This week in math 10 we learned how to solve equations using elimination. Elimination has fewer steps then substation that’s why I prefer this method of solving equations. Bellow are 2 examples of how to solve using elimination.

First example: 8x – y = 10

4x + y = 14

Step 1.  Add the x’s together, y’s together and the 10 and 14

now you have  12x = 24

Step 2. now divide 24 by 12

so now x = 2

Step 3. Now you have to get find y. Input 2 back into one of the equations above.

4(2) + y =14

8 + y = 14

now minus 8 on both sides to get y by itself

y = 6

Second example: x + 2y =3

-x + 3y = 2

5y = 5

y = 1

x +2(1) = 3

x =1

This week in math 10 we learned how to solve systems by substitution. I will be going over the steps of how to do substitution and I will provide examples.

Steps:

1. Pick one equation and rearrange so it is x = or y =
2. Substitute this into the other equation (use brackets)
3. Solve equation
4. Substitute the number just found, back into the rearranged equation
5. Verify solution

Examples:

2.

This week in math 10 we learned how to write and equation of a line in general form form. Bellow are some examples and descriptions of how to do general form.

1. y = -2x + 9

So you have this equation and basically all you need to do is get a zero on one side of the equation by itself.

So subtract the y and you get           0 = 2x – y + 9

2.   y – 4 = 7(x – 1)

y – 4 = 7x – 7                     so you have to distribute first, now get zero on opposite of the equation

y = 7x + 4 – 7

0 = 7x -y – 3

This week in math 10 we started a new unit and one section we learned about was the distance formula. I will be showing you how to find the distance between each pair of points.

1. (2,0) and (7,12) – First you must subtract the y coordinates which are 0 and 12. So 0 – 12 = -12. Then you must subtract the x coordinates which are 2 – 7 which equals -5.

Now you must square -5 and – 12 and that will become 25 and 144. now add 25 and 144 together and you get 169. The last step is to square root 169 which is 13. So now you know the distance between (2,0) and (7,12) which is 13.

This week in math 10 we continued to learn more about functions and relations. The newest thing we learned was function notation and problem solving.  I will be showing an example below.

1a) H(t) = 5/3t + 2

H(3)

5/3 * 3/1 = 15/3 + 2

15/3 + 6/3 – It became 6 over 3 because the +2 needed to have the same denominator as 15 over 3

15/3 + 6/3 = 21/3 – so 3 goes into 21 seven time so the answer is 7

This week in math 10 we learned about domain and range. The domain of a relation is the set of all possible values which can be used for the independent variable (x).The range of a relation is the set of all possible values of the dependent variable (y).

This week in math 10 we learned about Function notation. Function notation is used to replace the mapping notation.

Examples of function notation:

1. g(x) = 6 – x^2

g(4) = 6 – 16

g(4) = -10

2. g(x) = 6 – x^2

g(-6) = 6 – 36

g(-6) = -30

This week in math 10 we learning how to factor ugly trinomials.

I will be showing 6 different examples.