How do you graph using the intercepts for #5x+7y=77#?

1 Answer
Nov 3, 2015

Answer:

Plot the points for the x and y intercepts then draw a line through those points.

Explanation:

Given the equation
#color(white)("XXX")5x+7y=77#

The x-intercept is the value of #x# when #y=0#
#color(white)("XXX")5x+7(0)=77#
#color(white)("XXX")rArr x= 15.4#

The y-intercept is the value of #y# when #x=0#
#color(white)("XXX")5(0)+7y=77#
#color(white)("XXX")rArr y= 11#

So the intercept points are at #(15.4,0)# and #(0,11)#

If you plot these two points and draw a straight line through them, your graph should look something like:
graph{((x-15.4)^2 +y^2-0.1)(x^2+(y-11)^2-0.1)(5x+7y-77)=0 [-6.41, 25.63, -3.47, 12.55]}