Explanation

To find the equation of the tangent to the curve \( y = \frac{x - 1}{2x + 1} \) at the point (1, 0):

First, verify the point lies on the curve:  
Substitute \( x = 1 \): \( y = \frac{1 - 1}{2(1) + 1} = \frac{0}{3} = 0 \). Yes, it does.

Next, find the slope of the tangent by computing the derivative \( \frac{dy}{dx} \):  
\(y = \frac{x - 1}{2x + 1}\)
Using the quotient rule: 
\(\frac{dy}{dx} = \frac{(1)(2x + 1) - (x - 1)(2)}{(2x + 1)^2} = \frac{2x + 1 - 2x + 2}{(2x + 1)^2} = \frac{3}{(2x + 1)^2}\)

At \( x = 1 \): 
\(m = \frac{3}{(2(1) + 1)^2} = \frac{3}{9} = \frac{1}{3}\)

Equation of the tangent line at (1, 0):  
\(y - 0 = \frac{1}{3}(x - 1)\)
\(y = \frac{1}{3}x - \frac{1}{3}\)
Multiply through by 3: \( 3y = x - 1 \)  
Rearrange: \( 3y - x + 1 = 0\)

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