This week we reviewed systems of equations and i re-learned how to do elimination and substitution because during those 3 weeks of break i mostly forgot how to do them.

Here is how you do elimination when using systems!

Say you have the problem…

x  + 2y = 12

x + 3y = 17

So the first thing you’ll want to do is make some zero pairs! So I’ll times the top equation by -1, to make the x’s cancel each other out.

-x – 2y = -12

x + 3y = 17

The x’s then cancel each other out so the equation is now…

-2y = -12

3y = 17

Then you add the equations and youll get…

y=5

You then must take the original equation once more and cancel the y’s out so that you get…

(x  + 2y = 12)-3

(x + 3y = 17)2

=

-3x – 6y = -36

2x + 6y = 34

The y’s then cancel each other out and it becomes…

-3x = -36

2x = 34

You then add them together and get…

(-x = -2)-1

=  x = 2 and y = 5

You can then verify your answer by putting it back into the original equation.

2  + 2(5) = 12 which is correct and therfore means the answer is right.

2 + 3(5) = 17 which is also correct and therfore means the answer is right.

This week I learned how to use elimination to get to the “special” point on a graph (the point where 2 lines intercept)

You take 2 equations

2x+y=10

5x-y=4

and then you add them together! Which equals to (7x=14) or simplified is (x=2), because positive “y” and negative “y” are zero pairs.

And then you do the same thing to find the (y=?) but this time there isn’t a zero pair to easily get rid of the “x” so you can multiply both equations to make them zero pairs.

5(2x+y=10) =  10x+5y=50

-2(5x-y=4) =  -10x+2y=-8

Which equals (7y=42) and then simplified equals to (y=6)

So the “special” point on this graph would be (2,6).

This week I learned how to distribute and then add like terms to find the answer of a ploynomial equation.

Like with this equation (h)     

As you can see, I first distributed the -2 into the brackets, and then after I got that answer I added the like terms which got me to the answer which is -5 +22.

This week I learned a more simple and quick way to find the answer to an algebra question.

You do this by taking a question like –> (x-4)(x+2)

Then you take the x and times it by the other x which makes it

After that you take the -4 and add it to the +2 which makes it –> -4 + 2 = -2x (you add the x to the answer when doing this part)

And then you once again the the  -4 and the +2 but this time you multiply them together –> -4 x 2 = -8

Then finally you put them in order of powers which is -2x -8

his week, I learned that sometimes it really helps to draw and not just write when doing math! (Especially when it comes to Trigonometry) The reason why this is something important I learned this week is because all my life I have really disliked linking drawing to math because I get distracted very easily and so when I start to draw I will get a bit too into it and next thing I know I’ve been drawing a triangle for 30 minutes and truly not have realized it’s even been 2 minutes. But with this chapter of Trig I don’t think I could have done as good as I have without drawing the actual triangles and placing the angles and the side lengths in the right places on the triangle.

This week I learned how to find the exact length of a side of a right angle triangle using SOH CAH TOA. (Sine= opposite/hypotenuse, Cosine= adjacent/hypotenuse, Tangent= opposite/adjacent) Here’s a picture of the work/notes I did to explain it better.

This week in Math, 1 thing that I learned was how to divide numbers with exponents in a much easier way.

Last year they taught me that the way to simplify questions like / is to use zero pairs. So you would write the question like 5 5 5 5/5 5. And by doing that you could cancel out any number that has a twin on the numerator/denominator, and whichever amount you had left would be the simplified version of the question. So in the case of 5 5 5 5/5 5 the answer is .

But now I know that you can also just subtract the second exponent from the first one. (4-2) and it will take a lot less time and be much easier to use when you have large number exponents.