How do you write #y+7=-3/2(x+1)# in standard form?

1 Answer
Aug 28, 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

Therefore, we can first multiply each side of the equation by #color(red)(2)# to eliminate the fractions so all of the coefficients and the constants are integers:

#color(red)(2)(y + 7) = color(red)(2) xx -3/2(x + 1)#

#2(y + 7) = cancel(color(red)(2)) xx -3/color(red)(cancel(color(black)(2)))(x + 1)#

#2(y + 7) = -3(x + 1)#

Next, we can expand the terms on each side of the equation:

#(2 xx y) + (2 xx 7) = (-3 xx x) + (-3 xx 1)#

#2y + 14 = -3x - 3#

Now, we can subtract #color(red)(14)# and add #color(blue)(3x)# to each side of the equation to convert it to Standard Linear form:

#color(blue)(3x) + 2y + 14 - color(red)(14) = color(blue)(3x) - 3x - 3 - color(red)(14)#

#3x + 2y + 0 = 0 - 17#

#color(red)(3)x + color(blue)(2)y = color(green)(-17)#