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OpenStudy (anonymous):

solve equaion

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OpenStudy (anonymous):

\[\sqrt{7-r}=\sqrt{r-1}\]

OpenStudy (asnaseer):

if:\[\sqrt{x}=\sqrt{y}\]then you can say:\[x=y\]use this to solve your problem.

OpenStudy (anonymous):

i dont get it

OpenStudy (asnaseer):

if square root of one thing is equal to the square root of another thing, then the two things must be equal.

OpenStudy (anonymous):

still need help?

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OpenStudy (anonymous):

yes

OpenStudy (asnaseer):

ok, what is \(\sqrt{16}\)?

OpenStudy (anonymous):

4

OpenStudy (asnaseer):

correct, and what is \(\sqrt{8+8}\)?

OpenStudy (anonymous):

16

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OpenStudy (asnaseer):

no

OpenStudy (anonymous):

\[\sqrt{16}\]

OpenStudy (asnaseer):

correct - which is 4

OpenStudy (asnaseer):

so we can say:\[\sqrt{16}=\sqrt{8+8}\]which means:\[16=8+8\]

OpenStudy (asnaseer):

now, in your case, we have:\[\sqrt{7-r}=\sqrt{r-1}\]therefore we can say:\[7-r=r-1\]

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OpenStudy (asnaseer):

can you solve from here now?

OpenStudy (anonymous):

yes

OpenStudy (asnaseer):

great!

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