Subtracting rational expression

S

stephm071

New member
Joined
Mar 3, 2006
Messages
1
Can someone help me with this question. Sorry so messy.... I have left my answer on here and my work leading up to it, but I don't think it is correct....

X
---------------
X^2-10x+25

MINUS

x-4
-----
2x-10

___________________________________________________________________


first....

x
-----
(x-5)(x-5)

subtract

x-4
------
2(x-5)
___________________________________________________________________

2x (x-4)(x-5)
------------- - -----------
2(x-5)(x-5) 2(x-5)(x-5)
_____________________________________________________________

2x(x-4)(x-5)
--------------
2(x-5)(x-5)


Result:

2x(x-4)
---------
2(x-5)


THANK YOU!
 
That's not correct...plus I can't follow what you're doing...

We have:
x / [(x-5)(x-5)] - (x-4) / [2(x-5)]

let a = x-4 and b = x-5
so we now have:
x / b^2 - a / (2b)
= (2bx - ab^2) / (2b^3) : applying LCD
= b(2x - ab) / (2b^3)
= (2x - ab) / (2b^2)

Substitute back in:
[2x - (x-4)(x-5)] / [2(x-5)^2]
 
stephm071 said:
Can someone help me with this question. Sorry so messy.... I have left my answer on here and my work leading up to it, but I don't think it is correct....

X
---------------
X^2-10x+25

MINUS

x-4
-----
2x-10

___________________________________________________________________


first....

x
-----
(x-5)(x-5)

subtract

x-4
------
2(x-5)
___________________________________________________________________

2x (x-4)(x-5)
------------- - ----------- CORRECT TO HERE
2(x-5)(x-5) 2(x-5)(x-5)
_____________________________________________________________

2x(x-4)(x-5)
--------------
2(x-5)(x-5)


Result:

2x(x-4)
---------
2(x-5)


THANK YOU!
Click to expand...

Once you have the denominators the same, then you need to SUBTRACT the numerators:

2x - (x - 4)(x - 5)
-------------------
2(x - 5)(x - 5)

2x - (x<SUP>2</SUP> - 9x + 20)
---------------------
2(x - 5)(x - 5)

2x - x<SUP>2</SUP> + 9x - 20
-----------------------
2(x - 5)(x - 5)

-x<SUP>2</SUP> + 11x - 20
-------------------
2(x - 5)(x - 5)

The numerator cannot be factored over the integers, so the fraction is in simplest form.
 
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