### Instruction: Enter all the solutions to
\[ \sqrt{4x-3}+\frac{10}{\sqrt{4x-3}}=7,\]separated by commas.### Response: I notice that the expression inside the square root is the same in both terms, so I can let \[y=\sqrt{4x-3}\] and work with a simpler equation.
Substituting y for \(\sqrt{4x-3}\) gives me \[y+\frac{10}{y}=7.\]
To get rid of the fraction, I can multiply both sides by y, which gives me \[y^2+10=7y.\]
This is a quadratic equation, so I can rearrange it and try to factor it or use the quadratic formula.
Subtracting 7y from both sides gives me \[y^2-7y+10=0.\]
I look for two numbers that multiply to 10 and add to -7, and I find that -2 and -5 work.
So I can factor the equation as \[(y-2)(y-5)=0.\]
To solve for y, I can use the zero product property and set each factor equal to zero.
This gives me \[y-2=0\] or \[y-5=0.\]
Solving for y, I get \[y=2\] or \[y=5.\]
But remember, y is not the original variable I want to solve for.
y is \(\sqrt{4x-3}\), so I need to substitute that back in and solve for x.
For y=2, I have \[\sqrt{4x-3}=2.\]
Squaring both sides gives me \[4x-3=4.\]
Adding 3 to both sides gives me \[4x=7.\]
Dividing both sides by 4 gives me \[x=\frac{7}{4}.\]
So one solution is \[x=\frac{7}{4}.\]
For y=5, I have \[\sqrt{4x-3}=5.\]
Squaring both sides gives me \[4x-3=25.\]
Adding 3 to both sides gives me \[4x=28.\]
Dividing both sides by 4 gives me \[x=7.\]
So another solution is \[x=7.\]
But wait, I'm not done yet.
I need to check if these solutions make sense in the original equation, because sometimes squaring can introduce extraneous solutions that don't work.
For x=7/4, I plug it into the original equation and get \[\sqrt{4\cdot\frac{7}{4}-3}+\frac{10}{\sqrt{4\cdot\frac{7}{4}-3}}=\sqrt{4}+\frac{10}{\sqrt{4}}=2+\frac{10}{2}=7.\]
So x=7/4 works.
For x=7, I plug it into the original equation and get \[\sqrt{4\cdot 7-3}+\frac{10}{\sqrt{4\cdot 7-3}}=\sqrt{25}+\frac{10}{\sqrt{25}}=5+\frac{10}{5}=7.\]
So x=7 also works.
Therefore, both solutions are valid and I can write them as my final answer.