How do you write #y = 7/2x – 6# in standard form?

1 Answer
Apr 2, 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

First, multiply both sides of the equation by #color(red)(2)# to eliminate the fraction while keeping the equation balanced. The standard form of a linear equation must have coefficients which are integers.

#color(red)(2) xx y = color(red)(2)(7/2x - 6)#

#2y = (color(red)(2) xx 7/2x) - (color(red)(2) xx 6)#

#2y = (cancel(color(red)(2)) xx 7/color(red)(cancel(color(black)(2)))x) - 12#

#2y = 7x - 12#

Next, subtract #color(red)(7x)# from each side of the equation to move the #x# term to the left side of the equation while keeping the equation balanced. The standard form of a linear equation has the #x# and #y# terms on the left side of the equation.

#-color(red)(7x) + 2y = -color(red)(7x) + 7x - 12#

#-7x + 2y = 0 - 12#

#-7x + 2y = -12#

Now, multiply each side of the equation by #color(red)(-1)# to put the equation in standard form while keeping the equation balanced. In a standard form linear equation the coefficient of the #x# term should be a positive integer.

#color(red)(-1)(-7x + 2y) = color(red)(-1) xx -12#

#(color(red)(-1) xx -7x) + (color(red)(-1) xx 2y) = 12#

#color(red)(7)x - color(blue)(2)y = color(green)(12)#