How do you write #y - 2 = - 1/2 (x - 4) # in standard form?

1 Answer
Mar 16, 2017

See the entire 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

To start, multiply each side of the equation by #color(red)(2)# to eliminate the fraction and keep the equation balanced:

#color(red)(2)(y - 2) = color(red)(2) xx -1/2(x - 4)#

#(color(red)(2) xx y) - (color(red)(2) xx 2) = -color(red)(2)/2(x - 4)#

#2y - 4 = -1(x - 4)#

#2y - 4 = (-1 xx x) - (-1 xx 4)#

#2y - 4 = -1x - (-4)#

#2y - 4 = -1x + 4#

Now, add #color(blue)(4)# and #color(red)(1x)# to each side of the equation to put the #x# and #y# terms on the left side of the equation and the constant on the right side of the equation while keeping the equation balanced:

#color(red)(1x) + 2y - 4 + color(blue)(4) = color(red)(1x) - 1x + 4 + color(blue)(4)#

#color(red)(1x) + 2y - 0 = 0 + 8#

#color(red)(1)x + color(blue)(2)y = color(green)(8)#