How do you write #y = 1/2x-1# in standard form?

1 Answer
Jul 19, 2017

See a solution process 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

First, multiply each side of the equation by #color(red)(2)# to eliminate the fraction and ensure all of the coefficients are integers as required by the formula while keeping the equation balanced:

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

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

#2y = 2/2x - 2#

#2y = 1x - 2#

Next, subtract #color(red)(1x)# from each side of the equation so the #x# and #y# terms are on the left side of the equation while keeping the equation balanced:

#-color(red)(1x) + 2y = -color(red)(1x) + 1x - 2#

#-1x + 2y = 0 - 2#

#-1x + 2y = -2#

Now, multiply each side of the equation by #color(red)(-1)# to make the #x# coefficient non-negative as required by the formula while keeping the equation balanced:

#color(red)(-1)(-1x + 2y) = color(red)(-1) * -2#

#(color(red)(-1) * -1x) + (color(red)(-1) * 2y) = 2#

#color(red)(1)x + color(blue)(-2)y = color(green)(2)#

#color(red)(1)x - color(blue)(2)y = color(green)(2)#