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Find z for which
\[\frac{2 \sqrt{z} - 3}{\sqrt{z} - 1} + 1 = \frac{3 \sqrt{z} - 1}{1 - \sqrt{z}}.\]

 Oct 8, 2021
 #1
avatar+35058 
+1

Multiply R side by   -1/-1   then eliminate the denominator from both sides to get:

 

 

2 sqrt z - 3  +sqrtz -1   = 1-3 sqrt z

6 sqrt z = 5

sqrt z = 5/6

z = 25/36

 Oct 8, 2021
 #2
avatar+12393 
+1

Find z.

 

Hello Guest!

 

\(\dfrac{2 \sqrt{z} - 3}{\sqrt{z} - 1} + 1 = \dfrac{3 \sqrt{z} - 1}{1 - \sqrt{z}}\\ \dfrac{2 \sqrt{z} - 3}{\sqrt{z} - 1} + 1 = \dfrac{1-3 \sqrt{z} }{\sqrt{z}-1}\\ 1= \dfrac{(1-3 \sqrt{z})-(2\sqrt{z}-3) }{\sqrt{z}-1}\\ 1=\dfrac{4-5\sqrt{z} }{\sqrt{z}-1} \)

\(\sqrt{z}-1=4-5\sqrt{z}\\ 6\sqrt{z}=5\\ \sqrt{z}=\dfrac{5}{6}\)

\(z=\dfrac{25}{36}\)

laugh  !

 Oct 8, 2021
edited by asinus  Oct 8, 2021

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