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

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, subtract #color(red)(1)# and #color(blue)(x)# from each side of the equation to have the #x# and #y# term on the left side of the equation and the constant on the right side as required by the Standard Form for a linear equation while keeping the equation balanced:

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

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

#-x + y = 1#

Now, multiply each side of the equation by #color(red)(-1)# to transform the coefficient of the #x# variable to a positive integer as required by the Standard Form for a linear equation while keeping the equation balanced:

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

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

#1x + (-1y) = -1#

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