Find z for which
\[\frac{2 \sqrt{z} - 3}{\sqrt{z} - 1} + 1 = \frac{3 \sqrt{z} - 1}{1 - \sqrt{z}}.\]
+36916 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
+14915 Find z.
Hello Guest!
\(\dfrac{2 \sqrt{z} - 3}{\sqrt{z} - 1} + 1 = \dfrac{3 \sqrt{z} - 1}{1 - \sqrt{z}}\\ \dfrac{2 \sqrt{z} - 3}{\sqrt{z} - 1} + 1 = \dfrac{1-3 \sqrt{z} }{\sqrt{z}-1}\\ 1= \dfrac{(1-3 \sqrt{z})-(2\sqrt{z}-3) }{\sqrt{z}-1}\\ 1=\dfrac{4-5\sqrt{z} }{\sqrt{z}-1} \)
\(\sqrt{z}-1=4-5\sqrt{z}\\ 6\sqrt{z}=5\\ \sqrt{z}=\dfrac{5}{6}\)
\(z=\dfrac{25}{36}\)
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