SOLUTION: I am not sure when I subtract and when I add. I also get confused about where the signs (postive/negative) go. the question is Slope 1/3 through (1,6) y6= 1/3(x,1) 3 y

Algebra ->  Linear-equations -> SOLUTION: I am not sure when I subtract and when I add. I also get confused about where the signs (postive/negative) go. the question is Slope 1/3 through (1,6) y6= 1/3(x,1) 3 y      Log On


   

Question 146260: I am not sure when I subtract and when I add. I also get confused about where the signs (postive/negative) go.
the question is
Slope 1/3 through (1,6)
y6= 1/3(x,1)
3 y6= 3x 1/3 (x,1)
3y 18= 1x 3
3y +18-18= 1x 3-18
3y = 1x 21
x -3 =21
What am I doing wrong

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Are you trying to find the equation of a line with a slope of 1%2F3 that goes through (1,6)?




If you want to find the equation of line with a given a slope of 1%2F3 which goes through the point (1,6), you can simply use the point-slope formula to find the equation:


---Point-Slope Formula---
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is the given point

So lets use the Point-Slope Formula to find the equation of the line

y-6=%281%2F3%29%28x-1%29 Plug in m=1%2F3, x%5B1%5D=1, and y%5B1%5D=6 (these values are given)


y-6=%281%2F3%29x%2B%281%2F3%29%28-1%29 Distribute 1%2F3

y-6=%281%2F3%29x-1%2F3 Multiply 1%2F3 and -1 to get -1%2F3

y=%281%2F3%29x-1%2F3%2B6 Add 6 to both sides to isolate y

y=%281%2F3%29x%2B17%2F3 Combine like terms -1%2F3 and 6 to get 17%2F3 (note: if you need help with combining fractions, check out this solver)


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Answer:


So the equation of the line with a slope of 1%2F3 which goes through the point (1,6) is:

y=%281%2F3%29x%2B17%2F3 which is now in y=mx%2Bb form where the slope is m=1%2F3 and the y-intercept is b=17%2F3

Notice if we graph the equation y=%281%2F3%29x%2B17%2F3 and plot the point (1,6), we get (note: if you need help with graphing, check out this solver)

Graph of y=%281%2F3%29x%2B17%2F3 through the point (1,6)
and we can see that the point lies on the line. Since we know the equation has a slope of 1%2F3 and goes through the point (1,6), this verifies our answer.