Question

How do you determine whether a linear system has one solution, many solutions, or no solution when given 5x+ 4y= -18 and 2x+3y=-24?

Answer 1

These have one solution

Explanation:

One can answer this in several ways!

Method 1

We can try directly solving the system of equations

`5x+ 4y= -18`
`2x+3y=-24`

by multiplying the second one by `4/3` and subtracting from the first one :

`(5-2 times 4/3)x=-18+24times 4/3 qquad implies`

`7/3 x = 14 qquad implies qquad x = 6`

which in turn implies
`5times 6+4y=-18 implies 4y = -48 implies y = -12`

Thus, there is only one solution `x=6,y = -12`

Method 2

A bit more sophisticated method involves the coefficient matrix

`A = ((5,4),(2,3))`

The determinant of this matrix is

`det A = |(5,4),(2,3)| = 5times 3-2 times 4 = 7 ne 0`

Hence this matrix is non-singular and so `A^-1` exits. Thus the system has a unique solution

`((x).(y)) = ((5,4),(2,3))^-1 ((-18),(-24))`

Carrying out the explicit calculation will lead to the same answer as the first method (however, since all we need here is whether the solution exists or is unique, this last step is unnecessary - the existence of `A^-1` is enough to say that a unique answer exists)

Method 3

The two equations can be represented graphically by two straight lines, and the question of whether solutions exist then boils down to whether the lines intersect.

Since the slope of the line representing the first equation is `-5/4` , and that of the second is `-2/3`, the two lines are not parallel. So, they will intersect at exactly one point, and hence the solution is unique.

Note

• an explicit solution is easy in this case because there are only two equations in two unknowns - larger systems of equations are difficult to handle explicitly.
• the geometric picture (method 3) is easy to use because the graphical version is two dimensional
• method 2 will involve more effort for larger systems , and moreover, it only works when there are equal numbers of unknowns and equations (leading to a square matrix).
• for larger systems, a more advanced method like Gauss- elimination would be preferred.