## Algebra 1

Graph I has a line that goes through the points (0,0) and (3,1). The slope of the line is $\frac{1-0}{3-0}$, or $\frac{1}{3}$. The equation of a line is y=mx+b, and we can substitute the slope and the point (0,0) into the equation to solve for b. $0 = \frac{1}{3}*0 + b$ $0 = 0 + b$ $0 = b$ Thus, graph I has the equation y = 1/3 * x. Graph II has a line that goes through the points (0,1) and (-3,0). The slope of the line is $\frac{1-0}{0--3}$, or $\frac{1}{3}$. The equation of a line is y=mx+b, and we can substitute the slope and the point (0,1) into the equation to solve for b. $y = mx + b$ $1 = 1/3 * 0 + b$ $1 = 0 + b$ $1 = b$ Thus, graph II has the equation y = 1/3 * x + 1. By process of elimination, graph III has the equation y = 1/3 * x -1.