Math 10 Week #17

This week,  we covered systems of linear equations. Systems give us two lines, and we are supposed to find the point at which these two lines intersect. We learned about three different methods to solve systems. One of these ways was to simply graph the two lines given and see where they intersect.

Another solution is to substitute one equation for another. Say you have two equations:

  1. 3x+2y=4
  2. x-3y=5

x is currently by itself. Simply isolate x, and the equation is now x=3y+5

Substitute x into the first equation, and you now have 3(3y+5)+2y=4. Expanded, it is 11y=-11. We now know y=-1

Substitute y=-1 into one of the equations, x-3(-1)=4

x+3=4. This means x=1

The lines intersect at (1,-1)

Math 10 Week #16

This week we discussed how to write different equations representing lines on a graph. There are three kinds of equations: slope-intercept form, general form, and point-slope :  form.

Slope intercept form: y=mx+b, where m represents the slope, and b is the y-intercept.

General form: Ax+By+C=0, where A>0, and A,B, and C are integers

Point-slope form: m(x-x1)=y-y1

Slope intercept form is the form that is used most of the time. Sometimes General form is used for an equation to look “nice”, because everything is on one side of the equation. If you have an equation in slope intercept form, that equation can be converted into general form.

y=\frac {2}{5}x-1          Converted into general form, the equation will say

2x-5y-5=0

Math 10 Week #15

This week we learned about slopes on a graph. One thing that we learned this week was how to find the slope of a line on a graph. To find the slope of a line on a graph, use this formula: 

Take two points on a graph, say (5,6) and (4,5). Input both coordinates into the formula, and you’re left with the slope.

 

Math 10 Week #12

This week we started learning about Relations and Functions. We’ve learned how to graph specific points on a graph, along with finding x and y-intercepts.

To find the x-intercept of say y=2x+2, you would substitute y for 0. You’re then left with 0=2x+2.

Subtract 2 from both sides to get -2=2x.

Divide -2 from both side and x=-1

You can do the same thing with finding the y intercept. To find the y-intercept substitute x for 0. The equation comes to 2x=2.

Divide both side by two to get y=1

Math 10 Week #11

We continued learning about factoring this week. We talked about difference of squares problems. Factor of squares problems can be written as (a^{2}-b^{2}) = (a+b)(a-b) or (a-b)(a+b).

An easy one is (x^{2}-25). That can be factored into either (x+5)(x-5) or (x-5)(x+5)

Another more complicated one is (x^{3}-16x). It is factored into x(x+4)(x-4), and then factored further into x(x+4)(x+2)(x-2).

 

 

Math 10 Week #10

We continued our polynomial unit this week. An important concept that we covered recently was division with polynomials. We’ve added, subtracted, and multiplied polynomials, but we haven’t divided. With polynomials, it’s not called division, it’s called factorization. We’ve factored before, but that was with only single numbers. This time it’s with polynomials.

In the first question, you can factor out a x, because there is an x in both terms. So the x factored out, and is placed on the outside of the brackets. In the second question, you can factor out a 2 from both terms, so the equation goes from 4x+6 to 2(2x+3).

 

Math 10 Week #9

This week, we talked about multiplying two binomials together. We used the distributive property to find the products of two binomials. To do this, we used the claw method, which helps us visualise the distributive property. It’s called the claw method, because, well, it looks like a claw. It’s funny. I’ve written an example in the picture below. The x in the first bracket is distributed to both terms in the second bracket. Then the 3 from the first bracket is also distributed to both terms in the second bracket.