Test - Wednesday

We will be having a test this Wednesday the 23rd of September. It will be on linear algebra (so plotting lines as well as solving for unknown values). It will be an open book test, however you will not have access to any devices throughout the test. You may use a calculator, however it cannot be a calculator on your phone.

I strongly suggest that you try and be in class for this test, because if you are not then you will have to do it in your last lesson before the end of term, or on your first day back from term break.

As always, if you do have any questions about this feel free to comment on this post, or email or chat to me in person.

Assignment + Test

Today we started our assignment on linear algebra. You can find the assignment in the "Assignments" tab. It is due next week on Wednesday the 23rd of September. All the information you need for this assignment can be found in old posts on here.

We will also will have a test coming up. This will also be on the Wednesday 23rd of September. I suggest that you all try and attend on this day, because if you do not you will have to do the the test in your last lesson of term or after the term break. This test will be an open book test, but you will not be allowed to use phones or any other devices.

As always, if you do have any questions feel free to comment on this post, or email or chat to me in person.

Inequalities

We are now looking at solving inequalities. These are very similar to our standard equations, but they have < or > instead of = signs.
The < or the > stand for greater/less than. This depends on the way the sign is facing. I like to think of the < or the > as a mouth, with the larger open end the opening of the mouth. The mouth always wants to eat the larger meal, so it always goes for the larger part of our inequality. So if we have 3x +2 > 2x - 1, the "mouth" is trying to eat the "3x + 2" so this must be the larger part. So we can rewrite this equation as in words as "3x + 2 is greater (larger) than 2x - 1".
Now there are slight modifications that can happen to these. We can write them as ≥ or ≤. The change is the line underneath. This changes it from being "greater/less that" to "greater/less than or equal to". So for example; x ≥ 4 means that x could take any value greater than OR equal to 4. 


We can USUALLY solve these in the same way as we would solve standard equations. So with the following example:

Solve 3x + 4 > 7

We want x on one side of our > and everything else on the other side. So, we will first subtract 4 from both sides of our equation. We get;

3x + 4 - 4 > 7 - 4
3x > 3

Now we divide both sides by 3. We get;

3x/3 > 3/3
x > 1

So for the equation 3x + 4 > 7 to be true, x must be greater than 1 (x>1)


Now you would see that I said we only USUALLY solve the equations in the same way. The exception to this is when we divide or multiply by a negative number. We have to do something a little different with these. If we do ever have to multiply or divide by a negative, we need to switch the way the > or < sign is facing. So, < becomes >, or visa versa.

For example;

Solve 7 - 4x > 15

First we will get rid of the 7, by subtracting 7 from both sides. We get;

7 - 4x - 7 > 15 - 7
-4x > 8

Now, we will divide by -4, and because this is a negative we must switch the direction of the > (so it will become <). We get;

-4x/-4 < 8/-4
x < -2

So we have now solved this, giving us a solution of x being any number less than -2 (or x < -2)


Now, we also need to know how to plot these on a number line. To do these, I have found that we can simply follow 3 steps.

1. Draw an open circle at the number on our number line, i.e. if we had x>1, we would draw an open circle at the point 1 on our number line. 
2. Colour this circle in if it is a ≥ or a ≤ 
3. Draw a line with an arrow in the correct direction. 

Just as a reminder, this is what a number line looks like:

So for example, plot x > 2 on the number line; 

First we draw our open circle at the point 2 on our number line like so;

Next, we look to see if our sign is a ≥ or a ≤. In this case it isn't, so we don't have to colour our circle in. 

Now, we want to draw a line with an arrow in the correct direction. In this case, the line needs to represent x, which is any number larger than 2 (x>2), so our line and arrow has to cover any number greater than 2. So it will look like; 

 Done!

Another example could be x ≤ -1

First, we draw a circle on our number line at -1. 

Now since this one does have a ≥ or a ≤, we have to colour in the circle. This gives us; 

Finally, we draw the line and arrow in the direction required. In this case, the equation is telling us that x is any number less than or equal to -1, so we have to draw our line and arrow to represent this. We then get; 

A great website for more information and examples on this is this one: 

http://www.mathsisfun.com/algebra/inequality-solving.html

I have uploaded a worksheet on this topic into the "Work Sheets" tab. You should now be able to work through these questions. They need to be completed by Wednesday next week. 

As always, if you have any questions about this, feel free to comment on this, or email or chat to me in person. 

Simultaneous Equations

Today we had a look as solving simultaneous equations. These are equations that have 2 variables (letters, i.e. x, y, a, b, etc). Previously we have looked at solving standard equations that only have 1 variable, such as 2x + 6 = 10. We will be using these same techniques, but solving equations with 2 variables. If we are solving for 2 variables, we do need 2 equations to start with, so all of our problems with these will give us 2 equations.

