writing (5/4x^2 - 25) - (2x+3) / ((4x^2+16x+15)/7) as a single fraction

bumblebee123

Junior Member
Joined
Jan 3, 2018
Messages
200
Hi, I have been working on understanding this question for a few days. I have figured out a bit but am at a loss with the end bit. So, here goes:

Write

(5/4x^2 - 25) - (2x+3) / ((4x^2+16x+15)/7)

as a single fraction in its simplest form.

The answer is :

(40-14x)/(4x^2-25)

I hope I have written it right.

In short, although it's taken me ages to get there, I have

looked at the right-hand side and dealt with division first:

(2x+3) / ((4x^2+16x+15)/7)
to become
7/(2x+5)
I was able to factorize the (4x^2+16x+15) to (2x+5)(2x+3) so that I could then divide the (2x+3) to get rid of it from top and bottom of the fraction to leave 7/(2x+5)

Then I had to deal with subtracting it:

(5/4x^2 - 25)- 7/(2x+5)

But I thought I had to have the denominators the same so I tried multiplying left side with (2x+5) and right side with this 4x^2 - 25 and ended up with
(5(2x+5)-7(4x^2-25))/((2x+5)(4x^2-25))

but its not right.

Any help would be great, thank you in advance.
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,723
Write
(5/4x^2 - 25) - (2x+3) / ((4x^2+16x+15)/7)
as a single fraction in its simplest form.
The answer is :
(40-14x)/(4x^2-25)
I hope I have written it right.
That is not correct.
You can check yourself by assigning a value to x,
then substituting in both expressions.

The 1st expression probably requires more bracketing.....
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,442
Hi, I have been working on understanding this question for a few days. I have figured out a bit but am at a loss with the end bit. So, here goes:
Write (5/4x^2 - 25) - (2x+3) / ((4x^2+16x+15)/7) as a single fraction in its simplest form.
The answer is : (40-14x)/(4x^2-25)
That is almost impossible to read.
Is it \(\displaystyle \frac{{\frac{5}{{4{x^2} - 25}} - (2x + 3)}}{{\frac{{4{x^2} + 16x + 15}}{7}}}~???\)
I ask because you write of a RHS but there are no sides.
Moreover, if that is correct then the given answer does not seem right.
Please advise.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,698
If you think of this expression as having two sides you probably will think that you can multiply,divide, add or subtract the same quantity on both sides. In reality there are two sides ONLY when you have an equal sign and the equal sign divides the equation into two parts.

For example if you 5+2, you can not say that now you have 5*3 + 2*3 (I multiplied both sides by 3. The problem is that 5+2 = 7 while 5*3 + 2*3 = 21.

Now if you had 5+2 = 7, then you can say 5*3 + 2*3 = 3(5+2) = 3*7. This says 21 = 21. This is valid!
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,442
Hi, I have been working on understanding this question for a few days. I have figured out a bit but am at a loss with the end bit. So, here goes:

Write

(5/4x^2 - 25) - (2x+3) / ((4x^2+16x+15)/7)

as a single fraction in its simplest form.

The answer is :

(40-14x)/(4x^2-25)

I hope I have written it right.

In short, although it's taken me ages to get there, I have

looked at the right-hand side and dealt with division first:

(2x+3) / ((4x^2+16x+15)/7)
to become
7/(2x+5)
I was able to factorize the (4x^2+16x+15) to (2x+5)(2x+3) so that I could then divide the (2x+3) to get rid of it from top and bottom of the fraction to leave 7/(2x+5)

Then I had to deal with subtracting it:

(5/4x^2 - 25)- 7/(2x+5)

But I thought I had to have the denominators the same so I tried multiplying left side with (2x+5) and right side with this 4x^2 - 25 and ended up with
(5(2x+5)-7(4x^2-25))/((2x+5)(4x^2-25))

but its not right.

Any help would be great, thank you in advance.
I believe you intended to write 5/(4x^2 - 25) - (2x+3) / ((4x^2+16x+15)/7).

If so, your work is correct through 5/(4x^2 - 25) - 7/(2x+5).

What you need now is a common denominator. What you did would not make the LOWEST common denominator, requiring some extra simplification. (Also, you are probably expected to expand the numerator of your final answer - distribute and combine like terms - rather than leave it as you show it.)

