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solve equaion
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\[\sqrt{7-r}=\sqrt{r-1}\]
if:\[\sqrt{x}=\sqrt{y}\]then you can say:\[x=y\]use this to solve your problem.
i dont get it
if square root of one thing is equal to the square root of another thing, then the two things must be equal.
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yes
ok, what is \(\sqrt{16}\)?
4
correct, and what is \(\sqrt{8+8}\)?
16
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no
\[\sqrt{16}\]
correct - which is 4
so we can say:\[\sqrt{16}=\sqrt{8+8}\]which means:\[16=8+8\]
now, in your case, we have:\[\sqrt{7-r}=\sqrt{r-1}\]therefore we can say:\[7-r=r-1\]
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can you solve from here now?
yes
great!
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