First, the conditions for a unique solution. These are the constraints. Let’s illustrate the constraints using two examples:
EXAMPLE 1
3x+2y=3 and 6x+4y=6
The second equation is a multiple of the first so there’s only one equation really but there are two variables. So there are many solutions, no unique solution.
EXAMPLE 2
x+y=2, 3x+3y=5
In this case there are no solutions because of inconsistency. The second equation is the same as 3(x+y)=5. But the first equation says x+y=2 therefore the second equation says 3×2=5, which is never true.
So if we take the coefficients and compare a₁/a₂ with b₁/b₂ we find that with each example the two ratios are the same 3/6=2/4 (=1/2) and 1/3=1/3. To get a unique solution, these ratios must be different.
The rest of the text puts x=-1 and y=3 into two linear equations which when solved as a system will give the unique solution. Each equation has two coefficients and a constant, represented by letters a, b and c with a subscript 1 or 2 showing which equation they belong to. So if we invent numbers for the a’s and b’s we can find out what the c’s must be. We have an infinite number of coefficients to choose from so we have an infinite system of equations. As an exercise try putting in any numbers for the a’s and b’s and work out the c’s. Remember the constraints when you’re picking numbers.
The graph they show is two lines y=-2x+4 and y=2x-4 corresponding to equations y+2x=4 and y-2x=-4. The solution to these is where the lines intersect at (2,0), when x=2 and y=0. This doesn’t seem to bear any relation to x=-1 and y=3.
Here is a system which would give the solution (-1,3):
x+y=2, 2x+y=1
