### Instruction: Enter all the solutions to
\[ \sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=7,\]separated by commas.### Response: I see that this equation involves square roots and fractions, so I want to get rid of them as much as possible.
I notice that the same expression, $\sqrt{4x-3}$, appears both inside and outside the fraction, so maybe I can use that to my advantage.
I could try multiplying both sides by $\sqrt{4x-3}$ to eliminate the fraction, but then I would get a quadratic equation with a square root on one side, which could be messy.
Instead, I will try to make a substitution: let $y=\sqrt{4x-3}$.
Then the equation becomes
\[ y+\frac{10}{y}=7.\]
This looks much simpler, and I can multiply both sides by $y$ to get a quadratic equation:
\[ y^2+10=7y.\]
To solve for $y$, I will rearrange the equation and factor it:
\[ y^2-7y+10=0.\]
\[ (y-5)(y-2)=0.\]
This gives me two possible values for $y$:
\[ y=5 \text{ or } y=2.\]
But remember, $y=\sqrt{4x-3}$, so I need to solve for $x$ in terms of $y$.
I can square both sides of the substitution and isolate $x$:
\[ y^2=4x-3.\]
\[ x=\frac{y^2+3}{4}.\]
Now I can plug in the values of $y$ that I found and simplify:
\[ x=\frac{5^2+3}{4}=\frac{28}{4}=7 \text{ or } x=\frac{2^2+3}{4}=\frac{7}{4}.\]
These are the two solutions for $x$, but I need to check if they are valid.
Sometimes, when we square both sides of an equation, we introduce extraneous solutions that don't work in the original equation.
To check, I will plug in each value of $x$ into the original equation and see if it is true.
For $x=7$, I get
\[ \sqrt{4(7)-3}+\frac{10}{\sqrt{4(7)-3}}=\sqrt{25}+\frac{10}{\sqrt{25}}=5+\frac{10}{5}=7,\]
which is true.
For $x=\frac{7}{4}$, I get
\[ \sqrt{4(\frac{7}{4})-3}+\frac{10}{\sqrt{4(\frac{7}{4})-3}}=\sqrt{4}+\frac{10}{\sqrt{4}}=2+\frac{10}{2}=7,\]
which is also true.
So both solutions are valid, and I can enter them as my final answer.