Here's a super-detailed explanation, because it sounds like you really want to understand what's going on with your problem. Let me know if it doesn't quite make sense, and I hope this helps!
Think first about linear equations in two variables. Any equation of the form Ax + By = C defines a single line in the plane. Any point on that line is said to be a "solution" of that equation: it makes the equation true. So, a line is a set of solutions for its equation. Since there are an infinite number of points on a line, there are an infinite number of solutions for its equation.
For example, x + y = 0 has (1,-1) as a solution, but also (0,0), (2,-2), (3,-3), etc. There are an infinite number of points that solve the equation.
A system of two equations in two variables defines a pair of lines in the plane. Any point on both lines is said to be a "solution" of that system of equations.
If the lines are parallel, like x + y = 0 and x + y = 1, they never intersect, so the system will have no solutions. Trying to solve it would lead to a contradiction (here, x + y can't possibly equal both 0 and 1).
If the lines intersect, like x + y = 0 and 2x + y = 3, they will intersect at only one point (because lines are linear; they can't curve and re-intersect) and the system will have exactly one solution, which is what you're usually asked to solve for.
If the lines wind up being the same, like x + y = 1 and 3x + 3y = 3, the system will have an infinite number of solutions. Trying to solve it would lead to x and y disappearing, leaving you with an equation that's always true:
x = 1 - y (from the first equation)
3(1 - y) + 3y = 3 (plugging into the second equation)
3 - 3y + 3y = 3
3 = 3
Picturing this situation in the plane makes sense. If two lines are actually the same line, they intersect literally everywhere. Every point on one line is a solution for the other line as well.
Those three situations (no solutions, one solution, infinitely many solutions) are the only three cases for a system of linear equations in two variables. Lines can't interact in any other ways: they intersect never, once, or an infinite number of times.
Now, back to your question. Any equation of the form Ax + By + Cz = D defines a single plane in three-dimensional space. (Graph one on Wolfram Alpha if you can't quite picture that.) Any point lying in that plane is said to be a "solution" of that equation, just like a point on a line is a "solution" of its linear equation. So, a plane is a set of solutions for its equation. Since there are an infinite number of points on a plane, there are an infinite number of solutions.
A system of two equations in three variables defines a pair of planes in three-dimensional space. You can think through your problem by thinking about the different ways two planes can intersect, just like we thought about lines.
For example, two planes can be parallel. If your two planes are parallel, they will never intersect and you will have no solutions.
Two planes can also be exactly the same plane. If your two planes are exactly the same, they will intersect an infinite number of times.
But what if the planes aren't parallel but aren't the same? That's the case you're actually in here. In that case, the two planes intersect along a single line (just like two lines intersect at a single point). (You can play around with two pieces of paper to visualize this.) So, how many solutions are there to your system of equations? The answer is, however many points lie on the line of intersection. And, like before, a line has an infinite number of solutions.
IN CASE YOU'RE WONDERING:
It might seem weird that there's not a corresponding case for the one-solution system of equations involving lines. There actually is a corresponding case, but it involves having three equations in three variables. When three planes intersect, they can define no points, a single point, or an infinite number of points, just like two equations in two variables. The same is true for four equations in four variables, etc.
In general, in order to define a single point, you'll need as many equations as you have variables.
AND THE SPECIFICS OF YOUR PROBLEM:
Because of that whole it-defines-an-infinite-number-of-points thing, you can find a handful of points that solve your system by picking any value for x, plugging it in, and then solving the system as though it were a system of two equations in two variables. For x = 2, for example:
30(2) + 20y + 10z = 410
60 + 20y + 10z = 410
20y + 10z = 350
10z = 350 - 20y (solve first equation for z)
z = 35 - 2y
40(2) + 25y + 15z = 545
80 + 25y + 15z = 545
25y + 15z = 465
25y + 15(35 - 2y) = 465 (plug expression for z into second equation)
25y + 525 - 30y = 465
-5y = -60
y = 12
z = 35 - 2(12)
z = 35 - 24
z = 11
(2, 12, 11) is one solution for your system.
Marietta H.
05/04/15