# System of Equations Given One Equation

7=3x+2y-z

How many more equations would you need to solve x, y, and z? In which variables can the additional equations be? Give examples of equations that would help solve these variables. (Hint: Select values for x, y, and z first; then multiply them by a constant and add and subtract them to create your equations.)

If you are solving for 3 variables, can you solve a system of equations when each equation only has 2 variables? If so, how many equations would be required? In general, how many equations are required to solve a system of equations in several variables? Why?

• $n$ variables needs $n$ equations! Nov 19, 2014 at 16:11
• The variables are x, y, and z. Nov 19, 2014 at 16:12
• So you need to have three equation. Nov 19, 2014 at 16:13
• Great, thank you! :) Now, is it 3 additional equations or 3 all together? And can you help me figure out what the equations will be? Nov 19, 2014 at 16:15
• 3 all together. You may look at link for everything. Good Luck! Nov 19, 2014 at 16:18

Answers given to you are partial. You don't need 3 equations for 3 variables to be sure to determine a solution point. You need at least 3 equations. 3 equations can be enough but it can't also be useless in your resolution. In $\mathbb{R}^{3}$, your equation is the representation of a plane.

Imagine that I give you two more equations (that is to say two more planes), but those planes are parallel to the initial plane. Will you be able to determine an intersection between those planes? The answer is no.

That's why you need at least 3 equations but it can't be more if you want you system not to be inconsistent.

Think about equations in 2 variable. Each equation represents a line, and the intersection of 2 lines defines a point (draw a graph with 2 lines crossing).

Next, think about 2 equations in 2 variables, where each equation represents a plane.

The intersection of 2 planes represents a line (think of a Christmas card; the fold represents the intersection of the front and back planes of the card). To pick a single point from the line of intersection, we would need 1 more equation (stand up the card on a table saw and saw it in half; the saw blade represents the 3rd plane).

I leave an image a block of wood for you to consider. Notice how it takes at least 3 faces (or equations) to define a single corner or vertex, and that the intersection of 2 planes is just sufficient to define an edge.