The first thing we want to do when we are faced with simultaneous equations is to rearrange one of our equations so that it is "x = something" or "y = something". We do know how to do this from previous work, however I will run through a few examples.
Once we have done this, we can substitute this value for x or y into our other equation and solve this. This should give us a value of one of our variables. We can then use this value to find the other variable.

I know this all sounds complicated, however after a few examples I hope it should make more sense.

Example: Solve the following simultaneous equations

2y + x = 8   -> We will call this equation [1]
1 + y = 2x   -> We will call this equation [2]

First, we will rearrange equation [2] so we have "y = something". We will do this by getting rid of the 1 on the left hand side of this equation. This then gives us:

1 + y - 1 = 2x - 1
y = 2x - 1     -> We will now call this equation [3]

Now, we know that y = 2x - 1, so we can now take this knowledge and help us solve equation [1]. We then substitute (2x - 1) wherever there is a "y" in equation [1]. This is because equation [3] has told us that y = 2x - 1
We then get:

2(2x - 1) + x = 8

Now, expanding the brackets we get:

4x - 2 + x = 8

We can then collect like terms, and get:

5x - 2 = 8

Now solving this, we get:

5x - 2 + 2 = 8 + 2 
5x = 10
5x/5 = 10/5
x = 2

We now know that x = 2, so we have solved half of these equations. We can then substitute this value of x = 2 into equation [3], and solve this to get a value for y.
We the get:

y = 2(2) - 1
y = 4 - 1
y = 3

So now we know a value for y as well. So we have now successfully solved our equation. We have found that:
x = 2
y = 3


If you want further example of this, the following websites are good:

http://www.khanacademy.org/math/algebra-basics/core-algebra-systems/core-algebra-systems-tutorial/v/solving-linear-systems-by-substitution -> Watch the video, but only up to the 4min mark.

http://www.purplemath.com/modules/systlin4.htm -> This is handy, but some of the language they use and some of the examples are a little comple, but it still has some good information.


I have uploaded a new worksheet to the "Worksheets" tab. You should be able to work through the number style questions (i.e. question 1). Do not worry too much about the worded questions. I will go through those on Wednesday.

As always, if you have any questions feel free to comment on here, or email or chat to me in person.

More complex equation solving

Today we looked at some more complex equations and how we solve these. The worksheet we looked at can be found in the "Worksheets" tab.

These problems have pro-numerals on both sides of the equals sign in the equations. We want to get rid of the pro-numerals on one side of this equation. For example:

4x - 7 = 2x + 1

The first thing we want to do with these is to get rid of the '2x' on the right hand side of the equation. We do know how to do this from previous examples. We want to subtract 2x, remembering to do this to both sides of our equation. We the get:

4x - 7 - 2x = 2x + 1 - 2x

Doing these sums (by collecting like terms) we get:

2x - 7 = 1

We can now solve this like we would have with our previous examples.

2x - 7 + 7 = 1 + 7
2x = 8
2x ÷ 2 = 8 ÷ 2
x = 4

So the techniques we use are very similar to what we have previously done, however there is the extra step at the beginning where we remove the pro-numeral from one side of the equation.

Another example could be:

2(2x + 4) = 3(6 - 2x)

The first step with these is to expand the equation. This then gets us:
4x + 8 = 18 - 6x

Next, we want to get rid of the "-6x" term from the right hand side of the equation by adding 6x, again remembering to do this to both sides of our equation. We then get:

4x + 8 + 6x = 18 - 6x + 6x
10x + 8 = 18

Now we can solve it

10x + 8 - 8 = 18 - 8
10x = 10
10x ÷ 10 = 10 ÷ 10
x = 1

Finally, we may get examples like these:

a/3 + a/4 = 7

To solve these, we want to get rid of the fractions. To get rid of the first fraction, we first want to multiply everything by 3 (i.e. the denominator of this first term). We get:

a/3 x 3 + a/4 x 3 = 7 x 3
a + 3a/4 = 21

Now that we have got rid of one of the fractions, we want to get rid of the other one. To do this we again will do the same, multiply everything by 4 (the denominator of the second term). We then get:

a x 4 + 3a/4 x 4 = 21 x 4
4a + 3a = 84
7a = 84

We can then solve this easily. We get:

7a ÷ 7 = 84 ÷ 7
a = 12


Using these new skills we should be able to work through the whole worksheet. There are a lot of questions, so I have cut a few out. I only want you to do the following questions:
Question 1 - a, b, d, e, g, h
Question 2 - a, b, c
Question 3 - a, c, e
Question 4 - a, c
Question 5 - a, d

As always, if you have any questions feel free to comment on this post, or email or chat to me in person.