Question

How many solutions does this equation have?

Answer 1

9/x-5 + 4/x^2-3x-10 = 6/x+2

Answer 2

I kind of know how to do this but not really

Answer 3

I know I need a common denominator first

Answer 4

I think there would be 2 solutions because the highest exponent tells you how many solutions there.

Answer 5

which would be (x-5)(x+2)

Answer 6

It does? I didnt know that.

Answer 7

Yeah I learned that the exponents show how many solutions there are

Answer 8

Ok then... lol

Answer 9

Haha yeah i learned that from some earlier lesson I had in school a couple months ago.

Answer 10

cool beans

Answer 11

Is the following the problem expression?\[\frac{4}{x^2-3 x-10}+\frac{9}{x-5}=\frac{6}{x+2} \]

Answer 12

YES

Answer 13

Move the RHS to the LHS.\[\frac{4}{x^2-3 x-10}+\frac{9}{x-5}-\frac{6}{x+2}=0 \]Combine the fractions.\[\frac{52+3 x}{(-5+x) (2+x)}=0 \]The numerator has to be zero. There is only one solution,\[x=-\frac{52}{3} \]

Answer 14

ok then 1 it is lol

Answer 15

wait could you show me the last steps that lead to the final answer?

Answer 16

you there???

Answer 17

Yes.

Answer 18

I used Mathematica and now I have to reverse engineer the answer.

Answer 19

Ok.

Answer 20

\[\frac{4}{x^2-3 x-10}+\frac{9(x+2)}{(x-5)(x+2)}-\frac{6(x-5)}{(x+2)(x-5)} \]\[\frac{4}{x^2-3 x-10}+\frac{9(x+2)}{\left(x^2-3 x-10\right)}-\frac{6(x-5)}{\left(x^2-3 x-10\right)} \]\[\frac{4+9(x+2)-6(x-5)}{x^2-3 x-10} \]\[\frac{(52+3 x)}{x^2-3 x-10} \]

Answer 21

ok thats what i have. your final answer was different before. that is why i was asking.

Answer 22

As it turns out the denominator can be factored.\[\frac{3 x+52}{x^2-3 x-10}=\frac{3 x+52}{(x-5) (x+2)} \]Mathematica frequently changes the order of terms and as a result it's expression forms "don't look right" and can be judged invalid by the casual observer. Takes a while to adjust to interpreting it's expressions, but then it becomes, second nature.