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.

Solving equations - part 2

So last lesson we looked at solving equation. We had a look at some simpler equations and how we solve these. This lesson we will look at a few more complicated example. The ideas behind what we do to solve these are still the same, however the equations themselves are more complicated, or involve more steps.

• The first example we will look at is 6 - 4x = 18

First, we want to get rid of the 6 on the left hand side of the equation. At the moment it is a positive 6. To get rid of it we want to take 6 away. This gives us:

6 - 4x - 6 = 18 - 6
-4x = 12

Now, we want to remove the -4 from the left hand side. At the moment we have -4 multiplied by x, so we want to do the opposite, so divide by -4. Doing this gives us:

-4x ÷ -4 = 12 ÷ -4
x = -3

So we have now solved this equation to get x = -3

• The next example we will look at is (x + 4) / 6 = 3

First, we want to get rid of the 6. In this equation, the 6 is being divided, so to get rid of it we want to multiply by 6. This gives us:

(x + 4)/6 x 6 = 3 x 6
x + 4 = 18

Now we want to get rid of the 4. To do this we subtract 4 from each side. this gives us:

x + 4 - 4 = 18 - 4
x = 14

• One last example for us to look at is 3(2x - 4) = 12

First we want to get rid of the 3. At the moment we have 3 multiplied by (2x - 4). To get rid of it, we want to divide by 3. This gives us:

3(2x - 4) ÷ 3 = 12 ÷ 3
2x - 4 = 4

Now we want to get rid of the 4 and we will do this by adding 4 to both sides. We then get:

2x - 4 + 4 = 4 + 4
2x = 8

Now to get rid of the 2, we divide both sides by 2. We then get:

2x ÷ 2 = 8 ÷ 2
x = 4 


You should now be able to do questions 1-5 on the worksheet on solving equations. If you feel like some extra work, have a look at questions 6, but don't worry about question 7. Try and get questions 1-5 finished by Monday, so we can move onto some new stuff. 

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

Solving Equations

Today we had a look at how we solve equations. By solving equations, what I really mean is we want to find a value for our pro-numeral (letter). So we want to manipulate our equation in such a way that we have our pro-numeral on one side of the equals sign, and a number on the other side of our equals sign.

To do this, we want to use a few techniques to "free up" the pro-numeral. Basically, what we want to do is move the numbers away from the pro-numeral by doing the opposite to what the number is actually doing. Hopefully after a few examples,  this will become clearer.

So for example, if we have the expression x + 5. We want to somehow get rid of the 5 so we are just left with x. To get rid of the 5, we must do the opposite of what 5 is currently doing. In the expression, we are adding 5 to x. So to get rid of the 5, we must subtract 5 from the expression, since the opposite of adding is subtracting.
So, we then get: x + 5 - 5
We then see that + 5 - 5 is equal to zero, so we are just left with x.

Similarly, if we have 4x, we want to get rid of the 4 so we are just left with x. In this expression, the 4 is multiplied by x, so to get rid of the 4, we want to do the opposite of multiplying by 4, which is dividing by 4.
So we get 4x ÷ 4  
This then simply becomes x

This is quite easy for simple expressions like these, however we will usually have to solve a lot more complex equations. I will go through a few examples of these, hopefully demonstrating what we need to do.

Example: x + 6 = 8

We want to get rid of the 6, so we are just left with x on one side of the equals sign, and numbers on the other side.
To get rid of the 6, we need to do the opposite of what it is currently doing. In this equation, we have "plus 6" so to get rid of it we must "minus 6", since this is the opposite.
We also need to remember that whatever we do on one side of our equals sign, we must do it on the other side as well. In this equation we need to subtract 6 from the left hand side, to get rid of the 6 that is already there. This means we must subtract 6 from the right hand side too.
So we are left with:
x + 6 - 6 = 8 - 6
If we work these sums out, we get
x = 2
So we have now solved the equation x + 6 = 8, and found that x = 2.

The do get more complicated than this. For example, we may have:

4x - 7 = 13

We want to get rid of the 7 and the 4, so this one will require 2 steps. First we will look at getting rid of the 7.
In this equation we have "minus 7", so to get rid of this, we need to "plus 7", remembering to do this to both sides. We then get:
4x - 7 + 7 = 13 + 7
If we do these sums we get:
4x = 20

Now, we want to get rid of the 4, so we will divide both sides by 4. This gives us:
4x ÷ 4 = 20 ÷ 4
Again, working out these sums we get:
x = 5

So we have now solved the equation 4x - 7 = 13, and found that x = 5. 

For a little more information on this, see this website: http://www.mathsisfun.com/algebra/equations-solving.html 

There is a new worksheet on this in the "Worksheets" tab at the top of the page. Work through questions 1 - 3 on this by Friday. 

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

Finding an Equation from a Line

Today we had a look at finding equations from a line. We do already have the skills to do these, but we will do some examples as practice.

If you remember earlier in the unit, we looked at finding an equation when we were given just two points. These examples can be found in the booklet from Monday the 17th of August. We also looked at some of these questions in section 4 of the worksheet. If you do not remember this, go back and have a look at these.

