Question

Solve the equation. Check your solution.

Answer 1

\[\frac{ 1 }{ 2 } + \frac{ x }{ 6 } = \frac{ 18 }{ x }\]

Answer 2

I'm not sure on where to start with this one. I did one before this but it didn't have the third piece..

Answer 3

since \(x\) cannot be equal to zero, maybe the best first step would be to multiply both sides by \(6x\) cancelling as you go

Answer 4

\(6x\left(\frac{ 1 }{ 2 } + \frac{ x }{ 6 }\right) =6x\left( \frac{ 18 }{ x }\right)\)

Answer 5

good from there, or you need more help?

Answer 6

Wouldn't you times by \[\frac{ x }{ 6 }\]

Answer 7

instead of 6x?

Answer 8

you could but that would not get rid of the denominators on the left hand side
it would make them worse

Answer 9

Oh, what about \[\frac{ 6 }{ x }\]

Answer 10

you mean "what about multiplying by \(\frac{6}{x}\)?"
that would make the right hand side much worse

Answer 11

the denominators for the equation \[\frac{ 1 }{ 2 } + \frac{ x }{ 6 } = \frac{ 18 }{ x }\] are \(2,6,\) and \(x\)
the least common multiple of these denominators is \(6x\)

Answer 12

I see.. Umm, can you show me the next step so I can see where it will go because I'm a bit confused as to where it will even out..

Answer 13

ok first since you are multiplying both sides by \(6x\) you have to multiply EVERYTHING by \(6x\)
first \[6x\left(\frac{ 1 }{ 2 } + \frac{ x }{ 6 }\right) =6x\left( \frac{ 18 }{ x }\right)\] then
\[6x\times \frac{1}{2}+6x\times \frac{x}{6}=6x\times \frac{18}{x}\] then comes the "cancelling" part

Answer 14

\[6x\times \frac{1}{2}+6x\times \frac{x}{6}=6x\times \frac{18}{x}\]
\[\cancel{6}^3x\times \frac{1}{\cancel2}+\cancel6x\times \frac{x}{\cancel6}=6\cancel x\times \frac{18}{\cancel x}\]\[3x+x^2=108\]

Answer 15

Okay, then divide everything by three?

Answer 16

no, since you do not want to divide \(x^2\) by \(3\)
set it equal to zero and solve the quadratic equation \[x^2+3 x-108 = 0\] which amazingly enough factors

Answer 17

Oh! Thank you, I see now.

Answer 18

Would it end with -12 and 9 being your answers?

Answer 19

Good luck with your studies, you were correct.

Answer 20

Thank you

Answer 21

@radar Can you help me?

Answer 22

I'm still here.

Answer 23

After you get down to \[x(x-12)(x+9)\]
Where do you go to find x for your check?

Answer 24

You actually get down to (x+12)(x-9)=0

Answer 25

It is a quadratic not a cubic.

Answer 26

Okay

Answer 27

I fixed it, now what?

Answer 28

Oh! wait

Answer 29

Is it the
(x-12) = 0
+12 both sides
x = 12

and

(x+9)=0
-9 on both sides
x = -9