How do you find the x and y intercepts of #1/2x+2y=-2#?

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Answer 1 · 2017-05-07T06:58:52

See a solution process below:

To find the x-intercept:

Substitute #0# for #y# and solve for #x#:

#1/2x + 2y = -2# becomes:

#1/2x + (2 * 0) = -2#

#1/2x + 0 = -2#

#1/2x = -2#

#color(red)(2) * 1/2x = color(red)(2) * -2#

#cancel(color(red)(2)) * 1/color(red)(cancel(color(black)(2)))x = -4#

#x = -4#

The x-intercept is #x =-4# or #(-4, 0)#

To find the y-intercept:

Substitute #0# for #x# and solve for #y#:

#1/2x + 2y = -2# becomes:

#(1/2 * 0) + 2y = -2#

#0 + 2y = -2#

#2y = -2#

#(2y)/color(red)(2) = -2/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = -2/color(red)(2)#

#y = -1#

The y-intercept is #y =-1# or #(0, -1)#