Question

My teacher gave me this problem:


3 + 2 = 1
x-3 x+3

The answer he gave us is:

x= -1 +/- (square root) 217
6

he wants us to show a check... please explain to me how to check this answer >(

Answer 1

drunk frog for encouragement >;3

Answer 2

oh my gosh :D Thanks, I will need It with the insane teacher I have.

Answer 3

Figured you did :3

Answer 4

My brain hurts...

Answer 5

\(\bf 3+\cfrac{2}{x-3}=\cfrac{1}{x+3}\quad ?\)

Answer 6

not sure what all of that means....

Answer 7

hmmm I see you're using ... Internet Explorer.... got any other browser handy?

try to check the same page in another browser

Answer 8

oh... that makes more sense... yes that is the correct problem

Answer 9

ok,let us move the right-hand side over

and you'd get \(\bf 3+\cfrac{2}{x-3}=\cfrac{1}{x+3}\implies 3+\cfrac{2}{x-3}-\cfrac{1}{x+3}=0\\ \quad \\
\textit{let us get the LCD, which will be} (x-3)(x+3)\qquad thus\\ \quad \\

\cfrac{3(x-3)(x+3)+2(x+3)-1(x-3)}{(x-3)(x+3)}=0\)

Answer 10

then multiplying by (x-3)(x+3) both sides we get \(\bf \cfrac{3(x-3)(x+3)+2(x+3)-1(x-3)}{(x-3)(x+3)}=0\\ \quad \\
3(x-3)(x+3)+2(x+3)-1(x-3)=0\\ \quad \\
\implies 3x^2-27+2x+6-x-3=0\implies 3x^2+x-24=0\)

as you can see is a quadratic, so you can either factor it, or use the quadratic formula

Answer 11

hmmm I have one tiny error... it should be -x +3

Answer 12

i understand how to find the answer. i just need to know how to show a check

Answer 13

\(\bf 3(x-3)(x+3)+2(x+3)-1(x-3)=0\\ \quad \\
\implies 3x^2-27+2x+6-x+3=0\implies 3x^2+x-18=0\)

Answer 14

ohhh just to check is correct
well that's simple, just take your values for "x" and plug them in the original equation

the left-side and the right-side must equate, thus is a EQUATion

Answer 15

I know that i have to substitute x for the answer that i found, but i have no clue where to start