# Section 2.3 Question 1

## Does every system have a unique solution?

For systems of two equations in two variables, it is possible to draw two lines that do not intersect. A system like this is called an inconsistent system.

Figure 2 – The graphs of parallel lines do not cross.

For an inconsistent system, there are no ordered pairs that simultaneously solve both systems since the graphs never cross. On a graph this may be difficult to diagnose. Two lines that are almost parallel might be indistinguishable from two lines that are parallel. Luckily when we attempt to find a solution with the substitution method or the elimination method, parallel lines are easy to distinguish from lines that are almost parallel.

### Example 1    Determine if a System of Equations is Inconsistent

Decide if the system of equations

is inconsistent by attempting to solve the system.

Solution This is the same system we solved in Section 2.1 Example 4. In that example, we graphed each equation and showed that lines were parallel. In this example, we’ll solve the equations by one of two possible strategies, substitution or elimination. We’ll look at each of these strategies to ensure that you understand how you could solve either way.

Substitution Method

It is easy to solve for x in the first equation by adding 3y to both sides of the equation. This yields

x = 3y + 27

Now replace the x in the second equation with 3y + 27:

-0.25 (3y + 27) + 0.75y = 6.75

We can simplify the left hand side of the equation to give us

Notice that the variable has dropped out of the equation. The resulting equation makes no sense since -6.75 cannot equal 6.75. This contradiction tells us that this system has no solutions.

Elimination Method

The leading coefficient in the first equation is a 1, we begin by eliminating x in the second equation. Either variable can be chosen, but in this system it is easier to eliminate x. Multiply the first equation by 0.25 and add it to the second equation:

Notice that y is eliminated at the same time x is eliminated. The sum of these equations results in a contradiction so the system has no solutions.

In each method, solving the system results in a contradiction. Contradictions such as these indicate that there are no solutions.

So far we have seen that a system may have a unique solution (one ordered pair solves the system) or no solutions at all. Other systems of equations have nonunique solutions. For these systems, there are many ordered pairs that satisfy each equation in the system. We may still use the Substitution Method or the Elimination Method to solve systems of equations with nonunique solutions.

### Example 2    Does the System of Equations have a Unique Solution?

Decide if the system of equations

has a unique solution by attempting to solve the system.

Solution As in the previous example, we can solve this system algebraically using the Substitution Method or the Elimination Method. In this example we’ll solve the system with both methods. In general, you only need to use one of these strategies to solve the system.

Substitution Method

This system is ideal for the Substitution Method since the first equation is already solved for one of the variables. Substitute -2x + 10  in place of y in the second equation to yield

Now simplify the left hand side to solve for x:

All the variables have dropped out and the resulting equation is always true. This signals that there is not a unique solution to the system. Many ordered pairs will satisfy both equations in the system of equations.

Elimination Method

To use the Elimination Method, we need to have all of the terms with variables on one side of the equals sign and the constant terms on the other side. The first equation does not have this format so add 2x to both sides of the equation to give

2x + y = 10

Replacing the first equation in the system with this equivalent equation leads to

We could multiply the first equation by 1/2 to make the leading coefficient a 1. But the second equation has a leading coefficient of 1 so we’ll interchange the equations to give an equivalent system,

To eliminate x from the second equation, multiply the first equation by -2 and add it to the second equation:

This new equation helps us to rewrite the system as the equivalent system

As in the Substitution Method, when all of the variables drop out and leave us with a true statement, there is not a unique solution. Since there is not an equation that specifies y to be a unique number, it can be chosen to be any number. Then the corresponding x value is found using the other equation in the system, . The easiest way to do this is to solve for x first to give .

We could also use the other equation in the original system, y = -2x+10, to find solutions to this system. In this case, we could specify x to be any number. Then the corresponding value of y is found from the equation y = -2x + 10.

In either case, the ordered pairs we find lie on a line. Any ordered pair on that line is a solution to the system. To locate points on this line, simply substitute a value for x in the equation y = -2x + 10 or substitute a value for y in the equation . We can pick any reasonable value for x and use the equation y = -2x + 10 to find the corresponding value for y:

We can also pick a value for y and use the equation  to find the corresponding value for x:

In either case, we find the exact same ordered pairs are solutions to the system. To write this solution formally, we say that the solution is all ordered pairs (x, y) where y = -2x + 10 and x is any real number. You can think of each ordered pair corresponding to an arbitrary x value and a y value that is calculated from the equation of the line. Or you can think of each ordered pair corresponding to an arbitrary y value and an x value that is calculated from the equation of the line.