Mathematics
OpenStudy (anonymous):

solve equaion

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?

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

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:$7-r=r-1$

OpenStudy (asnaseer):

can you solve from here now?

OpenStudy (anonymous):

yes

OpenStudy (asnaseer):

great!

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