How do you sketch the graph of #0=-5y+9x+25#?

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Answer 1 · 2017-06-08T11:04:14

See below.

#0=-5y+9x+25#

#5y = 9x+25#

#y = 1/5(9x+25)#

The equation of a straight line in slope #(m)# and intercept #(c)# form is #y = mx+c#

In this example; #m=9/5# and #c= 25/5 = 5#

Hence, the required graph is a straight line of slope #+9/5# passing through the point #(0,5)#

We can see this from the graph of #y# below.

graph{1/5(9x+25) [-16.02, 16.02, -8, 8.03]}