Ask your own question, for FREE!
Algebra 11 Online
OpenStudy (anonymous):

I need to know if I am did this right. please help me

OpenStudy (anonymous):

ok \[\sqrt{7r+2}+3=7 \] dose it have no solution

OpenStudy (loser66):

yes, it has

OpenStudy (anonymous):

it has one ?

OpenStudy (loser66):

may be, let see. -3 both sides. square both sides simplify done

OpenStudy (anonymous):

ok but dose it have a solution?

OpenStudy (anonymous):

yes

OpenStudy (jdoe0001):

I need to know if I am did this right <---- what did you get that may not be right?

OpenStudy (anonymous):

ok so what did you do

OpenStudy (loser66):

again, YES, it has

OpenStudy (anonymous):

\[\sqrt{7r + 2} +3=7\rightarrow \sqrt{7r+2}=4 \] \[7r+2=4^{2} \rightarrow 7r+2=16\] \[7r=14\rightarrow r=7\]

OpenStudy (jdoe0001):

ineedhelpnow08 \(\huge \checkmark\)

OpenStudy (jdoe0001):

well... wait... shoot.... spoke too soon anyhow

OpenStudy (jdoe0001):

\(\bf \sqrt{7r+2}+3=7\implies \sqrt{7r+2}=4\implies 7r+2=4^2 \\ \quad \\ 7r=16-2\implies r=\cfrac{14}{7}\)

OpenStudy (anonymous):

oh yeah my bad14/7 is 2

OpenStudy (jdoe0001):

yeap

OpenStudy (anonymous):

thx

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!