Q 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

ax + by = c

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:

4x + 5y = 40 x-y = 1

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).

System: 2x + y = 4 2x - y = 2 Solution: (1.5, 1) System: 2x + y = 4 x + y = 1 Solution: ( , )   System: x - 2y = -2 x - 2y = 2 No solution (The lines are parallel.) System: x - 2y = -2 -2x + 4y = 4 Solutions: There are infinitely many solutions (Both equations are represented by the same line.) Every point on this line gives a solution. A general formula for the solution can be obtained by solving either one of the equations for x in terms of y: x = 2y - 2 y is arbitrary     General Solution We can also write the general solution as (2y-2, y). Chosing a particular value for y gives a particular solution. For example choosing y = 3 gives the particular solution (4, 3), which is a point on the line. Q Another particular solution is ( , )

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.

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

3x = 6,   giving
x = 2.

To obtain y, we substitute x = 2 in either of the two equations (let us choose the first):

2(2) + 3y = 4,   giving
4 + 3y = 4,   so that
3y = 0,   or
y = 0.

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

13y = 13,   giving
y = 1

To obtain x, we substitute y = 1 in either of the two equations (let us choose the first):

2x + 3(1) = 3,   giving
2x + 3 = 3,   so
x = 0

Thus, the solution is (x, y) = (0, 1).

Consider the system

x - 2y = 3 3x - 3y = 3.

Q In order to eliminate x, we can multiply the first equation by and then add it to the second equation.
 

Q Doing this operation gives the first equation as .
 

Q Adding it to the second equation gives .
 

Q The solution to the system is therefore ( , ).
   

Consider the system

3x + 5y	=	8	... (1)
5x + 3y	=	0	... (2)

To eliminate y, we can:

	Multiply (1) by -3 and (2) by 5			Multiply (1) by -5 and (2) by 3
	Multiply (1) by -15			Multiply (2) by -15

Eliminating y and solving for x gives

	x = 1			x = 2/3
	x = -3/2			x = -1

Substituting the correct value of x gives

	y = -5/2			y = 5/2
	y = -2/5			y = 2/5

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.

Example of a System with Infinitely Many Solutions

3x - 5y	=	1
-6x + 10y	=	-2

We can eliminate x by multiplying the first equation by     and then adding it to the second.

Doing the above results in the equation 0 = 0, which actually tells us nothing. The reason this occurs is that both equations represent the same striaght line, (graph them to see why) and so they are infinitely many solutions (see above).

In the case we have here, we can solve either one of the equations for x in terms of y to obtain the general solution as:

x = .    
y is arbitrary

Example of a System with No Solutions

3x - 5y	=	1
-6x + 10y	=	2

Again, we can eliminate x by multipyying the first equation by 2 and then adding it to the second. However, this time, we wind up with the equation

0 = 4,

which is absurd. What this is telling us is the following:

If there was a solution (x, y) for the given system, then we could conclude that 0 = 4.

However, 0 is not equal to 4, and so there could not have been a solution to begin with! (This form of argument is called "proof by contradiction.") Geometrically, the lines representing the two equations are parallel lines, and thus do not intersect (see above for another example).

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.