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4_____________ _ 3____________
X2-6X+5                 x2-10x+25=

Hi Marie;
[4/(x2-6x+5)]-[(3/(x2-10x+25)]
Let's factor both denominators...
x2-6x+5
For the FOIL...
FIRST must be (x)(x)=x2
OUTER and INNER must add-up to -6x
LAST must be (5)(1) or (1)(5) and both numbers must be negative to render the sum of -6 and the product of +5.
(x-5)(x-1)

x2-10x+25
For the FOIL...
FIRST must be (x)(x)=x2
OUTER and INNER must add up to -10x
LAST must be (5)(5) or (25)(1) or (1)(25) and both numbers must be negative to render the sum of -10 and product of 25.
(x-5)(x-5)

{4/[(x-5)(x-1)]}  -  {3/[(x-5)(x-5)]}
Let's take the first bracketed equation and multiply it by (x-5)/(x-5).
Let's take the second bracketed equation and multiply it by (x-1)/(x-1)
{[(4)(x-5)]/[(x-5)2(x-1)]}  -  {[(3)(x-1)]/[(x-5)2(x-1)]}
Now, the denominators are the same.  Let's subtract...
[(4)(x-5)]-[(3)(x-1)]/[(x-5)2(x-1)]}
(4x-20-3x+3)/[(x-5)2(x-1)]
(x-17)/[(x-5)2(x-1)]

4/(x^2-6x+5) - 3/(x^2-10x+25)

= 4/((x-5)(x-1)) - 3/((x-5)(x-5)), x ≠ 1,5

= 4(x-5)/((x-5)(x-5)(x-1)) - 3(x-1)/((x-5)(x-5)(x-1)), x ≠ 1,5

= (4(x-5) - 3(x-1))/((x-5)(x-5)(x-1)), x ≠ 1,5

= (x-17)/((x-1)(x-5)^2), x ≠ 1,5