Sometimes we will often be given a line on a Cartesian plane and asked find the equation of this. To do this we need to pick 2 points and use these to find the equation in the same way as I mentioned in the paragraph above.
When we pick the two points, it is important to remember to pick easy points to work with. We don't want to be picking points that are decimals or anything like this. When we pick the points, try and make sure they are nice easy whole numbers, as this will make our lives a whole lot easier.

So for example, we can see on the graph below that I have chosen 2 nice easy points with whole numbers. There is one on the left, which is (-2, 4) and the other on the right which is (2, -1). We can then use our skills we have from previous lessons to find the equation of the line.

You should now have all the skills required to work through the whole first worksheet on linear equations. I need this finished by Wednesday the 26th of August, so we can then move onto the next topic in this unit on linear algebra. 

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

Plotting lines

Today we looked at plotting lines. To be able to plot a line, we need to have at least two points. If we know two points on the line, simply draw a line through these two points and continue it through.

There are a few different techniques to find two points. We need to look at the information we are given in our question, and then use this to find our two points.

Firstly, we may be told both our y and x intercepts (such as the questions in part 5 of our worksheet). If we know these, we already have our two points. Our x-intercept is the point where the line cuts through the x-axis, so we can mark this as a point. The y-intercept is the point where our line cuts through our y-axis, so we can also mark this point. Now that we have two points marked, we can draw a straight line through this and we have our solution.
For example: Sketch the line that has an x-intercept of 4 and a y intercept of -2

Firstly, we need to draw our Cartesian plane:

Next, we can go ahead and mark our x-intercept at the point 4 on the x-axis. 

Next, we can also mark our y-intercept at the point -2 on the y-axis. 

Now that we have our two points, simply use your ruler and draw a straight line through these two points, and we are done. 

Now sometimes we may have to do some more work to find our two points. For example, all we may be given is the equation of our line (like those in part 10 of the worksheet). If we have these, we do already know one of our points. Remembering that our equation for a line is y= mx + c where "m" is the gradient and "c" is the y-intercept. So if we have an equation given to us, we can easily just read off the value for "c" and use this as a point for our y-intercept, like we did above. 

The next step is a little more complicated. We need to find another point. We do this by giving our "x" in our equation a value, and then finding out what "y" is equal to at this x coordinate. I will usually use the point x=1, as this is often very easy to solve. 

I will go through an example, and hopefully this method will make a little more sense. 

Example: Sketch the line for the equation y = 4x - 1

Again, we first need to draw our Cartesian plane.

Next, we can simply read off our y-intercept from our equation. In this equation, our y-intercept ("c" value) is -1. We can now mark this as a point on our Cartesian plane. 

Now, we want to find another point. To do this, we will substitute in a value for x, and find out what y is equal to at this x-coordinate. For this example I will let x equal 1. 

So, substituting this value of 1 in wherever there is an x in our equation gives us:

y = 4 x 1 - 1 

y = 3

So this tells us that on this line, when x=1, y=3. So we now have another point (1, 3). We can now go ahead and mark this on our Cartesian plane. 

Now we have our two points, we can now go and draw our line through these. 

Done!

Now there is one more special case. If you see on your worksheet in question 6, we are asked to sketch graphs, but the equations for these graphs are in a slightly different style to what we are used to. They are written like x = # or y = # or something like this. 

If we look at, say, y = 4 for an example. What this is basically saying is that we want a line that at every single point y is equal to 4. This means we will have a straight line like this: 

Looking at this line we can see that at every single point, y is equal to 4. It doesn't matter what x is equal to, y is always equal to 4, which is what our equation asked. 

Similarly, if we get an equation of, say, x = -3, we want to draw a straight line so that at every single point, x is equal to -3. So it would look like this:

 Again, we can see that at every single point, x is equal to -3, which satisfies our equation. 

You should now have the skills to work through questions 5, 6 and 10 on the worksheet (which can be found in the "Worksheets" tab). 

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

New Unit - Linear Algebra

We have now started a new unit on linear algebra. This basically involves working with lines, and the algebraic equation of these lines, and finding key information out about these lines. I have made up a word document that explains the first part of what we need to work through. This can be found by clicking here. I have also put a worksheet in the "Worksheets" tab, so once you have read through the word document, you should be able to complete the worksheet.

The worksheet can be found by clicking here.

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

Test Postponed

Due to a large amount of student absences today (Wednesday) I have postponed the test till Monday the 10th of August. Keep working through your assignment as revision, as this is now due on Monday as well (at the latest).
As mentioned in the last post, you may bring an information sheet into the test. It must be A5 and single sided, however it can have any information you may find useful on it. You will also need a calculator, so bring one of these if you have one.

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

Snow Day!!

Snow day today so no maths, however we will still have our test on Wednesday. Enjoy your day off, but please do  still study if you get a chance. I will still be available to contact on here if you have any questions. Use the assignment as revision. If you do not have a copy of this it can be found under the "Assignments" tab.

For the test on Wednesday, you will be allowed to bring in an A5 information sheet, with any relevant info that you think will help you with the test.This could be any formula's or any examples from any classwork or assignments. It is up to you what you put on this sheet. You may also use a calculator, however it cannot be a calculator on your phone or any other device. 

As always, comment on here if you have any questions, or email or see me in person. Enjoy the snow!!