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## anonymous 5 years ago Can anyone help with this one 4/x +7 = 2/3x Thank you!!

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1. shadowfiend

So you have: $\frac{4}{x} + 7 = \frac{2}{3x}$ We can actually multiply through by $$x$$: $4 = 7x + \frac{2}{3}$

2. shadowfiend

Er, sorry: $4 + 7x = \frac{2}{3}$

3. shadowfiend

Can you solve that equation a little more easily?

4. anonymous

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

5. shadowfiend

Cool :)

6. anonymous

I got -10/21

7. anonymous

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

8. shadowfiend

That is right!

9. shadowfiend

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}

10. shadowfiend

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}

11. anonymous

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

12. shadowfiend

Awesome! :)

13. anonymous

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

14. anonymous

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

15. shadowfiend

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?

16. anonymous

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

17. shadowfiend

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

18. anonymous

x-5 = 8x/5

19. shadowfiend

Exactly. Do you think you can solve that one?

20. anonymous

Thanks, I'll try.

21. shadowfiend

Cool :)

22. anonymous

x= -25/3

23. shadowfiend

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.

24. anonymous

shadow can u help me in the question i asked earlier

25. anonymous

not the same 1

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