Question

Solving Rational Equations; Solve and show work for #14 A Medal will be rewarded! :D
Here's the link: http://www.kutasoftware.com/FreeWorksheets/Alg2Worksheets/Solving%20Rational%20Equations.pdf

Answer 1

# 12


\[\large \frac{r-4}{5r} = \frac{1}{5r} + 1\]


\[\large \frac{r-4}{5r} = \frac{1}{5r} + 1*\frac{5r}{5r}\]


\[\large \frac{r-4}{5r} = \frac{1}{5r} + \frac{5r}{5r}\]


\[\large \frac{r-4}{5r} = \frac{1+5r}{5r}\]


\[\large r-4 = 1+5r\]


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\[\large r = ???\]

Answer 2

12

\(\dfrac{r-4}{5r} = \dfrac{1}{5r} + 1\)

What's the trick?

Restriction: \(r \ne 0\)
Common Denominator: \(5r\)

Go!

Answer 3

@tkhunny Do #14, #12 has already been done for me

Answer 4

I'm not "DO"ing anything. It's not my homework. I can help YOU do it!

Answer 5

that's what I mean by do as in help

Answer 6

@marco26 You can help on 17 or 18

Answer 7

\(\dfrac{a-2}{a+3} - 1 = \dfrac{3}{a+2}\)

Restrictions: \(a \ne -3\;and\;a\ne -2\)
Common Denominator: \((a+3)(a+2)\)

Go!

Answer 8

I said 14, because that's the problem I am doing

Answer 9

@tkhunny

Answer 10

You show me #14. I have seen NONE of your work, so far, in this thread. That's not good enough. Please do better.

You should know SOMETHING about adding fractions. Now is the time to prove it.

Answer 11

@Mertsj On #14, I multiplied x^2 + 2x ---> x(x + 2) on both sides and I got: Is it right?

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

Answer 12

You CANNOT do that until you KNOW your multiplication is NOT zero. This requires that we state any restriction FIRST.

Please state the restrictions.

Answer 13

Anyway @Mertsj Is it right?

Answer 14

I am not trying to make people do all of my problems for me

Answer 15

I don't know man, geez take a chill pill anyway:

left side: x^2 + 2x

Answer 16

It gets cancelled out and no denominators except 1

Answer 17

1

Answer 18

@saifoo.khan & @satellite73 Only #14 and it's the only problem I will do

Answer 19

okay here we go

Answer 20

\[\frac{1}{x^2+2x}+\frac{x-1}{x}=1\] add up on the left
common denominator will be \(x^2+2x\) so we can add via
\[\frac{1}{x^2+2x}+\frac{(x+2)(x-1)}{x^2+2x}=1\]

Answer 21

so far so good?

Answer 22

I have been stuck for 45minutes now on that question #14

Answer 23

now add up in the numerator and get
\[\frac{1+(x+2)(x-1)}{x^2+2x}=1\] multiply out in the top , gives
\[\frac{1+x^2-x+2x-2}{x^2+2x}=1\]
\[\frac{x^2+x-1}{x^2+2x}=1\]

Answer 24

this means
\[x^2+x-1=x^2+2x\] or
\[x-1=2x\]and now you are almost done

Answer 25

subtract \(x\) from both sides, gives \(x=-1\) and that should do it unless i made an algebra mistake

Answer 26

no it seems perfect

Answer 27

yes, i checked, it is right

Answer 28

@satellite73 I didn't make you do this problem, because I didn't do it, Mertsj didn't help me and got me stuck on the problem, because he gave me a lecture. I tried to do it myself and I couldn't get to the solution.

Answer 29

hope it is clear now
good luck

Answer 30

Blame Mertsj! :D @satellite73