# Thread: Finding point of intersection algebraically

1. ## Finding point of intersection algebraically

Another question for a test tomorrow that I don't understand... I know how to find the point of intersection for two lines on a graph easily but I don't understand how you do it algebraically and the teacher refuses to explain it to anyone again since we've already done it though I was absent that day.

2. Originally Posted by ~NeonFire372~
Another question for a test tomorrow that I don't understand... I know how to find the point of intersection for two lines on a graph easily but I don't understand how you do it algebraically and the teacher refuses to explain it to anyone again since we've already done it though I was absent that day.
you set them equal to each other.

say one line is y = 2x + 5 and the other is y = 3x - 1, then the (x-coordinate for the) point of intersection is given by:

2x + 5 = 3x - 1

do you see how that makes sense? we want to see where their y-values match up, and then match up the x-values at that point. if the x- and y-values are the same on both lines, it means it is a point they both pass through, and hence they intersect there

3. 2x + 5 = 3x - 1

So I'd go:

2x -3x = 5 - 1
-1x = 4

... I'm really confused now.

4. Originally Posted by ~NeonFire372~
2x + 5 = 3x - 1

So I'd go:

2x -3x = 5 - 1
-1x = 4

... I'm really confused now.
that's because you did the wrong thing. first off, i'd move the x's to thee right hand side, to avoid the -x. it's no big deal really, but i like my math to look aesthetically pleasing.

so, 2x + 5 = 3x - 1 .............subtract 2x from both sides.

=> 5 = 3x - 1 - 2x ..............simplify

=> 5 = x - 1 ......................now add 1 to both sides

=> 5 + 1 = x ......................simplify

=> x = 6

thus we have found the x-coordinate for which the lines intersect. to find the y-coordinate, plug the x-value we found into either equation.

say we decided to use y = 2x + 5

plug in x = 6, we get:

y = 2(6) + 5

=> y = 17

say we decided to use y = 3x - 1

plug in x = 6, we get:

y = 3(6) - 1

=> y = 17 again, aha!

so when x = 6, y = 17

thus the point of intersection is (6,17)

5. Originally Posted by Jhevon
that's because you did the wrong thing. first off, i'd move the x's to thee right hand side, to avoid the -x. it's no big deal really, but i like my math to look aesthetically pleasing.

so, 2x + 5 = 3x - 1 .............subtract 2x from both sides.

=> 5 = 3x - 1 - 2x ..............simplify

=> 5 = x - 1 ......................now add 1 to both sides

=> 5 + 1 = x ......................simplify

=> x = 6

thus we have found the x-coordinate for which the lines intersect. to find the y-coordinate, plug the x-value we found into either equation.

say we decided to use y = 2x + 5

plug in x = 6, we get:

y = 2(6) + 5

=> y = 17

say we decided to use y = 3x - 1

plug in x = 6, we get:

y = 3(6) - 1

=> y = 17 again, aha!

so when x = 6, y = 17

thus the point of intersection is (6,17)
The two simplify steps - I don't get why the 5 would be = to the rest of it and why then it would be changed to =x.

6. Originally Posted by ~NeonFire372~
The two simplify steps - I don't get why the 5 would be = to the rest of it and why then it would be changed to =x.
i did not change the 5 to x. i flipped the equation around. if i had simplified, i would have 6 = x, that looks weird, so i turned it around to say x = 6

7. But if you didn't simplify or flip anything around, how would it work?

8. Originally Posted by ~NeonFire372~
But if you didn't simplify or flip anything around, how would it work?
what do you mean. nothing would change if we write 6 = x as opposed to x = 6. that's saying the same thing. i don't see your confusion i'm afraid. i simplify to make things look nicer and easier to use. like writing 6 instead of 5 + 1. they are the same thing, but one is simplified...