How do you graph the equation by plotting points #3x + 5y = 12#?

2 Answers
Jun 7, 2016

Answer:

First get the equation into standard form:
#y= -3/5x + 12/5#

Then, start plugging in values for #x# and plotting points.

Explanation:

Getting the equation to standard form

It will be much easier to plug in values if the equation is in standard form (so that #y# is isolated). First, subtract #3x# from each side. This gives you:

#5y = -3x + 12#.

Next, divide each side by 5 to isolate #y#.

#y = (-3x +12)/5#

You can do the distributive property to get it to truly standard form, but it'll be pretty easier from here.

Pluggin' in values

I always find it helps to make a table, but however you want, find a way to record. Pick a simple value for #x#. Let's say 1. Your equation would look like this:

#y = (-3(1) +12)/5#

Work that out... #-3 + 12 = 9#, #9 / 5 = 1 4/5#. So you have your first coordinate, #(1, 1 4/5)#.

Try another! Let's have #x = 4#. So...

#y = (-3(4) +12)/5#

Well, #-3 * 4 = 0#, and #0/5 = 0#, so your coordinate is #(4, 0)#. Now that you have two points, plot them on a graph, and draw a line connecting them!
graph{y = (-3x +12)/5 [-8.89, 8.89, -4.444, 4.445]}

Jun 7, 2016

Answer:

Express the equation in the point-slope form, i.e, y = mx + c.

Explanation:

#3x + 5y = 12#

#5y = -3x + 12#

#y = -3/5*x +12/5#

Now put values of x in the equation and get corresponding values of y. Plot the graph according to the values you get.

For example :

when #x = 0, y = 12/5#

when #x = 1, y = 9/5#

when #x = -1, y = 15/5 = 3#

graph{(-3/5)*x+12/5 [-7.174, 6.87, -1.386, 5.644]}