect nyo ung test ha….

GOD BLESS!!!!

 

SOLVING LINEAR EQUATION USING ELIMINATION METHOD… July 7, 2008

Filed under: MATH — jhano @ 12:54 am
Tags:

1) 3x – y = 5 and 5x + 2y =12

Using Elimination Method

Step 1. If the two equations doesn’t have any additive inverse, multiply one or both equation by a number so that it will have an additive inverse. In this example we will multiply the 1st equation by 2.

(3x – y = 5)2 ——-> 6x – 2y = 10

Step 2. Now that the two equations have an additive inverse, you can eliminate one of the variables by adding the two equations.

6x - 2y = 10 Note: The numbers and variable in bold will be added while the

5x + 2y = 12 variables in Italics will be eliminated.

11x = 22 Divide by 11

x = 2

Step 3. To solve for y, substitute 2 for x, using any of the equation.

3x – y = 5

3(2) -y = 5 Substitute x = 2

6 – y = 5

- y = 5 – 6 Transpose 6

- y = – 1 Divide both side by -1

y = 1

Step 4. Check.

3x – y = 5 l 5x + 2y = 12

3(2) – 1 = 5 l 5(2) +2(1) = 12

6 – 1 = 5 l 10 + 2 = 12

5 = 5 l 12 = 12

therefore the solution is (2, 1)

Post your questions…

 

SOLVING LINEAR EQUATION USING SUBSTITUTION METHOD. July 2, 2008

Filed under: MATH — jhano @ 4:55 am
Tags: ,

1) x + 2y = -7 and 5x – 3y = 30

By substitution method:

Step 1. Find an equation to be used as a substitute to the other equation.

x + 2y = -7 –> x = -7 – 2y

Step 2. Substitute the equation to the other equation.

x = -7 – 2y

5x – 3y = 30

5 (-7 – 2y) – 3y = 30

-35 – 10y – 3y = 30

-10y – 3y = 30 + 35

-13y = 65 Divide by -13

y = -5

Step 3. Substitute -5 to y and solve the other variable.

x = -7 – 2y

x = -7 – 2(-5)

x = -7 + 10

x = 3

Step 4. Check

(3, -5)

x + 2y = -7 l 5x – 3y = 30

3 + 2(-5) = -7 l 5(3) – 3(-5) = 30

3 – 10 = -7 l 15 + 15 = 30

- 7 = – 7 l 30 = 30

Therefore the solution is (3, -5).

IF YOU HAVE ANY QUESTION JUST PUT A COMMENT IN THIS POST..TNX

-MR. GONZALES

 

The Art in Math May 31, 2008

Filed under: MATH — jhano @ 1:55 am
Tags: , ,

Art of Math

another art of math

What can you say about the art of Math?

Do you know some Piece of Art that has a connection with Math?

Please feel free to e-mail it to me : jhano_g@yahoo.com.ph

THANKS….

 

JUST MATH May 31, 2008

Filed under: MATH — jhano @ 1:20 am
Tags: , ,

Many students nowadays are having difficulty in math. Many reasons they’re giving me is that it is hard to understand and they can’t apply it in their everyday livings. So I’ve decided to make some point of views or advice to all the students that are having difficulty with math.

  1. JUST pay attention to your MATH teacher.
  2. JUST apply all the concepts learned in your MATH class.
  3. JUST think of ways on how you can cope up with MATH.
  4. JUST love MATH as you love your family and friends.
  5. Always remember that it is “JUST MATH.

JUST remember this tip and you will enjoy MATH in your everyday living!!!