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**demha****Member**- Registered: 2012-11-25
- Posts: 195

I am having a hard time understanding how to do these correctly. Could someone please help explain how to do them step by step?

1. (3x + 6x)/3x

2. (21x3 + 14x) / 7

3. (4x5 + 8x2) / 4x

4. [2(x 1) + 3(x 1)] / (x 1)

5. [3(2x 3) x(2x 3)] / (2x - 3)

6. [x2(5x + 6) 3(5x + 6)] / (5x + 6)

7. [3x2(2x - 3) + 27x3(2x - 3)] / [3x(3 2x)]

"The thing about quotes on the Internet is you cannot confirm their validity"

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

hi demha

OK let's start with Q1

Now you could simplify the top first but, looking at the later ones, it is probably better to split into two fractions.

Can you finish this one from here?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Sorry I haven't answered sooner. I have been busy with other matters.

To finish from there. If I'm correct, we have to reduce it.

So:

3x/3x + 6x/3x

Reducing:

1x/1x + 3x/1x

For first fraction:

1x goes into 1x one time so it will become just x.

For second fraction:

1x goes into 3x three times so it will become 3x.

So the problem becomes x + 3x. So the answer is 4x?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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For first fraction:

1x goes into 1x one time so it will become just x.

No it will become 1

Similarly, 6x/3x = 2

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

So you don't reduce them?

So then:

3x/3x + 6x/3x

For first fraction:

3x goes into 3x 1 time.

For second fraction:

3x goes into 6x 2 times.

Then it becomes 1 + 2 and of course the answer is 3.

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

hi demha

That's correct. Shall we try Q2 ?

Try to split this one into two fractions and then simplfy each one.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

So that would be set as:

21x^3/7 + 14x/7

For the first fraction:

7 goes into 21x 3x times I suppose. This will leave it with 3x^3?

For the second fraction:

7 goes into 14x 2x times.

The problem then becomes:

3x^3 + 2x

They are not entirely alike so you can't add them. Therefor that is the answer I believe?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

hi demha,

Yes, that's correct, well done.

The others are similar, although the expressions get more complicated.

How about trying Q3 and Q4 and then I'll check them.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

3.

(4x^5 + 8x^2) / 4x

4x^5/4x + 8x^2/4x

For first equation:

4x goes into 4x 1 time so it will be 1^5

For second equation:

4x goes into 8x 2 times so it will be 2^2.

It will then be:

1^5 + 2^2

I hope I did this one correct.

Number 4 seems to be pretty complicated. I'm not too sure how to start it!

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

hi demha,

You shouldn't have lost the xs. Have a look at this:

Cancel the 4 top and bottom and one x top and bottom to leave

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Oh alright! And the: 8x^2/4x would become 2x?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,726

That is correct!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

So how would you do number 4?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

hi demha

Q4. Same method:

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

2(x 1) + 3(x 1) / (x 1)

2(x 1)/(x - 1) + 3(x 1)/( x - 1)

For the first fraction:

so (x - 1) goes into (x - 1) one time making it 2(1). Now I want to stop right here and make sure that I'm doing this right.

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
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yes. first part is 2.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Glad to know I'm on the right track.

Second part of the equation:

3(x 1)/( x - 1)

Same way, (x - 1) goes into (x - 1) 1 time making it 3(1).

So now the equation is:

2 + 3 and the final answer will be 5.

Am I correct?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,726

Hi;

Correct! Very good!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

Well done.

5, 6 and 7 are very similar to 4, so have a go and post your answers.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

#5

[3(2x 3) x(2x 3)] / (2x - 3)

First fraction:

3(2x - 3)/(2x - 3)

(2x - 3) goes into (2x - 3) 1 time.

3(1)

Second fraction:

x(2x - 3)/(2x - 3)

(2x - 3) goes into (2x - 3) 1 time.

x(1)

Answer:

3 - x

#6

[x2(5x + 6) 3(5x + 6)] / (5x + 6)

First fraction:

x^2(5x + 6)/(5x + 6)

(5x + 6) goes into (5x + 6) 1 time.

x^2(1)

Second fraction:

3(5x + 6)/(5x + 6)

(5x + 6) goes into (5x + 6) 1 time.

3(1)

x^2 - 3

#7

[3x^2(2x - 3) + 27x^3(2x - 3)] / [3x(3 2x)]

This one seems a little confusing.

---

I also have two more having some trouble with:

8. 15x^7 45x^5 / 3x4

19. (3x 2)(x - 4) (x - 4)(6 5x) / (4 x)(8x 1)

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

#5

[3(2x 3) x(2x 3)] / (2x - 3)

First fraction:

3(2x - 3)/(2x - 3)

(2x - 3) goes into (2x - 3) 1 time.

3(1)

Second fraction:

x(2x - 3)/(2x - 3)

(2x - 3) goes into (2x - 3) 1 time.

x(1)

Answer:

3 - x

correct!

#6

[x2(5x + 6) 3(5x + 6)] / (5x + 6)

First fraction:

x^2(5x + 6)/(5x + 6)

(5x + 6) goes into (5x + 6) 1 time.

x^2(1)

Second fraction:

3(5x + 6)/(5x + 6)

(5x + 6) goes into (5x + 6) 1 time.

3(1)

x^2 - 3

correct!

#7

[3x^2(2x - 3) + 27x^3(2x - 3)] / [3x(3 2x)]

This one seems a little confusing.

At this stage you can divide each part and then simplify.

(3-2x) goes into (2x-3) exactly minus 1 times

Let's get this one finished and then we can look at 8 and 9.

Bob

---

I also have two more having some trouble with:

8. 15x^7 45x^5 / 3x4

19. (3x 2)(x - 4) (x - 4)(6 5x) / (4 x)(8x 1)

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

[3x^2(2x - 3) + 27x^3(2x - 3)] / [3x(3 2x)]

First one:

3x^2(2x - 3)/3x(3 2x)]

So since it is (2x - 3) and (3 - 2x), it goies into it but it becomes a -1? And as for the 3x^2 and 3x, would this becomes x^2?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Wait, I think it just becomes x(-1)

27x^3(2x - 3) / 3x(3 2x)

(2x - 3) goes into (3 - 2x) -1 time. 3x goes into 27 9 times and takes down the ^3 to ^2 which makes it 9^2(-1)

-x + -9^2

Is that correct?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,548

Nearly

takes down the ^3 to ^2 which makes it 9^2(-1)

should be 9x^2(-1)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

So the equation is

-x + -9x^2 then?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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