Question

Algebra II: Will medal- solving rational equations (question in description)

Answer 1

\[\frac{ x^2-3x-4 }{x^3-x^2 }-\frac{ 1 }{ x^2 }=\frac{ x-2 }{ x^2 }\]

Answer 2

remember the other question, get rid of those fractions first, multiply everyting by a common denominator, i would factor the x^3-x^2 first to x^2(x-1) ... maybe?

Answer 3

I'm stuck at this part though:

Answer 4

Take x^2 common in the denominator of the first fraction on left hand side

Answer 5

or you can start by taking 1/x^2 on the right hand side of the equation and then simplifying the fractions on right hand side

Answer 6

\[\large\frac{ x^2-3x-4 }{x^2(x-1) }-\frac{ 1 }{ x^2 }=\frac{ x-2 }{ x^2 }\]

multiply everything by x^2(x-1) now

Answer 7

no that will make it complicated

Answer 8

\[\large\frac{ x^2-3x-4 }{x^2(x-1) }-\frac{ 1 }{ x^2 }=\frac{ x-2 }{ x^2 } \]\[\large\frac{ x^2-3x-4 }{x^2(x-1) }=\frac{ x-2 }{ x^2 }+\frac{ 1 }{ x^2 } \]\[\large\frac{ x^2-3x-4 }{x^2(x-1) }=\frac{ x-2+1}{ x^2 } \]

Answer 9

Start like this

Answer 10

\[\frac{ x^2-3x-4(x^3-x^2) }{x^3-x^2 }-\frac{ -1(x^3-x^2) }{ 1x^2 }=\frac{ x-2(x^3-x^2) }{ x^2 }\]

Answer 11

whichever way you want, try both

Answer 12

\[\large\frac{ x^2-3x-4 }{x^2(x-1) }=\frac{ x-1 }{ x^2 } \]
Now multiply x^2 on both sides
\[\large\frac{ x^2-3x-4 }{(x-1) }=x-1 \] Here you have completely reduced fraction

Answer 13

hm...

Answer 14

but I mean, how does one cancel the negative and positive relationship shown on the second fraction?

Answer 15

\[\large\frac{ x^2-3x-4 }{x^2(x-1) }-\frac{ 1 }{ x^2 }=\frac{ x-2 }{ x^2 }\]

multiply everything bythe greatest common denominator x^2(x-1)

\[(x^2-3x-4) - 1(x-1)=(x-2)(x-1)\]

Answer 16

okay, I see

Answer 17

can you expand all that out , distribute everything

Answer 18

remember the end goal is just to get , x , alone and equal someting... x...

Answer 19

I think I've almost got it

Answer 20

\[\large x^2-3x-4 - x + 1 = x^2-x-2x+2\]

Answer 21

easier, the x^2 cancel away

Answer 22

type what you doin, not sure

Answer 23

hmm..here is the dirstibution to expand
\[(x^2-3x-4) - 1(x-1)=(x-2)(x-1)\]

Answer 24

oh I see, so my answer would be -5?

Answer 25

i typed that wrong with the + - sines

x^2 - 3x - 4 - x + 1 = x^2 - 3x + 2

Answer 26

take an x^2 off both sides, it cancels out, good
combine all the like x terms and the constant number terms

-4x - 3 = -3x+2

Answer 27

add 4x both sides

-3 = x + 2

take off 2

-5 = x

yep

Answer 28

open study is glitching bad tonight and making me repeat previous conversation. Anyway, thank you so much!

Answer 29

welcome, just moving things around in an equation... a few practice probs and you have it down