Question

Ok, What am I doing wrong:

1/ (x^2 - 7x +10) = x/ (x-5) + 1/ (x-2)
1/ (x-5)(x-2) = x/ (x-5) + 1/ (x-2)

LCM: (x-5)(x-2) so

(x-5)(x-2)( 1/ (x-5)(x-2) = (x-5)(x-2)(x/ (x-5) + 1/ (x-2)

1= x(x-5) + 1(x-2)
1=x^2 - 5x +1x - 2
1= x^2 - 4x- 2 .... solve for x by factoring... ???

I know there's going to be two answers but I don't know what I'm doing wrong

Answer 1

Your left side is correct

Answer 2

But the right side is not correct

Answer 3

what is x/ (x-5) multiplied by (x-5)(x-2) ???

Answer 4

I have to multiply both sides by the LCM, simplify, then solve

Answer 5

Yes and after you multiply the right side by (x-5)(x-2), you distribute that through

Answer 6

so 1= x(x-2) + 1(x-5)?
1= x^2 -2x+1x-5
1=x^2-x-5?

Answer 7

better

Answer 8

then make is equal to zero so 0= x^2-x-6?

Answer 9

good

Answer 10

how would i factor that though?

Answer 11

Find two numbers that multiply to -6 and add to -1

Answer 12

Those two numbers are...???

Answer 13

-3 and 2

Answer 14

good

Answer 15

So

x^2-x-6

factors to

(x-3)(x+2)

Answer 16

which means

x^2-x-6 = 0

is the same as

(x-3)(x+2) = 0

Answer 17

yay! ok.. thanks! x= -2 or 3

Answer 18

perfect

Answer 19

thank you!

Answer 20

you're welcome