9.4.1 General Linear Systems
Previously we studied a system of linear equations with two variables. Visually, we interpreted this as the point of intersection on a two-dimensional graph.
We now open our minds to the possibility of having a system of linear equations with more than two variables and more than two equations.
Now that our eyes are open to the possibility of multiple variable equations, let's generalize a process that we began when studying linear equations. First we solved one equation with one unknown variable. As an example, to solve the equation:
2x+3=7
we just subtracted 3 from both sides and then divided both sides by 2 to get that x=2.
Then, we developed a technique to solve a system of linear equations. In particular our system of equations consisted of 2 equations with 2 unknown variables. For example, if we have the system:
y=2x+3
y=4x-1
then we can just substitute the second equation into the first equation to give us that
4x-1=2x+3
and then put all of the x's to one side and the non-x's to the other side, giving us:
2x =4
From this equation we can see that x =2 and y = 7 (after plugging back in the x value).
What if we had a linear system with more than 2 variables? Things could get confusing really quickly as we can see from the following example taken from chemistry.
x\,{\rm C}_7{\rm H}_8\ + y\,{\rm H}{\rm N}{\rm O}_3 \quad \longrightarrow \quad z\,{\rm C}_7{\rm H}_5{\rm O}_6{\rm N}_3\ + k\,{\rm H}_2{\rm O}If you haven't taken a chemistry class, the above formula will seem like Greek. The main idea is that the equation represents a chemical reaction. The left-hand side represents the combination of two different chemicals, {\rm C}_7{\rm H}_8 and {\rm H}{\rm N}{\rm O}_3. On the right-hand side we have the products of the reaction, {\rm C}_7{\rm H}_5{\rm O}_6{\rm N}_3, and \textrm{ H}_2 \textrm{O} — otherwise known as water. By a principle from Chemistry, in order to balance the chemical reaction, the number of atoms of each element on the left-hand side has to equal the number of atoms of each element on the right-hand side, where the four elements in this equation are (C)arbon, (H)ydrogen, (O)xygen, and (N)itrogen. If we substitute in x=2, y=3, z=4, and k=5, it won't be balanced:
2\,{\rm C}_7{\rm H}_8\ + 3\,{\rm H}{\rm N}{\rm O}_3 \quad \longrightarrow \quad 4\,{\rm C}_7{\rm H}_5{\rm O}_6{\rm N}_3\ + 5\,{\rm H}_2{\rm O}If we just count up the H's, we see that on the left-hand side we have 2 × 8 + 3 × 1 = 19 of them. But, on the right-hand side, we have 4 × 5 + 5 × 2 = 30 of them.
With 4 different variables, it's going to super tough to just "guess" which combinations of numbers will work. Instead, let's develop a set of equations that we can solve in order for there to be an equal number of hydrogen, carbon, oxygen, and nitrogen atoms. Starting with the hydrogen, we see that we'll need:
8x+y=5z+2k
But, the number of Carbon atoms have to match as well, giving us:
7x=7z
As do the number of Nitrogen atoms:
y=3z
Finally, from the number of Oxygen atoms, we see that we'll need:
3y=6z+k
Altogether, then we have 4 linear equations with 4 unknowns:
\begin{align*}8x+y&=5z+2k\\7x&=7z\\y&=3z\\3y&=6z+k \end{align*}
If we use the substitution method that we used earlier, things could get pretty confusing, pretty quickly!
In the next section, we're going to develop a way of approaching a linear system with n equations and n unknowns, which will enable us to handle any number of variables as well as any number of equations.
