1. ## Intersection Problem

I have two operations too chose from. Each operation is essentially a base added too a coefficient and a variable.

Since each operation appears to be linear and the two operations exist within the same system I decided to determine were the lines intercept.

This is not a problem out of a math book.

(500+.2142y)(2.06x)+(500+.2142y)(1-x)+(500+.6y)(2.06x)+(500+.6y)(1-x)

And

(872+.3858y)(1-x)+(872+.3858y)(2.06x)

Since I want to find the point where x and y are equal to each other I set the two as equal to each other. After Reducing and simplifying I get;

1.180548yx+924.32x+.3858y+872=.86305yx+1414.04x+.8 142y+1334

I decided it would be quicker to work the problem out with out all the long coefficients too see if I could get at a solution at all. For simplicities sake I've been writing it out on paper as;

xy+x+y+10=3xy+2x+3y+7
There are no fractions so I start by factoring.
x(y+1)+y+10=x(3y+2)3y+7
Then move terms over
x(y+1)+y-3y+10-7=x(3y+2)
Combine like terms
x(y+1)-2y+3=x(3y+2)

This is were i get stuck. If I divide away the 3y+2 on the right hand side and move it too the left, I still have an x sitting over on the left side.
Or;
x(y+1)-2y+3/(3y+2)=x

I can't just divide out the x on the left, since that would leave me with a 1 on the other side, X/X.
At first I thought it was a no solution type problem, but that doesn't seem to make sense. Have I just set the whole thing up wrong?

2. Originally Posted by bkbowser
x(y+1)-2y+3=x(3y+2)
This is were i get stuck.
Get the x terms together:
x(3y + 2) - x(y + 1) = 3 - 2y
Factor out the x:
x(3y + 2 - y - 1) = 3 - 2y
Simplify:
x(2y + 1) = 3 - 2y
Isolate the x:
x = (3 - 2y) / (1 + 2y)