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Solve the systems of equations (this example is also shown in our video lesson)

^$$\left\{\begin{matrix} x+2y-z=4\\ 2x+y+z=-2\\ x+2y+z=2 \end{matrix}\right.$$

First we add the first and second equation to make an equation with two variables, second we subtract the third equation from the second in order to get another equation with two variables. Now we have a system of two equations with two variables:

$$\left \{\begin{matrix} 3x+3y & = & 2\\ x-y & = & -4\\ \end{matrix}\right.$$

We then multiply the second equation with 3 on both sides and add that to the first equation:

$$\\ 6x=-10\\ \\ x=\frac{-10}{6}$$

We plug this value into the $3x+3y=2$ equation in order to determine our $y$-value:

$$\begin{array}{lcl} \\ 3\cdot \frac{-10}{6}+3y&=&2\\ \\ -5+3y&=&2\\ 3y&=&7\\ \\ y&=&\frac{7}{3}\\ \end{array}$$

Last we plug our x- and y-value into any equation in first system in order to determine our z-value:

$$\begin{array}{lcl} x+2y-z&=&4\\ \frac{-10}{6}+2\cdot \frac{7}{3}+z&=&2\\ 3+z&=&2\\ z&=&-1\\ \end{array}$$

But I'm confused, why do we choose only one equation to determine the value of $z$ maybe we could get different values for $z$ in other equations and therefore the first $system$ of equations have no solutions in the set of real numbers.

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  • $\begingroup$ Try solving for $z$ in the other two original equations. $\endgroup$ – John Habert Feb 20 '14 at 21:46
  • $\begingroup$ What? I don't understand what you mean $\endgroup$ – AMS Feb 20 '14 at 22:00
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I will try to help with your confusion by drawing on geometric ideas about which we can reason more intuitively.

"why do we choose only one equation to determine the value of z"?

You didn't really choose only one equation to determine the value of z. You combined the information from all three equations to determine the value of z. You combined them in a certain order, but your final answer will not depend on that order. To understand why, keep reading.

"maybe we could get different values for z in other equations"

Linear equations admit a small set of possibilities. Think of a two equation set for simplicity. Each equation can be visualized as a line. Either the lines cross at one and only one point, the lines are parallel, or the lines coincide. If you doubt this, try to draw two lines which do something other than that.

The first case is the usual case. This is just the nature of straight lines. Once they cross, they cannot cross again, so there is only one solution to be found, no matter whether you trace along line 1 to find the intersection point with line 2, or trace along line 2 to find the intersection with line 1.

If the lines are parallel, they cannot cross anywhere, and if they are coincident, they touch everywhere. These are the special cases, and you would find the same result no matter which order you examined the lines in.

Similar ideas apply to linear systems of any size. For three equations, you can visualize the equations as planes, which in the usual case of one solution, any two of the planes will intersect along a line and the third plane will cut the line at a single point. It does not matter which pair of planes we reduce to a line first (which equates to reducing the system from 3 variables to 2 by substitution). They all meet at the same point.

"and therefore the first system of equations have no solutions in the set of real numbers"

If the system of equations had no solution, you would not be able to find one by combining them in any order.

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