system question: For what value(s) of k does the system have

[mathfun]

The system:

. . .2x + ky = 8
. . .4x - y = 11

For what value(s) of k does the system have a unique solution? Find the unique solution.

I found the value of k that the system will have a unique solution:

. . .-2 - 4k /= 0
. . .-4k /= 2
. . .k /= -1/2

This part is correct. However, when i tried to find the solution, I got:

[x
 y]  =  1/(-2-4k) [-1 -k    [ 8   
                          -4  2]    11]

= [-1/(-2-4k)  -k/(-2-4k)   [8
     -4/(-2-4k    2/(-2-4k]    11]

= [-8 - 11k/(-2-4k)
   -10/-2-4k)]

=[-8 - 11k/(-2-4k)
   -5/-1-2k)]

The correct answer in the book was:

. . .x = 8 + 11k/(2+4k)
. . .y = 5 / (1+2k)

Where did the negative signs go?
_______________________
Edited by stapel -- Reason for edit: inserting "code" tags

 

[soroban]

Re: system question - matrix

Hello, mathfun!

My answers: \(\displaystyle \,\begin{array}{cc}x\:=\:\frac{-8\,-\,11k}{-2\,-\,4k} \\ y \:=\:\frac{-10}{-2\,-\,4k} \end{array}\)

Book's answers: \(\displaystyle \,\begin{array}{cc}x \:= \:\frac{8\,+\,11k}{2\,+\,4k} \\ y \:= \:\frac{5}{1\,+\,2k}\end{array}\)

Where did the negative signs go?

You know this . . . factor and cancel!

\(\displaystyle \L x\;=\;\frac{-8\,-\,11k}{-2\,-\,4k}\;=\;\frac{-(8\,+\,11k)}{-(2\,+\,4k)} \;= \;\frac{8\,+\,11k}{2\,+\,4k}\)

\(\displaystyle \L y \;= \;\frac{-10}{-2\,-\,4k}\;=\;\frac{-10}{-2(1\,+\,2k)} \;=\;\frac{5}{1\,+\,2k}\)

 

[mathfun]

AHHH you are right! AHHH that was a dumb question!
I have gone insane from math!!
I try to understand the complicated applications and now i don't seem to be able to see the easy methods!

 

[steve_b]

=[-8 - 11k/(-2-4k)
-5/-1-2k)]

The correct answer in the book was
x = 8 + 11k/(2+4k)
y = 5 / (1+2k)

where did the negative signs go?

==============

The fractions in your answer:

-11k/(-2-4k) and -5/(-1-2k)

can be multiplied by -1/-1 which will give the book's answer.

Steve