Changing Equations To Slope-Intercept Form

Date: 07/17/98 at 23:25:58
From: Holly
Subject: Slope-intercept form

I am having trouble understanding putting an equation in slope-
intercept form. One example is: 

   12(2x - 1) - 5(3y + 2) = 8. 

The answer is: 

   y = 8/5x - 2; slope = 8/5; y-intercept: (0, -2). 

I don't understand how this works. Please help!

Date: 07/19/98 at 19:42:25
From: Doctor White
Subject: Re: Slope-intercept form

Holly:

The slope-intercept form of the equation of a line is: y = mx + b, 
where m is the slope and b is the y-intercept. 

In order to solve equations for specific variables, you need to take 
several steps:

   1) simplify both sides of the equation
   2) combine like terms on each side
   3) move everything over to one side except the varibles you are   
      solving for
   4) solve for the variable

Now let's look at your problem:

   12(2x - 1) - 5(3y + 2) = 8

1) Simplify the left side by distributing to remove the parentheses:

   24x - 12  - 15y  - 10 = 8

2) Combine like terms:

   24x - 15y - 12 - 10 = 8
        24x - 15y - 22 = 8

3) Move terms to the right side except for the y (since you are solving 
   for y). You can move a term over by adding its opposite to the other  
   side of the equation:
 
   24x - 24x - 15y - 22 + 22  = 8 - 24x + 22

   Since you added -24x and +22 on the left side, then you must do the 
   same to the other side to keep the equality. Thus:

   0 - 15y + 0 = 30 - 24x
         - 15y = 30 - 24x

4) Now solve for y by dividing both sides by -15:

   -15y/(-15) = (30 - 24x)/(-15)
   y = 30/(-15) - 24x/(-15)   
   y = -2 + 8x/5

   You now have it in slope-intercept form:

   y = m   x +   b
   y = 8/5 x + (-2) 

   Thus m = 8/5, which is the slope, and b = -2, which is the 
   y-intercept, often written (0,-2).

I hope these steps help you to understand these type of problems. Let 
me know if you need further explanation. Come back to see us soon.
                             
- Doctor White, The Math Forum
Check out our web site! http://mathforum.org/dr.math/