How do you graph using the intercepts for #5x + 2y = -2#?

1 Answer
Nov 9, 2015

Answer:

Plot the intercept points and draw a straight line through them
graph{(5x+2y+2) ((x+2/5)^2+y^2-0.01)(x^2+(y+1)^2-0.01)=0 [-4.93, 4.934, -2.465, 2.465]}

Explanation:

The x and y intercepts give us two points on the line defined by the linear equation.

The x-intercept is the value of #x# when #y=0# (for all points on the X-axis #y=0#)

Given #5x+2y=-2#
when #y=0#
#rArrcolor(white)("XXX")x=-2/5#
which gives the point #(-2/5,0)#

Similarly for the y-intercept
we get the point #(0,-1)#