Linear equations with more than two unknowns

Task: I: x + 2y + z = 5, II: 4x - 8y = -12, III: 5x + 8y = 21.

When solving two linear equations with two unknowns, the basic principle was to obtain from the equations a single equation in one unknown. You follow the same idea now, that is, derive from these equation one equation in one unknown.

II and III yield 9x=9 or x=1. Substitution of this value for x into III yields y=2, and substitution of x = 1, y = 2 into I yields z = 0.

Task: I: 4x + 8y - 2z = 22, II: 2x - 8y + 5z = 5, III: 6x + 4y - 4z = 4.

Frequently, you cannot arrive directly at a single equation in one unknown. Since you can always obtain from two equations by the earlier methods (setting equal, etc.) a single equation, which contains one unknown less, you can at first reduce the equations to two equations in two unknowns and then solve them:

Naturally, you must every time eliminate the same variables so that the final equations contain only two unknowns. You now solve IV and V as before and find x = 2, z = 5, which you substitute in III to find y = 3.

If you have even more equations, for example, 5 equations in five unknowns, you must reduce them to four equations in four unknowns, etc. Frequently, if the equations favour it, you can abbreviate this process by obtaining 2 equations in two unknowns or even one equation in one unknown. The shorter the process, the better! For example, the equations of the next Task are quickly solved:

TASK:

Insoluble equations

Task: I: 2x - y + 3z = 11, II: 3x - 2y + 2z = 5, III: 5x - 3y + 5z = 16

Eliminate y from I and II and from II and III to obtain

IV: -x -4z = -17, V: -x - 4z= -17

and by subtraction of IV from V the equation 0·x + 0·z = 0. The equations are interdependent. III is the sum of I and II.

Given a larger system of equations, you must guard against deriving interdependent equations, as this does not lead to a solution.

If you obtain an equation like 0·x = a, the system of equations contain a contradiction (cf. The function of uniform growth y = mx + n).

It is frequently quite difficult to recognize at the start whether given equations meet the condition of independence and do not contradict, but the calculations will display this.

Simultaneous equations must not be contradictory and interdependent.

Text equations

Task: Siriporn (S) has 4 more Baht than Vichai (V). If she had twice as much as she really has, she would have 14 Baht more than Vichai. How many Baht have Siriporn and Vichai?

Solution involving one unknown: S has x Baht, V has x - 4 Baht. If S had twice as much, she would have 2x Baht and have 14 Baht more than V, whence

Siriporn has 10, Vichai 6 Baht.

Solution involving two unknowns: S has x, V has y Baht, whence

The setting up in the second approach is easier. In general, you should not try to involve too few unknowns. Often beginners find such tasks difficult, because they try to employ only a single variable. Later on, when you have become accustomed to this work, you can employ fewer unknowns by computing in your head relations between the unknowns and employing them.

Task: In a quadrangle, a is twice as large as b. How large is each angle?

There are infintely many solutions. The formulated problem will not allow an answer. The task is indeterminate. You must have as many definitions as you require equations. It can also happen that a task is overdetermined!

Task: The sum of two numbers is 8, their difference 2, the difference of three times these numbers 7:

This task is overdetermined. Equations II and III contradict each other. If you multiply both sides of II by 3, then you find that 3x - 3y = 6 which contradicts III: 3x - 3y = 7. There are two soluble systems: I and III, I and II.

The number of equations without contradiction and interdependence must equal the number of unknowns.

Exercises:

Answer

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