Question

Solve for X.
7/sqrt(33) + sqrt(33)/ 4 = 5x

Answer 1

Hey steph! \c:/
\[\Large \color{royalblue}{\text{Wecome To OpenStudy! :)}}\]

Answer 2

thanks

Answer 3

\[\Large \frac{7}{\sqrt{33}}+\frac{\sqrt{33}}{4}=5x\]Mmmmmm, I don't like fractions.
Let's multiply both sides by 4,\[\Large \frac{7\cdot4}{\sqrt{33}}+\frac{\sqrt{33}\cdot\cancel4}{\cancel4}=5x\cdot4\]Giving us,\[\Large \frac{28}{\sqrt{33}}+\sqrt{33}=20x\]Now let's multiply both sides by sqrt{33},\[\Large \frac{28\cdot\cancel{\sqrt{33}}}{\cancel{\sqrt{33}}}+\sqrt{33}\cdot\sqrt{33}=20x\cdot\sqrt{33}\]Giving usssssssss,\[\Large 28+33=\left(20\sqrt{33}\right)x\]From here it will be a tad easier to solve for x, now that we've dealt with the fractions.
Confused by any of that? :O

Answer 4

kinda

Answer 5

wouldnt it end up being someting like

Answer 6

61/22sqrt33=x

Answer 7

Dividing both sides by the coefficient on x got you there?
Yes good! :)\[\Large \frac{61}{20\sqrt{33}}=x\]

Answer 8

Hmm it's usually improper to leave a root in the denominator like that.
Let's multiply the top and bottom by sqrt33.

Answer 9

\[\Large \frac{61}{20\sqrt{33}}\cdot\frac{\sqrt{33}}{\sqrt{33}}=x\]Which simplifies to,\[\Large \frac{61\sqrt{33}}{20\cdot33}=x\]Understand what I did with that step there? :O

Answer 10

I mean unless you have multiple choice to choose from. In which case you should just stop when your answer matches :)

Answer 11

A.\[A.\frac{ 35+5\sqrt{13} }{ 6 }\]
B\[\frac{ 7+\sqrt{13} }{ 30 }\]
C.\[\frac{ 7+6\sqrt{13} }{ 210 }\]
D.\[\frac{ 35+30\sqrt{13} }{ 42 }\]
Those are the answer choices.

Answer 12

Mmm Ok lemme see if I made a boo boo somewhere.

Answer 13

Are you sure you've matched up the multiple choice to the right question?
Because we shouldn't end up with sqrt(13), or addition in the numerator either.
:(

Answer 14

wait ill check.

Answer 15

nvm i got the answer thanks for the help.

Answer 16

cool