when you're subtracting rationals, do you make it into a plus sign and make EVERYTHING in the numerator of the second negative, or just make 1 number negative? for examples x-1 4x+2 ---- - ----- x(x-3) x(x-3) would you make it into a plus sign, and then make 4x and 2 negative? -4x-2
yes
in the other equation you did how did you get positive 2?
the last one you answered
\[\frac{x-1-(4x+2)}{x(x-3)}\] \[\frac{x-1-4x-2}{x(x-3)}\] etc
well in the last one you caught me in a dumb algebra mistake that is why!
should be \[\frac{x+2}{x}=\frac{x-1}{x}-\frac{4x+2}{x(x-3)}\] \[(x+2)(x-3)=(x-1)(x+3)-4x-2\] \[x^2-x-6 = x^2-2 x-5\] \[-x-6=-2x-5\] \[x=1\]
wait what? i get x^2-x-6=x^2-8x+1 i keep on looking it over but i can't figure it out darn it
oh man did i mess up again? hold on
ok so, (x-3) and (x+2) equal x^2-x-6. (x-1) and (x-3) euqlas x^2-4x+3. then when i combine them x^2-x-6=x^2-4x+3x-4x-2
ok now you have caught me in two mistakes, so i am going to quit. you are right it is \[(x+2)(x-3)=(x-1)(x-3)-4x-2\]
then i combine like terms and get x^2-x-6=x^2-8x+1
no please don't this is a very confusing problem lol
then i get 2x^1+7x+7
ok but i am making mistake after mistake. lets see what we get now
i mean 2x^2+7x+7
did you get \[x^2-x-6 = x^2-8 x+1\]?
yes
ok subtract \[x^2\] from both sides
let me know what you get
7x+7
what happened to your equal sign?
0=7x+7
oh wait let me do this again
OMFG
\[x^2-x-6 = x^2-8 x+1\] \[\cancel{x^2}-x-6=\cancel{x^2}-8x+1\] \[-x-6=-8x+1\]
I GOT THE RIGHT ANSWER THANK YOU!!!!!!!!!!!!!! FINALLY THIS PROBLEM IS THE WORST THANK YOU!!!!!!!!!!!!!
YES!!!!!!!!!
took forever!!!!!!!
x=1
yw. now i have a question. what is vivistee?
my cat that ran away
never forget
listen thanks alright? my gosh that problem was so confusing and frustrating
:"}
it's finally over now :"}
i'm going to write that one down in my notes
glad you got it, sorry about your cat
Join our real-time social learning platform and learn together with your friends!