Question

algebra help? (question in desc.)

Answer 1

\[\frac{ 2}{ x-1 }+\frac{ 7 }{ x }=9\]

Answer 2

hint: multiply both sides by (x-1)

Answer 3

I did... for my finalish answer I got \[9x^2-11z-2\]

Answer 4

-11x*

Answer 5

hmmm one sec

Answer 6

hmm

Answer 7

\(\cfrac{ 2}{ x-1 }+\cfrac{ 7 }{ x }=9
\\ \quad \\
({\color{brown}{ \cancel{x-1}}})\cfrac{ 2}{ \cancel{x-1 }}+({\color{brown}{ x-1}})\cfrac{ 7 }{ x }=9({\color{brown}{ x-1}})
\\ \quad \\
2+\cfrac{7x-7}{x}=9x-9\implies \cfrac{7x-7}{x}=9x-9-2
\\ \quad \\
\cfrac{7x-7}{x}=9x-11\implies 7x-7=9x^2-11x
\\ \quad \\
0=9x^2-11x-7x+7\implies 0=9x^2-18x+7\)

Answer 8

haha so my teacher taught me wrong, he said to multiply be (x-1) and (x)

Answer 9

nope, is correct, I was ..... I kinda missed the "x" :/ it should have included all denominators, so x(x-1) is fine, to get rid of the denominators

Answer 10

first

Answer 11

\(\cfrac{ 2}{ x-1 }+\cfrac{ 7 }{ x }=9
\\ \quad \\
({\color{brown}{ x\cancel{(x-1)}}})\cfrac{ 2}{ \cancel{x-1 }}+({\color{brown}{ \cancel{x}(x-1)}})\cfrac{ 7 }{ \cancel{x} }=9({\color{brown}{ x(x-1)}})
\)

would have worked, yes, so he/she's correct

Answer 12

you'd still end up with \(\bf 0=9x^2-11x-7x+7\implies 0=9x^2-18x+7\) btw, so.. just factor that, keep in mind that 7 is prime

Answer 13

anyhow.. .9 factors to 3,3,1 and 7 is prime, so it's just 7,1
so.. you can get a middle term of 18 with \(\bf 0=9x^2-18x+7\implies 0=(3x+1)(3x-7)\)

Answer 14

wait a second I messed up... it's 4/x-1

Answer 15

what the?

Answer 16

\(\bf \cfrac{ 4}{ x-1 }+\cfrac{ 7 }{ x }=9?\)

Answer 17

\[\frac{ 4 }{ x-1 }+\frac{ 7 }{ x }=9\]

Answer 18

no biggie, just do the same \(\bf \cfrac{ 4}{ x-1 }+\cfrac{ 7 }{ x }=9
\\ \quad \\
({\color{brown}{ x\cancel{(x-1)}}})\cfrac{ 4}{ \cancel{x-1} }+({\color{brown}{ \cancel{x}(x-1)}})\cfrac{ 7 }{ \cancel{x }}=9({\color{brown}{ x(x-1)}})\)

Answer 19

right but that's where I got the 9x^2-11x-2

Answer 20

\(\cfrac{ 4}{ x-1 }+\cfrac{ 7 }{ x }=9
\\ \quad \\
({\color{brown}{ x\cancel{(x-1)}}})\cfrac{ 4}{ \cancel{x-1} }+({\color{brown}{ \cancel{x}(x-1)}})\cfrac{ 7 }{ \cancel{x }}=9({\color{brown}{ x(x-1)}})
\\ \quad \\
4x+7(x-1)=9(x^2-x)
\\ \quad \\
4x+7x-7=9x^2-9x\implies 11x-7=9x^2-9x
\\ \quad \\
-7=9x^2-9x-11x\implies 0=9x^2-20x+7\)

Answer 21

ooh I see I didn't carry the x for the -9

Answer 22

alright I got it from here, thank you!

Answer 23

yw

Answer 24

my final answer was 9x^2-20x+7

Answer 25

but that's not factorable... so no solution?

Answer 26

Hey how do you cross out equations like you just did @jdoe0001

Answer 27

@Directrix

Answer 28

... but that's not the question?

Answer 29

oh yeah I messed up. the question was \[\frac{ 4 }{ x-1 }+\frac{ 7 }{ x }=9\]

Answer 30

Use the quadratic formula to solve \(\Large 9x^2 - 20x + 7 = 0\)


Once you get each possible solution, check them back into the original equation. I recommend using the decimal approximations instead of the exact radical forms.

Answer 31

\(\Large 9x^2 - 20x + 7\) isn't factorable which is why you have to use the quadratic formula.

Answer 32

nvm I messed it up either way the first time around,

you can use the determinant to find the roots after you do this if you want.