What is the least value of $x$ for which $\frac{x+2}{3} = \frac{4}{x+1}$?
To solve for $x$, I need to get rid of the fractions on both sides. I can do that by multiplying both sides by the lowest common denominator of the fractions, which is $3(x+1)$. This gives me $(x+2)(x+1) = 4 \cdot 3$. I can simplify this by expanding the brackets and subtracting 12 from both sides. This results in $x^2 + 3x - 10 = 0$. Now I have a quadratic equation that I can factor. I look for two numbers that multiply to -10 and add to 3. Those numbers are 5 and -2. So I can write $x^2 + 3x - 10 = (x+5)(x-2)$. To find the values of $x$ that make this equation true, I set each factor equal to zero and solve for $x$. I get $x = -5$ or $x = 2$. But I want the least value of $x$, so I choose the smaller one. Therefore, the least value of $x$ that satisfies the given equation is -5. 