Algebraic Plane Solution Let’s solve the system of equations: 2x – 5y + 2z = 15 x + 3y - z = -4 2x - y - z = 2 First we make our assumption that they all do intersect at some common point ( a unique independent solution) and start by numbering our equations for clarity. The method will then follow that of algebraic elimination like in grade 10, except there is more than one variable. 2x – 5y + 2z = 15 (1) x + 3y - z = -4 (2) 2x - y - z = 2 (3) To start, look for a variable that you would like to eliminate. It doesn’t matter which, but lets get rid of the z in two equations. (2) – (3) -x + 4y = -6 (4) (1) + 2 (3) 6x – 7y = 19 (5) What do these new equations represent?

Algebraic Plane Solution Although these look like equations of lines, they are in fact equations of planes that have a common intersection point with the original 3 planes. But we can continue our elimination as if they were lines. -x + 4y = -6 (4) 6x – 7y = 19 (5) Now sub equation 6 into (4) to find x. 6 (4) + (5) 17y = -17 y = -1 (6) -x + 4(-1) = -6 x = 2 (7) Now sub equation 6 and 7 into any one of the original equations. x + 3y - z = -4 (6)&(7) into (2) (2) + 3(-1) - z = -4 z = 3 (8) What do the equations (6), (7) and (8) represent? Let’s start with what they aren’t: They aren’t the equations of three lines perpendicular to their corresponding axis. Individually they don’t represent a point either. So what are they?

Algebraic Plane Solution x = 2 represents a plane which has all x values equal to 2. In other words a plane parallel to the yz plane. Similarly y = -1 is a plane parallel to the xz plane and z = 3 is a plane parallel to the xy plane. All three of them together, however, represent a single intersection point for the three values. So the solution to the three equations is a point (remember we received no algebraic contradictions about our assumption) with coordinates (2, -1 3)