xeon123 said:
1 - So, you mean that exists many solutions. This is because of this specific equation, or all these type of equations have several solutions, and what I did is enough?
Yes, there are many (an infinite number of) solutions to your equation. Any equation of the form Ax + By = C, where A and B are not both zero, represents a straight line. Any point (x, y) on the line is a solution to the equation, and any solution to the equation is a point on the line.
As to what you did, all you did was write an equivalent equation y = 10 - (5/3)x. This is just another form for the same equation - you didn't find any solutions. You also wrote another equation in which you solved for x. That also is not finding a solution.
Toward the end of your 2nd post, you did some more work in which you concluded that y = y. This is true, but not at all useful. A variable is obviously equal to itself.
Any of the three equations can be used to find as many solutions as you want. For example, using the equation y = 10 - (5/3)x, if you let x = 0, it's easy to see that y = 10. So the point (0, 10) is a solution.
If you let x = 3, then y = 5, so (3, 5) is a solution. And so on for as many values of x as you want to put into the equation.
xeon123 said:
2 - If I draw a line in a plan between (6,0) and (0,10), it means that all the values that are inside the area of the line ( (6,0) -- (0,0) -- (0,10); points AOB) are solutions?
No. The only solutions to the equation 5x + 3y = 30 (or y = (-5/3)x + 10) are the points
on the line - not under it or over it.
The points (6, 0) and (0, 10) are solutions to the equation (and are points on the line) because each of these pairs of numbers makes 5x + 3y = 30 a true statement. The point (0, 0) is not a solution (and so is not on the line) because 5*0 + 3*0 \neq 30.