Question

Solve the equation. Check the solution.

Answer 1

\[\frac {6} {x^2-9} - \frac {1} {x-3} =1\]

Answer 2

I believe it to be -4..?

Answer 3

hmm have you covered quadratic factoring yet?

Answer 4

I'm not quite sure I understand. I may have but I think we learned it by a different name....can't quite remember..give me a sec ;)

Answer 5

Oh duh, yes I have lol isn't it like: \[ax^2+bx+c\]

Answer 6

so \(x^2-9\) is (x-3)(x+3)

Answer 7

well.. yes
one way to go about it would be using the LCD, which is (x-3)(x+3)
anotrher way we could go about it is, by simply multiplying all by that LCD
so let us use the one multiplying... one sec

Answer 8

I have gotten to:

\[\frac {-x+9} {(x-3)(x+3)}=1\] but that's pretty much it..

Answer 9

Do I multiply both sides by the denominator now?

Answer 10

\(\bf \cfrac {6} {x^2-9} - \cfrac {1} {x-3} =1\implies \cfrac {6} {(x-3)(x+3)} - \cfrac {1} {x-3} =1
\\ \quad \\
{\color{brown}{ {\color{brown}{\cancel{ (x-3)(x+3)} }}}}\cdot \cfrac {6} {\cancel{ (x-3)(x+3)}} - {\color{brown}{ {\color{brown}{ \cancel{ (x-3)}(x+3)}}}}\cdot
\cfrac {1} {\cancel{ x-3}}\\ \qquad ={\color{brown}{ {\color{brown}{ (x-3)(x+3)}}}}\cdot 1
% ------- above
\\ \quad \\
6-(x+3)=(x-3)(x+3)\implies 6-x-3=x^2-9\)

Answer 11

Ope, got that wrong.......then put it in:

\(ax^2+bx+c\) which is \(x^2+x-12\)
then into the quadratic formula (I like this formula :D )

\[x=\frac{ -(1) \pm \sqrt {(1)^2-4(1)(-12)} }{ 2(1) }\]

Answer 12

hmm actually, you don't need the quadratic formula

Answer 13

is just two plain integers

Answer 14

Because we can factor it....(x+4)(x-3)

Answer 15

yeap

Answer 16

So x=-4 and x=3

Answer 17

Thanks :)

Answer 18

"So x=-4 and x=3" \(\huge \checkmark\)

Answer 19

:D

Answer 20

yw