Where two numbers are unknown, we need two independent relations for those two numbers to calculate their respective values.

## Example:

When graphed, both of these relations form straight lines, however the methods here are not restricted to linear equations.

Each equation gives new information about how and are related to each other.

Take one equation on its own: . There are infinitely many values of and that work in this equation – eg, and . There are infinitely many ordered pairs that satisfy this equation.

However when graphed together, straight line graphs fall into one of three categories:

- parallel and never intersect – no common point;
- not parallel, intersect at one and only one point;
- parallel and always intersect – that is, the lines are coincident.

When graphed, it is obvious what category our set of equations fall into. However without the graph the algebra will let us know:

- if they are parallel and don’t intersect, our algebra has a final line such as which makes no sense. This is a contradiction informing us that no values will work in both equations.
- if they are not parallel and intersect at one and only one point, our algebra will result in the unique values of and of this point.
- if they are coincident, are algebra will end up with a linear relationship between and , such as – meaning that any point on this line is on both lines, all are solutions to the system. There are infinitely many solutions.

# Solve with Substitution

First, let’s isolate either or in one or the other of the equations. In this case, it is easy to isolate in the second equation.

Now we take the expression for in the second equation and substitute it to the first, as follows:

Finally, substitute this value of into the equation (or the other equation) to find .

Solution: , .

Check to see that this solution works in the both of our original equations.

# Solve with Elimination

First, it is helpful to align the unknowns:

Next, we examine to see if any unknown (either or ) has the same coefficient in both equations. This is not true in our example.

We could:

- Multiply the first equation by 4 and the second equation by 3 to have in both equations, or
- Multiply the second equation by 2 to have in both equations.

The latter is more efficient. Multiplying the second equation by 2 gives:

Subtract equation 2 from equation 1 to eliminate the term:

or, more simply:

Now substitute this value of into the equation (or the other equation) to find .

Solution: , .

# Videos

# Applet to practice both techniques:

# Modelling with Systems of Equations

Check out this selection of questions on Khan Academy.