Question

If I give someone an problem can you tell me if the answer I got is correct?

Answer 1

That depends.

Answer 2

Haha @ElmerHigglesworth true true...

Answer 3

I got x= -6

Answer 4

@mathstudent55

Answer 5

Hmm... this is a toughie... might take me a second.

Answer 6

Or -1/6

Answer 7

I am thinking -1/6

Answer 8

\[\eqalign{
&\frac{5}{4x}+\frac{2}{1}=\frac{1}{x} \\
&\frac{8x+6}{4x}=\frac{1}{x} \\
&8x^2+6x=4x \\
&8x^2-2x=0 \\
&2(4x^2-x)=0 \\
&2x(4x-1)=0 \\
}\]

Answer 9

I got \[x∈{{2/5,(−2)}}\]

Answer 10

So would that be NO SOLUTION?

Answer 11

If the answer was \(-\frac{1}{6}\), then:
\[\eqalign{
&\left.\left(\frac{5}{4x}+\frac{2}{1}=\frac{1}{x}\right)\right|_{x=-\frac{1}{6}} \\
&\left(\frac{5}{4\times-\frac{1}{6}}+\frac{2}{1}=\frac{1}{-\frac{1}{6}}\right) \\
}\]

Answer 12

Yeah, pretty much no solution.

Answer 13

Err wait yeah...
The solution to:
\[2x(4x+1)=0\]
Is \((x=0)\phantom{.}or\phantom{.}(x=-\frac{1}{4})\)

Answer 14

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

Answer 15

So how would I know what to put for x?

Answer 16

Wait. If this is another equation for another question, put it under a separate question and tag people to answer it. Don't put more than one question on one space.