Just a quick simple algebra question...?

Question

I can't beleive I don't remmeber this but just a quick question.

How do you reduce this:

x(squared) - x - 6/ x(squared) - 8x + 15

I know the two x (squared)'s cancel to male

x - 6 / 8x + 15

but can I reduce that farther?
and what are the restrictions for what x can equal?

Thank you! =)

Answer 1

I'm going to assume you mean:
(x^2 - x - 6) / (x^2 - 8x + 15)

It has a different meaning than:
x^2 - x - 6/x^2 - 8x + 15

The x squared do not cancel as you indicated. To see an example, consider:

(4 + 3) / (4+ 10)
You do not cancel the 4s to get 3/10. That's why you can't just cancel the x squared.

Factor both the numerator and the denominator to get:
(x-3)(x+2) / (x-3)(x-5)

You can now cancel x-3 from top and bottom to get:
(x+2) / (x-5)

That is about as reduced as you can get.

Answer 2

Pardon a correction, but you cannot do that, cancelling without justification, as there is no rule that permits that. (Counterexample: 12/11 = (10 + 2)/(10 + 1) =/= 2/1 = 2.)

Your original problem was to reduce (x^2 - x - 6)/(x^2 - 8x + 15). Here, the order of operations is observed, and ^2 was used to mean squared. Now, what you do is to factor both numerator and denominator and cancel factors, not terms. I tell you now that one factor cancels, and that's it.

Good luck.

Answer 3

Consider the numerator first.
x(squared) - x - 6 = x(squared) - 3x + 2x - 6
Taking common,
x ( x - 3 ) + 2( x - 3 ) = ( x - 3 )( x + 2 )
( x - 3 )( x + 2 ) is the numerator.

Consider the denominator.
x(squared) - 8x + 15 = x(squared) - 5x - 3x + 15
Taking common,
x ( x - 5 ) - 3 ( x - 5 ) =( x - 3 )( x - 5 )
( x - 3 )( x - 5 ) is the denominator
Given equation:
[ ( x - 3 )( x + 2 ) ] / [ ( x - 3 )( x - 5 ) ]

( x - 3 ) cancelled.
Answer is ( x + 2 ) / ( x - 5 )

You got it wrong earlier.

Answer 4

watch your brackets!
(x^2-x-6) / (x^2-8x+15) = (x+2)(x-3) / (x-5)(x-3) = (x+2)/(x-5)

which is as far as you should reduce it. Your restriction for x is that the denominator (bottom) of the fraction can not equal zero. so here, we cannot have x=5. apart from that there are no restrictions.

Answer 5

If you mean (x^2-x-6)/(x^2-8x+15), then
this =(x-3)(x+2) / [(x-7)(x-8)]
There is no factor common to top and bottom, and therefore nothing to cancel.
x cannot equal 7 or 8, since these make the bottom zero, and division by zero is impossible, but all other values are fine.

Answer 6

You must factorise this first-:

x^2 - x - 6 / x^2 - 8x + 15

(x - 3) (x + 2) / (x - 5) (x - 3)

(x - 3) can cancel leaving-:

(x + 2) / (x - 5)

Answer 7

x² - x - 6 / x² - 8x + 15

(x - 3)(x + 2) / (x - 3)(x - 5)

(x + 2) / (x - 5). . . .(x - 3) cancell

- - - - - - - - - - - -s-

Answer 8

replace the two x in the 1st equation with 2y-a million So that is 2(2y-a million) =3 Distribute the two. 4y-2=3 upload 2 to the two area. 4y=5 which means: y=5/4. positioned this back in to remedy for x. x= (10/4)-a million x= 6/4 adult males adult males... we are think to grant him the incorrect answer so he will learn his lesson

Answer 9

[x(sq) - x -6 ] / [x(sq) - 8x +15]

= [x(sq) - 3x+ 2x -6] / [ x(sq) - 3x-5x +15]

= [(x-3)(x+2)] / [ (x-3)(x-5) ]

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

is it what you are asking for??? if not then please specify...