Find the slope and \( y \)-intercept of the line and draw its graph. \[ 2 x+3 y-1=0 \]

To find the slope and \( y \)-intercept of the line given by the equation \( 2x + 3y - 1 = 0 \), we first need to rewrite it in slope-intercept form, \( y = mx + b \), where \( m \) is the slope and \( b \) is the \( y \)-intercept. Starting with the original equation: \[ 3y = -2x + 1 \] Now, divide everything by 3 to isolate \( y \): \[ y = -\frac{2}{3}x + \frac{1}{3} \] From this, we see that the slope \( m = -\frac{2}{3} \) and the \( y \)-intercept \( b = \frac{1}{3} \). To draw the graph, you would start at the point \( (0, \frac{1}{3}) \) on the \( y \)-axis, and use the slope of \( -\frac{2}{3} \) to find another point: go down 2 units and right 3 units. This will give you a line that slopes downwards from left to right.