All Answers 355948 Answers

x - 1          x  -  k

____  =   _______    

x - 2          x  - 6

Cross-multiply

x^2  - 7x  + 6  = x^2 - 2x -kx + 2k

When k  = 5  we  have no solutions because  we will end up with  

6 =10    which is impossible

Also...when k  = 6  the rational  function on the  right will  have the graph of the  line  y = 1  (with a "hole" at x = 6)

But the rational  function on the left will have  a horizontal asymptote at y =1.....so  these functions will never intersect when k  = 6

So....no solutions when k= 5 or k  = 6

The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.

x + y + 3z  =  10     --->     x -2     --->     -2x - 2y - 6z  =  -20

-4x + 3y + 5z  =  7       --->                      -4x + 3y + 5z  =  7

Add down the columns:                             -6x - z  =  -13   

Since  kx + z  =  3        --->         z  =  3 - kx

Substituting these:        -6x - (2 - 3kx)  =  -15

                                    -6x - 2 + 3kx  =  -15

                                      6x - 2 - 3kx  =  15

                                          6x - 3kx  =  1

                                          3x(2 - k)  =  17

If  2 - k  =  0,  there is no value for x that will result in a product of 10,

so  k = 2  results in no possible solution.

For every other value of k, there will be a value of x that can produce a product of 10.

So the answer is k = 2.

yea how can you say that to a kid? calling a kid an entitled punk and telling them to step down of their "little soapbox" sounds just wrong. i agree that the kid just sounds lonely and stressed, you should of just helped them and been on your way and don't assume their situation

Yes, 3 is the only correct answer, I found only 3 and 2/5 are the correct answer!

Yes you are right.

One of the answers is correct.

Now the answerer has given you a technique that you say you understand.

And he/she has given you one correct answer.

That correct answer makes it easier for you to solve it by yourself.  (I know because i have solve it)

So you have a technique and a correct answer.

Seems to me that the guest answerer has helped you are real lot!

Which one.  I am sure the answerer would like some positive feedback from you.

Thank you for the suggestions, I will follow your rules and try my best to "respect", by the way, My results were that only one answer the guest provided was correct.

The respondent does not need to check their answer if they do not want to.

It is up to the asker to accept, reject or alter an answer as they feel the need.

The asker should at the very least check answers by substitution if that is possible.

The respondents are here to offer their help or suggestions.

It is not the respondent's job to give polished answers that can be copied without thought.

👌 

Are you the question asker?

Yes the first three answers should have been discarded but did you bother to check the next 2 answers.

You still have not said that you have?  Are they right?  Are they wrong?

If you knew some things were correct then you should have said so.

Your response was very rude.

You can politely suggest to people that maybe they should check their answer by substitution without being rude about it.

Wait I thought of another one, if you use k = 6, then the equation will be x-1=x-2, that means -1=-2, which has no solutions, so k= 5, 6, and Idk if there are more

No, I tried 5, it is incorrect, there must me more.

Yes, I checked, since seeing the first three solutions I immediately saw and noticed, the strategy was alright, but the person needed to recheck and see if it was valid, I am not here to criticize. When I do my work, I always check.

Did you bother to check if the other answers are correct?

Or are you just here to criticize?

\(\frac{2x^2-3x}{x^2-x} + 5x -11 = \frac{3x^2+5x+2}{x^2-1}\\ \frac{2x^2-3x}{x(x-1)} + 5x -11 = \frac{3x^2+5x+2}{(x-1)(x+1)}\\ x\ne\pm1,\quad x\ne0\\ [\frac{2x^2-3x}{x(x-1)} + 5x -11]*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{(x-1)(x+1)}]*[x(x-1)(x+1)] \\ \frac{2x^2-3x}{x(x-1)} *[x(x-1)(x+1)]+ (5x -11)*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{1}]*[x] \\ (2x^2-3x)(x+1)+ (5x -11)(x(x^2-1))=x (3x^2+5x+2) \\ (2x-3)(x+1)+ (5x -11)(x^2-1)= (3x^2+5x+2) \\ (2x^2-x-3)+ (5x^3 -11x^2-5x+11)= (3x^2+5x+2) \\ 5x^3 -9x^2-6x+8= 3x^2+5x+2 \\ 5x^3 -12x^2-11x+6= 0 \\ (x-3)(5x-2)(x+1)=0\\ \text{x can't equal -1 so}\\ x=3 \;\;or\;\;x=0.4 \)

LaTex

\frac{2x^2-3x}{x^2-x} + 5x -11 = \frac{3x^2+5x+2}{x^2-1}\\
\frac{2x^2-3x}{x(x-1)} + 5x -11 = \frac{3x^2+5x+2}{(x-1)(x+1)}\\
x\ne\pm1,\quad x\ne0\\
[\frac{2x^2-3x}{x(x-1)} + 5x -11]*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{(x-1)(x+1)}]*[x(x-1)(x+1)] \\
\frac{2x^2-3x}{x(x-1)} *[x(x-1)(x+1)]+ (5x -11)*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{1}]*[x] \\
(2x^2-3x)(x+1)+ (5x -11)(x(x^2-1))=x (3x^2+5x+2) \\
(2x-3)(x+1)+ (5x -11)(x^2-1)= (3x^2+5x+2) \\
(2x^2-x-3)+ (5x^3 -11x^2-5x+11)= (3x^2+5x+2) \\
5x^3 -9x^2-6x+8= 3x^2+5x+2 \\
5x^3 -12x^2-11x+6= 0 \\
(x-3)(5x-2)(x+1)=0\\
\text{x can't equal -1  so}\\
x=3 \;\;or\;\;x=0.4

That's right, k = 5 is the only solution.

Distance  Iowa City to Ames  = √ [ (100^2 + 30^2  ] = √14500 minutes  =  20√15 minutes

Distance  Iowa City to Cedar Rapides  =√ [(100 - 120)^2  + 30^2 ]  = √1600  =  40 minutes

So...the difference in distance is  :   20√15 - 40  minutes

By the quadratic formula, the solutions are x = (5 \pm 2*sqrt(5))/2, so a + b + c + d = 12.

I hope this helps.

(By the way I got that one value of k is 5)

The probability is 3/10.

(2x^2 - 3x)/(x^2 - x) + 5x - 11 = (3x^2 + 5x + 2)/(x^2 - 1)

==> (2x^2 - 3x)(x^2 - 1) + (5x - 11)(x^2 - x)(x^2 - 1) = (3x^2 + 5x + 2)(x^2 - x)

==> 5x^2 - 19x^4 + 7x^3 + 19x^2 - 12x = 0

==> x(x - 1)(x + 1)(5x - 4)(x - 3) = 0

Solutions are 0, 1, -1, 4/5, 3

(2x  + 3.5)(2x  + 3.5  )

(2x+3.5)² =    4 x² +14x + 12.25

I think this is often referred to as the base change identity

log_c  b =   x       means     c^x = b

c^x=b                           now take log_a of both sides

x log_a c = log_a b         now divide both sides by log_a c

x =  log_a b / log_a c      see that red equation above?   x = log_c b   ....sub that in for x

log_c b  = log_a b / log_a c     which is your identity !