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Mathematics for Technology II (Math 2131)

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Study Force Academy
Durham College, Mathematics
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  • 36 lessons
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  • 14 week duration

Solving Systems of Equations with 3 Unknowns

There are practically countless methods devised over the past millennia that have enabled mathematicians to solve systems with three unknowns. In this section, we will focus on one method exclusively which is based on the methods that you’ve already learned when you solved systems with two unknowns. Technically, when a third variable first-degree variable is added to an equation, the graph of the equation gains a third dimension, so when graphed on an x, y, z-plane, it’s no longer a straight line, but a three-dimensional sheet (illustrated below).

The strategy here is to reduce a given system of three equations with three unknowns to a system of two equations with two unknowns, which you already know how to solve from previous lessons. This is done by taking any two of the given equations and, by addition-subtraction or substitution, eliminate one variable, obtaining a single equation with two unknowns. You then take another pair of equations (which must include one that’s not yet used, as well as one of those already used), and similarly obtain a second equation containing the same two unknowns as before. This pair of equations can then be solved simultaneously, and the values obtained substituted back to obtain the third unknown variable value. This is demonstrated in the following video:

Depending on the system of three equations you’re given, sometimes you may be given one or more equations where all three variables aren’t present (as shown in the next video). If that’s the case, you may be able to find the unknowns via substitution quicker than you would via elimination.

Important note:

If you happen to come across a system of three equations with three unknowns where the variables are found in the denominator, it’s advised that you reassign each variable with its reciprocal counterpart. For example, if the system contains the equation:

3x+45y1z=6

You’d set:

x=1a,   y=1b,   z=1c

Substituting these values makes each term a first degree, thereby making it easier to solve:

3a+4b5c=6

Once a, b, and c are found, don’t forget to reciprocate those numbers at the very end.

Of course, you could also find the lowest common denominator instead (for example, 5xyz), and multiple each term by this factor, but you’re more likely to make arithmetic errors along the way if you do so.

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