Question

Hi can someone walk me thru the steps:

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

I know that x^2-7x+10 can be factored to: (x-2)(x-5)
What to do next

Answer 1

\(\large\color{black}{ \displaystyle \frac{1}{x^2-7x+10}=\frac{x}{x-5}+\frac{1}{x-2} }\)

add the fractions on the right side (by finding the common denominator).

Answer 2

(note though, that you will have two restrictions for x
\(\large\color{black}{ \displaystyle x\ne2 }\)
\(\large\color{black}{ \displaystyle x\ne5 }\)
and if your solution is for some reason x=2 or x=5, then it is extraneous)

Answer 3

add the fractions on the right hand side for me please.

Answer 4

What would be the common denominator how do i add them
is the common denominator x^2-7x+10

Answer 5

yes

Answer 6

I still don't under I got
(1/x^2-7x+10)=(x^2-x-5)/(x^2-7x+10)

Answer 7

Sorry about that
I meant
I still don't understand it

Answer 8

I tried cross multiplying but I am stuck

Answer 9

Confused please help

Answer 10

is the problem to solve for x ?

Answer 11

\[ \frac{1}{x^2-7x+10}=\frac{x}{x-5}+\frac{1}{x-2} \\
\frac{1}{(x-5)(x-2)}=\frac{x}{x-5}+\frac{1}{x-2}
\]
multiply both sides by (x-5)(x-2) like this:
\[ (x-5)(x-2)\cdot \frac{1}{(x-5)(x-2)}=(x-5)(x-2)\frac{x}{x-5}+(x-5)(x-2)\frac{1}{x-2} \]
now "cancel" where you have the same thing in the top and bottom
\[ \cancel{(x-5)(x-2)}\cdot \frac{1}{\cancel{(x-5)(x-2)}}=\cancel{(x-5)}(x-2)\frac{x}{\cancel{x-5}}+(x-5)\cancel{(x-2)}\frac{1}{\cancel{x-2}} \]

all of that mess becomes
\[ 1 = (x-2)x + x-5 \]
or
\[ 1 = x^2 -2x +x -5 \]
can you finish ?

Answer 12

Yes thanks I think i can