Given:#" "color(brown)( -2x+y=3)#
Change this into the standard format of #y= mx+c#
Add #color(blue)(2x)# to both sides
#color(brown)(color(blue)(2x)-2x+y" "=" "3color(blue)(+2x)#
#0+y=2x+3#
#y=2x+3#............................(1)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find the y intercept")#
If you look at the graph the plotted line crosses the y-axis when #x=0#
So substitute #color(green)(x=0)# into equation (1)
So #y=2x+3" becomes " y= 2(color(green)(0))+3#
that is:# " "y=(2xxcolor(green)(0))+3" " =" " 0 + 3#
So #color(red)(y_("intercept") = 3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find the x intercept")#
If you look at the graph the plotted line crosses the x-axis when #y=0#
So substitute #color(green)(y=0)# into equation (1)
So #color(brown)(y=2x+3)" becomes " color(brown)(color(green)(0)= 2x+3)#
Subtract #color(blue)(3)# from both sides
#color(brown)(color(green)(0)color(blue)(-3)= 2x+3color(blue)(-3))#
#-3=2x+0#
#color(brown)(2x=-3)#
Divide both sides by 2 which is the same as #color(blue)(xx1/2)#
#color(brown)(color(blue)(1/2xx) 2x=color(blue)(1/2xx)(-3)#
#2/2 x=-3/2#
But #2/2 = 1# giving:
#x=-3/2 -> -1 1/2 -> -1.5#
#color(red)(x_("intercept")=-1.5)#
