Geometrical interpretation of systems of linear equations

Consider the following three system of linear equations, and solve them:

Exercise 1: System 1

Solution

Let's replace the $y$ in the second equation by $y=2x+1$ . We get

-3x+2=2x+1-2

and thus $x=\underline{0.6}$ . It follows $y=2\cdot 0.6+1=\underline{2.2}$ .

\begin{array}{|lll|} 2x+1 & =&y \\ -3x+2 & =&y-2 \end{array}

Exercise 2: System 2

Solution

Let's replace the $y$ in the second equation by $y=2x+1$ . We get

2=2(2x+1)-4x=2

So we get

2=2

So every $x$ fulfils this equation, which means that there are infinitely many solutions.

\begin{array}{|lll|} 2x+1 & =&y \\ 2 & =& 2y-4x \end{array}

Exercise 3: System 3

Solution

Let's replace the $y$ in the second equation by $y=2x+1$ . We get

1=2(2x+1)-4x=2

So we get

1=2

So every $x$ we try ends up with a contradiction. Thus there is no $x$ that can fulfil the equation. So no solution.

\begin{array}{|lll|} 2x+1 & =&y \\ 1 & =& 2y-4x \end{array}

Is there a better way to see way a system of linear equation sometimes has one solution (and is there only one?), sometimes no solution, and sometimes infinitely many solutions?

Things become more clear if we take a geometrical view of what it means to solve a system of linear equations. Take system 1, which we copy here again:

\begin{array}{|lll|} 2x+1 & =&y \\ -3x+2 & =&y-2 \end{array}

Let us rewrite the equations as follows:

\begin{array}{|lll|} y & =&2x+1 \\ y & =&-3x+4 \end{array}

Now, one way to solve this system is to write

2x+1 =y = -3x+4

thus to find $x$ we have to solve the equation

2x+1 = -3x+4

We could do that, and would find again $x=0.6$ and inserting this value in one of the equations for $y$ , say $y=2\cdot 0.6+1=2.2$ , we get $y=2.2$ . But the point we want to make here is a different one. Forget for the moment about solving system of linear equations, and lets go back to finding the point of intersection, let's call it $S$ , between the two functions

\begin{array}{lll} f(x)&=2x+1\\ g(x)&=-3x+4 \end{array}

To find the $x$ -coordinate of $S$ , we have to find an $x$ such that

f(x)=g(x)

2x+1=-3x+4

Observe that we solve exactly the same equation as above, so $x=0.6$ , and the $y$ -coordinate is $f(0.6)=2\cdot 0.6+1 = 2.2$ .

In other words:

Theorem 1

The solution of a system of two linear equations (with two unknowns) is the point of intersection between two straight lines.

Thus, a system of linear equations has exactly one solution (if the lines intersect), infinitely many solutions (if the lines overlap), and no solution (if the lines are parallel).

Exercise 4

Determine the number of solutions of systems 2 and 3 by investing if the corresponding straight lines are intersecting, parallel or overlapping.

Solution

For system 2

$y=2x+1=f(x)$ and $y=2x+1=g(x)$ . So the straight lines (the graphs of $f$ and $g$ ) are overlapping. So there are infinitely many solutions (all points on the line $f$ ). One solution, for example, is $x=0$ and $y=f(0)=2\cdot 0+1=1$ . Another one is $x=1$ and $y=f(1)=2\cdot 1+1=3$ , and so on.

For system 3

$y=2x+1=f(x)$ and $y=2x+0.5=g(x)$ . So the straight lines (the graphs of $f$ and $g$ ) are parallel but not overlapping (they have the same slope but different $y$ -intercepts). So there is no solution.