+25 Find $z$ for which $\dfrac{2\sqrt{z}-3}{\sqrt{z}-1} - 1 = \dfrac{3\sqrt{z}-1}{1-\sqrt{z}}$.
Hi Fytonic!
Let's simplify the left-hand side first:
\(\frac{2\sqrt{z}-3-\sqrt{z}+1}{\sqrt{z}-1}=\frac{\sqrt{z}-2}{\sqrt{z}-1}\)
Now, let's equate this to the right-hand side as given:
\(\frac{\sqrt{z}-2}{\sqrt{z}-1}=\frac{3\sqrt{z}-1}{1-\sqrt{z}} \)
Now, notice: \(\frac{\sqrt{z}-2}{\sqrt{z}-1}=\frac{3\sqrt{z}-1}{-(-1+\sqrt{z})} =-\frac{3\sqrt{z}-1}{\sqrt{z}-1}\) (we factored a negative from the denominator). So when we multiply both sides by: \(\sqrt{z}-1\), there is no longer a denominator:
\(\implies \sqrt{z}-2=-(3\sqrt{z}-1)\)
\(\implies \sqrt{z}-2=-3\sqrt{z}+3 \\ \implies 4\sqrt{z}=5 \\ \implies \sqrt{z}=\frac{5}{4} \\ \implies z=(\frac{5}{4})^2=\frac{25}{16}\)
I hope this helps!
Oh! I am sorry, I distributed the bracket wrong in the first answer (at -(3sqrt(z) -1) I make this: -3sqrt(z)+3 while it should have been -3sqrt(z)+1.
Ok here is the correct way:
\(\dfrac{2\sqrt{z}-3-\sqrt{z}+1}{\sqrt{z}-1}=-\dfrac{3\sqrt{z}-1}{\sqrt{z}-1}\)
So:
\(\frac{\sqrt{z}-2}{1}=-3\sqrt{z}+1 \implies \sqrt{z}=-3\sqrt{z}+3 \implies 4\sqrt{z}=3 \\ \sqrt{z}=\frac{3}{4} \\ z=\frac{9}{16}\)
