(Based on Section 3.1 in Finite Mathematics and Finite Mathematics and Applied Calculus)
Just what is a "system of linear equations in two unknowns?"
A First, a linear equation in two unknowns x and y is an equation of the form
where a, b, and c are numbers, and where a and b are not both zero.
Examples: Linear Equations:
|4x + 5y = 0||This has a = 4, b = 5, c = 0|
|x - y = 11||This has a = 1, b = -1, c = 11|
|4x = 3||This has a = 4, b = 0, c = 3|
Second, a system of linear equations is just a collection of these beasts. To solve a system of linear equations means to find a solution (or solutions) (x, y) that simultaneously satisfies all of the equations in the system.
Example: System of Linear Equations:
|This is a system of two linear equations with solution x = 5, y = 4. We can also write the solution as (5, 4)|
Solving a System of Two Equations Graphically
The solutions to a single linear equation are the points on its graph, which is a straight line. For a point to represent a solution to two linear equations, it must lie simultaneously on both of the corresponding lines. In other words, it must be a point where the two lines cross, or intersect.
Thus, to locate solutions to a system of two equations in two unknows, plot the graphs, and locate the intersection points (if any).
Solving a System of Two Equations in Two Unknowns by Elimination
Q Do we really need another method of solving a system of linear equations?
A The problem with the graphical approach is that it only gives approximate solutions; locating the exact point of intersection of two lines would require perfect accuracy, which is impossible in practice.
The method of elimination is an algebraic way of obtaining the exact solution(s) of a system of equations in two unknowns by manipulating the equations in such a way as to eliminate of the variables (x or y). The best way to understand this method is through some examples.
|(a) Solve||2x + 3y = 4
x - 3y = 2
If we simply add these equations (add the left-hand sides and the right-hand sides) the y's cancel out, and we get
To obtain y, we substitute x = 2 in either of the two equations (let us choose the first):
Thus, the solution is (x, y) = (2, 0).
|(b) Solve||2x + 3y = 3
3x - 2y = -2
This time, adding (or subtracting) the equations does not result in either x or y being eliminated. However, we can eliminiate x by multiplying the first equation by 3 and the second by -2:
|2x + 3y = 3||3||6x + 9y = 9|
|3x - 2y = -2||-2||-6x + 4y = 4|
Now if we add them, we get
To obtain x, we substitute y = 1 in either of the two equations (let us choose the first):
Thus, the solution is (x, y) = (0, 1).
Q In the graphical approach to solving linear systems of equations with two unknowns, we saw cases where there were infinitely many solutions (both equations representing the same line) and no solutions (the equations representing parallel lines). How can we tell if this is happening in the algebraic approach?
A Let's illustrate these possibilities with examples.
You can now try some of the exercises in the textbook (Section 3.1) or go on to the tutorial for Section 3.2 by pressing "Next Tutorial" on the sidebar.