How do you write #y = 1/4x - 5# in standard form?

1 Answer
Jan 13, 2017

See full answer below.

Explanation:

The standard form of a linear equation is:

#color(red)(A)x + color(blue)(B)y = color(green)(C)#

where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

The first step is to eliminate all fractions by multiplying both sides of the equation by #color(red)(4)#:

#color(red)(4) xx y = color(red)(4) xx (1/4x - 5)#

#4y = (color(red)(4) xx 1/4x) - (color(red)(4) xx 5)#

#4y = (cancel(color(red)(4)) xx 1/color(red)(cancel(color(black)(4)))x) - 20#

#4y = 1x - 20#

Next step is to move the #x# term to the left side of the equation by subtracting #color(red)(1x)# from each side of the equation:

#-color(red)(1x) + 4y = -color(red)(1x) + 1x - 20#

#-1x + 4y = - 0 - 20#

#-1x + 4y = -20#

Now, to make the coefficient of #x# positive we need to multiply both sides of the equation by #color(red)(-1)#

#color(red)(-1) xx (-1x + 4y) = color(red)(-1) xx -20#

#(color(red)(-1) xx -1x) + ((color(red)(-1) xx 4y) = 20#

#1x - 4y = 20#

or

#color(red)(1)x - color(blue)(4)y = color(green)(20)#