How to solve a double incognite equation?

[KRiBaH]

Hello! I am new here! And i really like this site! So here's is my problem:

Well i know how to solve a simple equation with 2 incognites like:

x+y=2

x+4=2y

So in this case i do a substitue y=x-2 and use it in the second equation and it would be like:

x+4=2(x-2)

But the problem is if they only give me one equation with 2 incognites how can i solve it?

Example:

3x-4y+2=2(x+1)


Thanks!

 

[statdad]

If all you are given is one equation in two unknowns - like your example

3x-4y+2=2(x+1)

and no other information, then you have only two choices

1. Solve for x in terms of y: x = material involving only y and constants

2. Solve for y in terms of x: y = material involving only x and constants

Typically version 2 is selected, as we are accustomed to seeing equations with y isolated.
What is the setting for this question?

 

[HallsofIvy]

A single equation in two unknown numbers, x and y, has an infinite number of (x,y) solutions: The graph of something like 3x-4y+2=2(x+1) on an xy- coordinate system would be a straight line. Every point on the line gives an (x,y) pair that satifies the equation.

 

[KRiBaH]

So that means that x or y can't have an exact number because it can be any number, like it can't be x=2, it must be like x € (-infinite,infinite) like the inequations is that what you mean?

 

[HallsofIvy]

Either x or y can be any number. Once one of them is chosen, the other is fixed.