## Algebra 1

a) $y=1/2*x+4.5$, Slope-intercept chosen since slope needed to be computed. b) $y=2x-4$, Slope-intercept chosen since slope and intercept were given. c) $x+2y = 6$, Standard form chosen since both intercepts were given. d) $y-2 = -5/3*(x-1)$, Point-slope form chosen since a point and the slope were given
a) $m = (4-2)/(-1--5)$ $m = 2/4 = 1/2$ $y=mx+b$ $2 = -5*1/2 +b$ $2 = -2.5 +b$ $2+2.5 = -2.5 + b + 2.5$ $4.5 = b$ $y=1/2*x+4.5$ b) $m = 2$, y-intercept = -4 $y=mx+b$ $y=2x-4$ c) (6,0) and (0,3) are the intercepts $m = (3-0)/(0-6)$ $m = 3/-6 = -1/2$ $y=mx+b$ $3 = 0*-1/2 +b$ $3 = 0 + b$ $3= b$ $y=-1/2*x+3$ $2*y=2*(-1/2*x+3)$ $2y = -x+6$ $2y+x = -x + 6 +x$ $x+2y = 6$ d) passes through (1,2) and $m=-5/3$ $y-y_{1} = m*(x-x_{1})$ $y-2 = -5/3*(x-1)$