What is the gradient and #y#-intercept of the line #12x-5y=15#?

1 Answer
Jun 20, 2017

See a solution process below:

Explanation:

This equation is in the Standard Linear form. 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

#color(red)(12)x + color(blue)(-5)y = color(green)(15)#

The slope or gradient for an equation in Standard Linear form is:

#m = (-color(red)(A))/color(blue)(B)#

Substituting the coefficients from the equation in the problem gives:

#m = (-color(red)(12))/color(blue)(-5) = 12/5#

The #y#-intercept can be found by substituting #0# for #x# and calculating #y#:

#color(red)(12)x + color(blue)(-5)y = color(green)(15)# becomes:

#(color(red)(12) * 0) + color(blue)(-5)y = color(green)(15)#

#0 + color(blue)(-5)y = color(green)(15)#

#color(blue)(-5)y = color(green)(15)#

#(color(blue)(-5)y)/-5 = color(green)(15)/(-5)#

#(cancel(color(blue)(-5))y)/color(blue)(cancel(color(black)(-5))) = -3#

#y = -3#

The #y#-intercept is #-3# or #(0, -3)#