### Instruction: Solve \[\frac{5x+1}{2x^2+5x-3}=\frac{2x}{2x-1}\]for $x$.### Response: To solve this equation, I want to clear the fractions by multiplying both sides by the lowest common denominator (LCD) of the two fractions.
To find the LCD, I need to factor the denominators of both fractions.
I see that the denominator of the first fraction, $2x^2+5x-3$, can be factored as $(2x-1)(x+3)$ by looking for two numbers that multiply to -6 and add to 5.
I also see that the denominator of the second fraction, $2x-1$, is already in factored form.
So the LCD is the product of the two different factors, $(2x-1)(x+3)$.
Multiplying both sides by $(2x-1)(x+3)$, I get \[(2x-1)(x+3)\frac{5x+1}{(2x-1)(x+3)}=(2x-1)(x+3)\frac{2x}{2x-1}\]
Simplifying, 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 can be factored as \[0=(2x-1)(x+1)\]
Setting each factor equal to zero, I get \[2x-1=0 \quad \text{or} \quad x+1=0\]
Solving for $x$, I get \[x=\frac{1}{2} \quad \text{or} \quad x=-1\]
These are the two possible solutions for $x$.