Math 1220 Linear Algebra

Subsection 2.1.1 The standard method for finding the solution

Suppose we want to find all solutions to the equations

The standard technique is to manipulate one or both of the equations until either \(x\) has the same coefficient in both equations or \(y\) has the same coefficient, and then subtract to eliminate one variable. Since there is only one variable left, its value can be found; with this information, the value for the other variable can be found.

In the case above, we can multiply both sides of the second equation by \(\frac23\text{,}\) to get

and subtracting from the first equation gives

or

Using the first equation and the (now) known value of \(y\text{,}\) we find that

Hence we see that there is exactly one pair of values for \(x\) and \(y\) that simultaneously satisfy both equations: \(x=\frac{11}5\) and \(y=\tfrac15.\) We can also write this as \((x,y)=(\frac{11}5,\frac15).\) In this case we say that there is a unique solution (in mathematics, the term unique means exactly one).

Subsection 2.1.2 The geometric method of finding the solution

The set of equations solved in Subsection 2.1.1 can also be viewed geometrically. The points in the \(x\)-\(y\) plane satisfying one of the equations lie on a straight line, and any point satisfying both of the equations must lie on both lines. The plot of the two lines looks like this:

The point of intersection is \((x,y)=(\frac{11}5,\frac15)\text{,}\) so this is the only solution, just as before.