How do you graph using slope and intercept of #-5x+7y=1#?

1 Answer
Jul 10, 2018

Refer to the explanation.

Explanation:

Given:

#-5x+7y=1#

To graph this equation using slope and y-intercept, convert the linear equation from standard form to slope-intercept form by solving for #y#.

Add #5x# to both sides.

#7y=5x+1#

Divide both sides by #7#.

#y=5/7x+1/7#

The y-intercept is the value of #y# when #x=0#. Substitute #0# for #x# and solve for #y#.

#y=5/7(0)+1/7#

#y=1/7# or #~~0.143#

The y-intercept is #(0,1/7)# or #(0,~~0.143)#. Plot this point.

The slope is rise (change in #y#) over run (change in #x#). Starting at the y-intercept, move up five spaces and to the right seven spaces. The new point will be #(7,5 1/7)# or #(7,~~5.143)#. Plot the point. You can also go in the negative direction. Again, start at the y-intercept and move down five spaces and to the left seven spaces. The point will be #(-7, -4 6/7)# or #(-7, ~~-4.857)#. Plot the point. You can keep doing this with each point until you get as many points as you want, although you only really need two points.

You now have three points. Draw a straight line through the points.

graph{-5x+7y=1 [-10, 10, -5, 5]}