Numerical solution of linear equations in two unknowns.

As a rule, graphical solutions cannot have the same accuracy as numerical solutions, because 1) their curves can only be executed with a limited accuracy and 2) the measurement of segments, that is the extraction of results, is inaccurate. We will now solve numerically two linear equations in two unknowns.

Method of equating equations

Example 1: y = 2x + 3, II: y = 4x + 5.

These equations are satisfied by arbitrarily many pairs of values. If x and y are such that both equations are fulfilled, then the left and right sides will also be equal, whence

You find the value of y by substituting x = -1 into one of the equations; I yields: y = -2 + 3 = 1.

x = -1, y = +1 are the solution of the given equations.

As a rule, you must first obtain equations in the form above and it does not matter whether you first find x or y.

Example 2:

Once you know the value of one of the unknowns, you can obtain that of the other one from either of the original or reformed equations.

y = 4x - 8 yields y = 12 - 8 = 4, whence the solutions are x = 3, y = 4.

You test your result substitute it in each of the given equations, just in case you have made a mistake during the rearrangement.

Substitution method (lat. substituere = substitute)

You are given I: 2y = 4x - 8 and II: 3x + y = 11. It is often simpler to find y or x from one of the equations; for example, use I: y = 2x - 4 and substitute this expression in II, to obtain 3x + 2x - 4 = 11; hence x = 3 and substitution in I yields y = 2.

Example:

x = 7 - 2y yields for y = 2: x = 7 - 4 = 3. The solutions are: x = 3, y = 2.

Addition method

Given I: 2x + 3y = 23 and II: 4x - 3y = 1. You have seen that the essence of the two preceding methods was to obtain from two equations in two unknowns one equation in one unknown. You can do this still in another way. You apply the Rule that addition of equal terms yields equal results. Addition of the left hand sides of the equations yields 6x = 24, whence 6x=24 or x = 4. Substitution of this value into I yields:

The reason why one variable drops out was that in these two equations the one unknown (y) had the same coefficient with opposite signs. Should this not be the case at the start, you can manipulate with this goal in mind. For example, in the following task, you can achieve this for the unknown y by multiplication of the first equation by 3, of the second equation by -2. Apart from signs, you must expand the equations and reduce the coefficients of the respective variables to their smallest common denominator.

I yields with x = 3:

These three methods can always be used to solve two linear equations in two unknowns.

Example: I: 2x + 4y = 22, II: 4x - y = -1.

Equations which cannot be solved

Task: I: y - x = 3, II: y - x = -3

If you compute y from these two equations and equate the two values, you will find x + 3 = x - 3 or 0x = - 6, that is for every value of x you find 0 = -6, which is false.

How can this be explained? Since the coefficient of x in both equations is the same, you know that you are dealing with parallel straight lines, that is with lines with the same inclination which cannot intersect. Hence there cannot be a solution! The difference of two numbers y and x also cannot be simultaneously +3 and -3.

The equations contradict.

Task: I: x + y = 2, II: 3x + 3y = 6

y=2-x substituted in II yields 3x+3(2-x)=6 or 0·x=0.

This equation holds for any value of x.

In order to understand this conclusion graphically, you reduce the equations to their normal form: I: y=-x+2, II: y=-x+2. The equations represent the same line. Equation II is 3 times Equation I!

The two equations are interdependent

Introduction of new unknowns

Task: I: 2(x + 4)/5 + 3(y - 3)/2 = 3, II: (x + 4)/5 + (y - 3)/2 = 3.

These equations can be solved by any of the three methods above. However, it is simpler to introduce new variables: x + 4 = u, y - 3 = v, when the equations become I: 2u/5 + 3v/2 = 5, II: u/5 + v/2 = 2, for which the addition method yield u = +5, v = +2. Since you were to find x and y, you must still solve the equations: I: x + 4 = u, II: y - 3 = v or I: x + 4 = 5 and II: y - 3 = 5, to obtain the solutions x = 1, y = 5.

Task: I: 1/x + 2/y = 5/2, II: 3/x - 1/y = 1/2.

Let 1/x = u and 1/y = v.Then I: u + 2v = 3/2, II: 3u - v = 1/2, whence u = 1/2, v = 1, and therefore x = 2, y = 1.

The equations of this task are not of the first degree. Multiplication of both sides of Equation I by xy yields y + 2x = 5xy/2 - a quadratic equation.

Exercises: Use different methods to solve each of the systems of two equations and compare their efficiencies:

Answers

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