How do you find the x and y intercepts of #4x=-1/2y-1#?

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Answer 1 · 2017-07-07T22:19:47

See a solution process below:

#x#-intercept:

To find the #x#-intercept, substitute #0# for #y# and solve for #x#:

#4x = -1/2y - 1# becomes:

#4x = (-1/2 xx 0) - 1#

#4x = 0 - 1#

#4x = -1#

#(4x)/color(red)(4) = -1/color(red)(4)#

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

#x = -1/4#

The #x#-intercept is: #-1/4# or #(-1/4, 0)#

#y#-intercept:

To find the #y#-intercept, substitute #0# for #x# and solve for #y#:

#4x = -1/2y - 1# becomes:

#(4 * 0) = -1/2y - 1#

#0 = -1/2y - 1#

#0 + color(red)(1) = -1/2y - 1 + color(red)(1)#

#1 = -1/2y - 0#

#1 = -1/2y#

#1 xx color(red)(-2) = 1/-2y xx color(red)(-2)#

#-2 = 1/color(red)(cancel(color(black)(-2)))y xx cancel(color(red)(-2))#

#-2 = y#

#y = -2#

The #y#-intercept is: #-2# or #(0, -2)#