4_____________ _ 3____________

X2-6X+5 x2-10x+25=

4_____________ _ 3____________

X2-6X+5 x2-10x+25=

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Middletown, CT

Hi Marie;

[4/(x^{2}-6x+5)]-[(3/(x^{2}-10x+25)]

Let's factor both denominators...

x^{2}-6x+5

For the FOIL...

FIRST must be (x)(x)=x^{2}

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)

x^{2}-10x+25

For the FOIL...

FIRST must be (x)(x)=x^{2}

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)]

Westford, MA

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

Bonney Lake, WA

Answer is: (x-17)/((x-1)(x-5)^2)

Factor the denominator in both terms:

First term: 4/(x^2-6x+5) = 4/((x-5)(x-1))

Second term: 3/(x-5)^2

LCD is (x-1)(x-5)^2

So we get: (4(x-5) - 3(x-1))/((x-1)(x-5)^2)

the numerator simplifies to x-17

the denominator is already factored into (x-1)((x-5)^2)

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