Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

help, 57 and 58. medal and fan

OpenStudy (anonymous):

OpenStudy (anonymous):

Ok, starting with 57, they give you an equation and they want to know if r=9 is a solution. All you have to do is plug in 9 for r on both sides of the equation and solve. If both sides balance out then it's a solution, if they don't it's not a solution.

OpenStudy (anonymous):

so 2-9/5?

OpenStudy (anonymous):

\[\frac{ 2-r }{ 5 }+\frac{ 3r }{ 5 }=\frac{ r+2 }{ 3 }\] \[\frac{ 2-9 }{ 5 }+\frac{ 3\times9 }{ 5 }=\frac{ 9+2 }{ 3 }\]

OpenStudy (anonymous):

i got 5.6=9.66

OpenStudy (anonymous):

that's not what I got...work on the numerators first...2-9=? and 3x9=?

OpenStudy (anonymous):

-7 and 27

OpenStudy (anonymous):

perfect, so you have -7/5+27/5...what do you get?

OpenStudy (anonymous):

common denominators means you can add the numerators...denominator stays the same.

OpenStudy (anonymous):

20/5=4

OpenStudy (anonymous):

good, now what about the right side of the equation...(r+2)/3?

OpenStudy (anonymous):

11/3

OpenStudy (anonymous):

exactly, so now you just need to ask yourself if 4=11/3. Then you'll have your answer!

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

what about 58?

OpenStudy (anonymous):

it's been awhile since I've done a problem like that. I'm solving it now so I can hopefully help you :)

OpenStudy (anonymous):

thank you !!!!!

OpenStudy (anonymous):

ok, so in theory I have an answer. I'm just going to check quick to be sure!

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

my original work gave me an answer of 4/3, but when I plug it in it doesn't work out right.

OpenStudy (anonymous):

I'm sorry, it's just been too long...I can't remember how to do it correctly. :(

OpenStudy (anonymous):

thats okay. its jst one question! ill jst make a educational guess

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!