# Math Help - Equations:

1. ## Equations:

I'm a bit confused as to where to put this, because as far as i know there's no section for "Equations"
But i've been reading through an introduction to Equations! That's the booklet i have started studying and it has: y + 2 and 5f - 1 are expressions. y and f are called variables. I know all that!
But maybe it's just the case that i've just come off my school holidays, and i'm a bit slow! But i can really remember how to do the following! well i don't really have to do it, i'm not mean't to! Right now it's just stating fact, but when i read i like to be ableto understand what their talking about: y + 2 = 11, and 5f - 1 = 14! i do understand that 5 - 1 is 4! but if it was to live upto the result of 14! it all depends on what the "f" stands for! so in this case would the f stand for 10? But how would y + 2 equal 11??

2. and i'm totally confused because i'm not used to be doing these things!
Erm, it's just like a quiz to see wheather you can get the number, but i've spend 30 minutes trying to get it! I dare say you can get it!
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Think of a number!
1. Half of a number
2. When 4 is added, the result is 12
3. when doubled, the result is 22
4. When 5 is subtracted from it, the result is 3
5. When it is subtracted from 5, the result is 3
6. When squared, the result is 36
7. Add 7 to it, then multiply by 2. The result is 24
8. a quarter of it is 12.
9. The square route of it is 4
10. Add half of it to 11 to give 15

3. Hi Sazza! I'm glad that you have an interest. I will most gladly explain these basic equations for you.

So the first one is: $y\,+\,2\,=\,11$

This means that some number plus 2 is 11. Just by looking at it I see y = 9.
But you can find y another way until you get skilled.

When solving for a variable in a simple equation, you want to get y = something. So we want to get rid of the 2. Since the 2 is positive, we subtract it from both sides, called the inverse. That's how you solve for a variable. Do the inverse(s )to the term(s) you are trying to get rid of, resulting in getting the variable by itself.

Inverse of multiplication is division
Inverse of a square is square root
There are many others

4. So now we have a little more complication problem with $5f\,-\,1\,=\,14$

So 5 times a number minus 1 is equal to 15.

So we want to get rid of the -1 and the coefficient 5 of f. So the 1 is negative so add it to both sides of the equation getting rid of it.

Now we have: $5f\,=\,15$

We can either say, what times 5 (f) gives me 15? 3

Or we could see 5f is 5 times f. We need to get rid of 5, so we do the inverse, division of both sides of equation by 5.

That get rid of the 5, getting us: $f\,=\,\frac{15}{5}\,=\,3$

Does this make sense?

5. Originally Posted by Jonboy
Does this make sense?

Yes thanks =] but i'm kinda confused here:
i'm totally confused because i'm not used to be doing these things!
Erm, it's just like a quiz to see wheather you can get the number, but i've spend 30 minutes trying to get it! I dare say you can get it!
==============
Think of a number!
1. Half of a number
2. When 4 is added, the result is 12
3. when doubled, the result is 22
4. When 5 is subtracted from it, the result is 3
5. When it is subtracted from 5, the result is 3
6. When squared, the result is 36
7. Add 7 to it, then multiply by 2. The result is 24
8. a quarter of it is 12.
9. The square route of it is 4
10. Add half of it to 11 to give 15

6. Well do some practice. I think you'll find this explanation helpful and it has practice problems: Purplemath

7. I'll do the first 5, see if you can figure out the rest from what i did
Originally Posted by Sazza

Yes thanks =] but i'm kinda confused here:
i'm totally confused because i'm not used to be doing these things!
Erm, it's just like a quiz to see wheather you can get the number, but i've spend 30 minutes trying to get it! I dare say you can get it!
==============
Think of a number!

Okay, let me call the number x

1. Half of a number
more info needed here. as far as we know, we just have x/2

2. When 4 is added, the result is 12
okay, so x + 4 = 12

=> x = 8 here

3. when doubled, the result is 22
2x = 22
=> x = 11 here

4. When 5 is subtracted from it, the result is 3
x - 5 = 3

=> x = 8 here

5. When it is subtracted from 5, the result is 3
this is the same as the last

8. 5. $5-x=3\Rightarrow x=2$
6. $x^2=36\Rightarrow x=6$ (Suppose x is a positive integer)
7. $2(x+7)=24\Rightarrow x=5$
8. $\frac{x}{4}=12\Rightarrow x=48$
9. $\sqrt{x}=4\Rightarrow x=16$
10. $\frac{x}{2}+11=15\Rightarrow x=8$

9. Originally Posted by Jhevon
I'll do the first 5, see if you can figure out the rest from what i did

Okay, let me call the number x

more info needed here. as far as we know, we just have x/2

okay, so x + 4 = 12

=> x = 8 here

2x = 22
=> x = 11 here

x - 5 = 3

=> x = 8 here

this is the same as the last

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The first number would be 14, because 7+7= 14, but it doesn't really makes sense when i read it, i mean if you were adding 4 onto that number, wouldn't it be 18? but then i relised the number 14 was what the number was before half was taken away so then it would be 8? thanks for all your help

10. and just a quick comparison and question: the aim is to Simplify the following algebraic fractions:
for example 16x/20 (it's in fraction form like the 16x is over the 20, i have not yet mastered how to do those little diagrams) would i: divide [15 divided by 20 = 8x] or [16 x 20 = 320x] i personally am inclined to believe the divided by sum, because that seems a more logical answer, but i'm just checking! so we divide? because i just rememeber in some you multiply!
and also one more question, that i just want to verify, because i'm kind of slow after the holidays! When were writing the answer we have to put the variables in Alpha Order, and for example a question like this: 15xyz/5yz how would we put the variables?? the answer plus xyyzyz?? or what? Thanks =]

11. Sazza can you just simply tell us the problem? Lol I'm lost with all the wordiness.

Is it $\frac{16}{20}x$ ?

If so you should review quotient properties: Link