### Instruction: Solve \[\frac{5x+1}{2x^2+5x-3}=\frac{2x}{2x-1}\]for $x$.### Response: To solve a rational equation like this, I want to clear the fractions by multiplying both sides by the lowest common denominator of all the fractions.
The denominator of the left side is $2x^2+5x-3$, which I can factor as $(2x-1)(x+3)$.
The denominator of the right side is $2x-1$.
So the lowest common denominator is $(2x-1)(x+3)$.
Multiplying both sides by $(2x-1)(x+3)$, I get \[5x+1=2x(x+3)\].
Expanding the right side, I get \[5x+1=2x^2+6x\].
Subtracting $5x+1$ from both sides, I get \[0=2x^2+x-1\].
This is a quadratic equation, which I can solve by factoring, completing the square, or using the quadratic formula.
I notice that this equation has integer coefficients, and the constant term is -1, so I try to factor it by looking for two numbers that multiply to -2 and add to 1.
I see that those numbers are 2 and -1, so I can write \[0=(2x-1)(x+1)\].
By the zero product property, either $2x-1=0$ or $x+1=0$.
Solving for $x$, I get $x=\frac{1}{2}$ or $x=-1$.
These are the possible solutions for $x$, but I need to check if they make any of the original denominators zero, because that would make the equation undefined.
Plugging in $x=\frac{1}{2}$, I see that the denominator of the right side becomes zero, so this is not a valid solution.
Plugging in $x=-1$, I see that none of the denominators are zero, so this is a valid solution.
Therefore, the only solution for $x$ is $x=-1$.