How do you find the slope and y intercept for: #8x-12y = 24 #?

2 Answers
Oct 9, 2015

Answer:

Slope: #2/3#

y-intercept: #(-2)#

Explanation:

Given
#color(white)("XX")8x-12y=24#

#color(white)("XX")12y= 8x-24#

#color(white)("XX")y=2/3x+(-2)#
which is is "slope-intercept form" (#y=mx+b#)
with slope #m=2/3#
and
y-intercept #= -2#

Oct 9, 2015

Answer:

Gradient #+2/3# .......... Intercept #x=0 , y=-2# written as (0,2)

Explanation:

To cater for all levels of understanding I have given an 'over the top' explanation. The reader will have to filter stuff out if their understanding is a bit higher!!!

With most mathematics you break the whole into elements or parts and solve the parts. This is what I am going to do and show each step so that you can see what is happening:

Before we start there are several things that need to be understood about this context.

Slope ( gradient ) is the amount of up as you move from left to right for the amount of along. The 'amount' of up can be either positive or negative. If positive then it is indeed going upwards so it is a positive gradient. On the other hand if negative it is going downwards and it is a negative gradient (slope).

Tradition is that:
1. The vertical axis designates the values of the variable of 'y'.

  1. The horizontal axis designates the value of the variable of 'x'.

  2. The value 'y' is the consequence of what ever value is assigned to 'x'. So y depends on the value of 'x' which is why 'y' is called the Dependant Variable. As you may assign any value you chose to 'x' so it is called the Independent Variable.

  3. The usual standardised format of this equation type is to write the dependent variable on it's own to the left of the equals sign and everything else to the right. so you would write #8x-12y=24# as:

#12y=8x-24 #
#y=(8x-24)/12 #
# y=2/3x - 2#

To find the #y# intercept:
The intercept is where the plotted line crosses the #y"-axis"#. In this case it only does it once. The value of the #x"-axis"# at this point is zero. So we may substitute the value of zero for #x# in #y=2/3x-2# giving

#y=2/3(0)-2#
#y = 0 -2#
#y = -2#

To find gradient:
With this type of context it is simply the coefficient of #x# ( the number in front of it. So for every 1 along the #x#-axis you go upwards #1/3# on the #y#-axis. So the gradient is #(1/3)/1 = 1/3#

This is positive so it is a positive gradient (upwards),