SOLUTION: Graphs/81664 (2007-05-11 11:05:43): I can't seem to fully understand how to get through the whole process to the answer. in the book the problem states as follows: 5y-3x=15 le

Algebra ->  Linear-equations -> SOLUTION: Graphs/81664 (2007-05-11 11:05:43): I can't seem to fully understand how to get through the whole process to the answer. in the book the problem states as follows: 5y-3x=15 le      Log On


   

Question 81665: Graphs/81664 (2007-05-11 11:05:43): I can't seem to fully understand how to get through the whole process to the answer.
in the book the problem states as follows:
5y-3x=15
let y=0 5y-3x =15
5(0)-3x =15
-3x =15
x =15
let x=0 5y-3x =15
5y-3(0)=15
5y =15
y =3
let y=6 5(6)-3x =15
30-3x =15
-3x = - 15
- 15
x = -15/-3 or 5
okay, this is the example in the book. I don’t understand where the uses of the 6 as y comes in. Please help. thank you.
0 solutions
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
When you plug in any y value and then solve for x, you are a point that is on the line 5y-3x=15. So when you plug in y=6 and find x=5, that means the point (5,6) is on the line 5y-3x=15. What's probably throwing you off is that you're doing this backwards where you plugged in x to find y before.