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(7 over x-3) - (4 over x^2 - 9)the problem just says "perfrom the indicated operation...

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j1195059 | Student, Undergraduate | eNotes Newbie

Posted January 31, 2010 at 5:45 AM via web

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(7 over x-3) - (4 over x^2 - 9)

the problem just says "perfrom the indicated operation and simplify"

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guitarhero811 | Student, Grade 11 | eNotes Newbie

Posted January 31, 2010 at 8:31 AM (Answer #1)

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What you do is find the common denomonator which is simply (x-3)(x^2-9). Then you multiply that by each fraction.

(7 over x-3) times (x-3)(x^2-9) is 7x^2-63 because the (x-3)'s cancel out and you distribute the 7 to the (x^2-9).

(4 over x^2-9) times (x-3)(x^2-9) is 4x-12 (the (x^2-9)'s cancel out and distribute the 4 to the x-3)

so then you have 7x^2-63-4x-12 which simplifies to 7x^2-4x-75!

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neela | High School Teacher | Valedictorian

Posted January 31, 2010 at 2:03 PM (Answer #2)

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To perform the operation,

(7 over x-3) - (4 over x^2 - 9):

Solution:

7 over x-3 = 7/x -3: The  reason is  a o b - c means  subtract c from the result a o b.  Operation 'over' has no effect (or distributive effect) on c. Moreover 7 over x -3  are two terms : the first is 7 over x connetced by - sign to the 2nd term 3.

In a similar manner 4 over x^2-9 = 4/x^2-9.

Therefore, the given expression , (7 over x-3) - (4 over x^2 - 9) is equal to:

(7/x-3)-(4/x^2-9)

=7/x-4/x^2+9-3

=7/x-4/x^2+6 or

= (7x-4+6x^2)/x^2 or

=(6x^2+7x-4)/x^2.

But if you intend to mean (7 over x-3) - (4 over x^2 - 9) to be  7/(x-3) - 4/(x^2-9), then you have to write your expression like:

7 over (x-3) - 4 over (x^2-9) which means same as [7 over (x-3)] - [4 over (x-9)], the rectangular brackets is only redundant but not necessary and won't spoil the inteded meaning also.

The denominators, x-3 and x^2-9 has the LCM x^2-9 and could be made common denominator.Also x^2-9 = (x-3)(x+3).

So, 7 over (x-3) - 4 over (x^2-9)  = 7(x+3)/[(x-3)(x+3)]  - 4/(x^2-9).

=(7x+21)/(x^2-9)-4/(x^2-9)

=(7x+17)/(x^2-9)

 

 

 

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giorgiana1976 | College Teacher | Valedictorian

Posted February 1, 2010 at 1:17 AM (Answer #3)

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(7 over x-3) - (4 over x^2 - 9) has to be "translated" into:

[7 / (x-3)] - [4 /( x^2 - 9)]

The second step is to notice that you have the advantage  that

x^2 - 9=(x-3)*(x+3)

So,

[7 / (x-3)] - [4 /( x^2 - 9)]=[7 / (x-3)] -[4 /(x-3)*(x+3)]

Now, in order to subtract the second ratio from the first one, all we have to do is to find out the common denominator, which is obviously, (x-3)*(x+3). It is easy to see that the first ratio has to be multiplied with (x+3).

[7(x+3)-4 / (x-3)*(x+3) = (7x+21-4)/(x-3)*(x+3)

(7x+17)/(x-3)*(x+3)

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Wiggin42 | TA , Grade 11 | Valedictorian

Posted March 23, 2014 at 5:50 PM (Answer #4)

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7 over x-3) - (4 over x^2 - 9)

`7/(x-3) - (4/(x^2-9))`

`<br>`

``The first fraction can be written as 7(x-3) / (x^2 - 9) 

Now that the denominators are same, we can subtract as normal. 

 ( 7(x-3) - 4 ) / (x^2 -9)

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