How do you find the x and y intercept of #0.2x+3.2y=12.8#?

1 Answer
Sep 30, 2016

Answer:

#x# intercept: #(64,0)#
#y# intercept: #(0,4)#

Explanation:

The axis intercepts are, as the name suggests, points on the given axis. So, the #x# intercept must be a point on the #x# axis, and the #y# intercept must be a point on the #y# axis.

By definition, a point lays on the #x# axis if its #y# coordinate equals zero, and vice versa. Let's use these conditions to find the results:

  • For the #x# intercept, we need to set #y=0#:

#0.2x+3.2y=12.8 \to 0.2x+3.2(0)=12.8\implies 0.2x=12.8#

From here, we can easily solve for #x#:

#x=12.8/0.2 = 64#

And since we already knew that #y=0#, if #x=64# our point will be #(64,0)#

  • For the #y# intercept, we do exactly the opposite: we know that our point will be of the form #(0,y)#, and we need to find #y#, setting #x=0#: after cancelling the #x# coefficient, the equation becomes

#3.2y = 12.8 \implies y=12.8/3.2 = 4#