Question

Can anyone help with this one
4/x +7 = 2/3x Thank you!!

Answer 1

So you have:

\[\frac{4}{x} + 7 = \frac{2}{3x}\]

We can actually multiply through by \(x\):

\[4 = 7x + \frac{2}{3}\]

Answer 2

Er, sorry:

\[4 + 7x = \frac{2}{3}\]

Answer 3

Can you solve that equation a little more easily?

Answer 4

Thank you so much...giving that a try.

Answer 5

Cool :)

Answer 6

I got -10/21

Answer 7

but that's not right...what am I doing wrong?

Answer 8

That is right!

Answer 9

Try plugging back into the original equation:

\[\begin{align}
\frac{4}{\frac{-10/21}} + 7 &= \frac{2}{3\frac{-10}{21}}\\
\frac{4\cdot 21}{-10} + 7 &= \frac{2\cdot 21}{-30}\\
\frac{84}{-10} + \frac{70}{10} &= \frac{42}{-30}\\
\frac{-84 + 70}{10} &= \frac{-14}{10}\\
\frac{-14}{10} &= \frac{-14}{10}
\end{align}\]

Answer 10

Whoops. Sorry about that.

\[\begin{align}
\frac{4}{\frac{-10}{21}} + 7 &= \frac{2}{3\frac{-10}{21}}\\
\frac{4\cdot 21}{-10} + 7 &= \frac{2\cdot 21}{-30}\\
\frac{84}{-10} + \frac{70}{10} &= \frac{42}{-30}\\
\frac{-84 + 70}{10} &= \frac{-14}{10}\\
\frac{-14}{10} &= \frac{-14}{10}
\end{align}\]

Answer 11

must have plugged it in wrong....thanks so much! I may be back : )

Answer 12

Awesome! :)

Answer 13

Hey, Do you think you could help with one more?

Answer 14

1/x - 5/x^2 = 8/5x

Answer 15

Ok, so generally you want to multiply by the greatest power in a denominator so that you can have no \(x\) in the denominators. What is that in this case?

Answer 16

Thanks. Sorry, Im multi-tasking so I keep walking away. So multiply by \[x^{2}\] ?

Answer 17

That's okay :) Yes, that's right. So what would that make the equation look like?

Answer 18

x-5 = 8x/5

Answer 19

Exactly. Do you think you can solve that one?

Answer 20

Thanks, I'll try.

Answer 21

Cool :)

Answer 22

x= -25/3

Answer 23

That's what I got too :) Remember to double-check it by plugging it into the original equation and making sure the equality holds true.

Answer 24

shadow can u help me in the question i asked earlier

Answer 25

not the same 1