SOLUTION: Consider the following equation of a line. 2x + 3y − 6 = 0 (a) Find the slope and y-intercept (if possible) of the line specified by the equation. (If an answer is undefi

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Consider the following equation of a line. 2x + 3y − 6 = 0 (a) Find the slope and y-intercept (if possible) of the line specified by the equation. (If an answer is undefi      Log On


   



Question 218681: Consider the following equation of a line.
2x + 3y − 6 = 0
(a) Find the slope and y-intercept (if possible) of the line specified by the equation. (If an answer is undefined, enter UNDEFINED.)
m = ?
(x,y)=?

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the following equation of a line.
2x + 3y − 6 = 0

(a) Find the slope and y-intercept (if possible) of the line specified by the equation. (If an answer is undefined, enter UNDEFINED.)

m = 2 This is the slope. See steps below
(x,y)= (0,2) This is the y-intercept. See steps below.

Step 1. The slope-intercept from is given as y=mx+b where m is the slope and b is the y-intercept at x=0 or at point (0,b).

Step 2. Put the given equation in slope intercept form.

2x%2B3y-6=0

Add 6-2x to from both sides of the equation

2x%2B3y-6=6-3x=0%2B6-2x

3y=-2x%2B6

Divide 3 to both sides of equation

3y%2F3=-2x%2F3%2B6%2F3

y=-2x%2F3%2B2

Step 3. The equation y=-2x%2F3%2B2 has a slope m=-2/3 and y-intercept b=2 or at point(0,2).

Now the slope of -2%2F3 for every 3 units you go the right, you go two units down.
To plot the line take the y-intercept at point (0,2). So starting at point (0,2) or x1=0 and y1=2 we go three units to the right means x2=0+3=3 and 3 units down means y2=2-2=0. So we have two points (0,2) and (3,0).

graph%28400%2C400%2C+-10%2C+10%2C+-10%2C+10%2C+-2x%2F3%2B2%29

I hope the above explanation was useful.

And good luck in your studies!

For free Step-By-Step Videos on Introduction to Algebra, please visit
http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit
http://www.FreedomUniversity.TV/courses/Trigonometry.

Respectfully,
Dr J