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)#