Instead, first factor the first denominator, 4x^2 - 25. Then you can multiply the second fraction's numerator and denominator by just one linear factor. Then you will be able to add. (Presumably when you wrote about left and right sides, which are terms we use for sides of an equation, you meant the first and second fractions.)

I haven't checked for other errors.
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,723
I was able to factorize the (4x^2+16x+15) to (2x+5)(2x+3) so that I could then divide
the (2x+3) to get rid of it from top and bottom of the fraction to leave 7/(2x+5).
You were ok up to this point.
Then I had to deal with subtracting it:
(5/4x^2 - 25)- 7/(2x+5) *****

But I thought I had to have the denominators the same so I tried multiplying
left side with (2x+5) and right side with this 4x^2 - 25 and ended up with:
(5(2x+5)-7(4x^2-25))/((2x+5)(4x^2-25)) but its not right.
(5/4x^2 - 25) - 7/(2x+5) *****

= (5/((2x+5)(2x-5))) - 7(2x+5)

= 1/(2x+5) * (5/(2x-5) - 7)

= 1/(2x+5) * ((5 - 7(2x-5))/(2x-5))

= (5-(14x-35)) / (4x^2-25)

= (5 - 14x + 35) / (4x^2 - 25)

= (40 - 14x) / (4x^2 - 25) = given solution

You owe me 2 Tylenols !
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,723
(5/4x^2 - 25) - (2x+3) / ((4x^2+16x+15)/7)
Would have been clearer if shown this way:

[5 / (4x^2 - 25)] - [(2x + 3) / ((4x^2 + 16x + 15) / 7)]

Remark: x <> +- 2.5 : do you see why?
 

bumblebee123

Junior Member
Joined
Jan 3, 2018
Messages
200
Hi,

Thanks everyone.

I wasn't sure if I had written it out right, so for confusion.

Denis and Dr Peterson you are quite right I mean't
[5 / (4x^2 - 25)] - [(2x + 3) / ((4x^2 + 16x + 15) / 7)]

Looking at your workings Denis, I now understand and follow the factorising
= (5/((2x+5)(2x-5))) - 7(2x+5)

Now I am lost again.

I would still try and get denominator the same so I can deal with numerators but I still don't get it. I can't see how you get anything past the factorising. Can you explain a little more?

Thank you.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
4,442
Denis' approach was to factor 1/(2x+5) from both terms, and then combine the remaining fractions using their least common denominator, (2x-5).

The more routine approach is to use the common denominator (2x+5)(2x-5), by multiplying the numerator and denominator of the second fraction by the "missing" factor 2x-5.

Are you familiar with using the LCD? As an example in numerical fractions, to add 2/15 + 4/5, you would change 4/5 to (4*3)/(5*3) = 12/15, and then add 2/15 + 12/15 = 14/15. You're doing the same thing here with variables.
 

bumblebee123

Junior Member
Joined
Jan 3, 2018
Messages
200
Thank you so much Dr Peterson.

I get it. I was vaguely familiar with what you described but haven't done it much so it didn't occur to me. Now you have explained it with numbers I can see what you mean.
I am now able to understand it to the end. Thanks again.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,577
Hi,

Thanks everyone.

I wasn't sure if I had written it out right, so for confusion.

Denis and Dr Peterson you are quite right I mean't
[5 / (4x^2 - 25)] - [(2x + 3) / ((4x^2 + 16x + 15) / 7)]

Looking at your workings Denis, I now understand and follow the factorising
= (5/((2x+5)(2x-5))) - 7(2x+5)

Now I am lost again.

I would still try and get denominator the same so I can deal with numerators but I still don't get it. I can't see how you get anything past the factorising. Can you explain a little more?

Thank you.
[5/((2x+5)(2x-5))] - 7/(2x+5)

=[5/((2x+5)(2x-5))] - [7*(2x-5)/{(2x+5) *(2x-5)}]

=[5/((2x+5)(2x-5))] - [(14x-35)/{(2x+5) *(2x-5)}]

={5 - (14x -35)}/{(2x+5)(2x-5)}]

={5 - 14x +35}/{(2x+5)(2x-5)}]

={40 - 14x}/{(2x+5)(2x-5)}
 
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