How Do I Solve A System Of Equations That Results In Two Equations And Three Variables?

When using Left to Right Elimination or Gauss’s Methods, a system of three equation in three variables often results in a unique solution. In other words, a single value for each of the three variables solves each of the three equations in the system. However when some systems are put into row echelon form, all of the constants and variables in one of the equations drop out. In a sense, one of the equations has disappeared and now you are left with a system of two equations in three variables.

Don’t confuse this with an inconsistent system. In an inconsistent system, the row operations lead to one of the equations being false…something like 0 = 6. This type of system has no solutions.

Let’s examine a system of three equations in three variables in which one of the equations drops out. Take the system

$latex \displaystyle \begin{matrix}
x&+2y&-4z &=5 \\
2x&+3y&-z &=3 \\
3x&+5y&-5z &=8 \\

If we use left to right elimination to solve this system, we get the equivalent system

$latex \displaystyle \begin{matrix}
&x&+2y&-4z &=5 \\
&&y&-7z &=7 \\
& & &0 &=0 \\

This new system contains only two equations since 0 = 0 does not describe a relationship between the variables. This new system has many solutions. We can show this by solving the second equation for y to give

$latex \displaystyle y=7z+7$

and substituting it into the first equation:

$latex \displaystyle x+2\left( 7z+7 \right)-4z=5$

Now solve this equation for x,

$latex \displaystyle \begin{matrix}
x+14z+14-4z=5 \\
x+10z+14=5 \\
x=-10z-9 \\

The solution to the system is

$latex \displaystyle \begin{matrix}
x&=&-10z-9 \\
y&=&7z+7 \\

This may also be written as an ordered triple $latex \left( -10z-9,7z+7,z \right)$.  We have solved for x and y, but z is not specified. We can pick it to be anything. For instance, if  z = 0, then x = -0 and y = 7. This means that the ordered triple (-9, 7, 0) solves the original system of equations. If z = -1, then x = 1 and y = 0, So (1, 0, -1) also solves the system.

How Do I Set Up A System Of Linear Equations?

For basic problems, a good starting point for writing out a system of equations is to define the variables. If you don’t know what the variables represent, it is almost impossible to write out the equations. Once this is accomplished, locate the different totals in the problem…these are often where the equations come from.

Problem 1 A safe investment earns 4% per year. A risky investment earns 6% per year. An investor has 20,000 dollars to invest. How much should be invested in each investment to earn 1090 dollars annually?

Solution The two totals in the problem are the total amount invested, 20,000 dollars, and the total interest earned, 1090 dollars. The 20,000 dollars total is simply the sum of the variables. To calculate interest, multiply the percentage earned times the amount.


Once the equations are written, solve them by applying the Elimination Method.

Problem 2 Coffee is usually blended before being roasted. Suppose a roaster has Guatemalan beans worth 5.50 dollars per pound and Panamanian beans worth 8.00 dollars per pound. How many pounds of each should be mixed to obtain 100 pounds of beans worth 6.25 dollars per pound?

Solution The sum of the variables is related to the total amount of beans, 100 pounds. The other total is not so obvious. Using the numbers in the problem, we can calculate the total worth of the mixture,

$latex \displaystyle \left( 6.25\frac{\text{dollars}}{\text{pound}} \right)\left( 100\text{ pounds} \right)=625\text{ dollars}$

We can find the worth of the beans that make up the mixture by multiplying the cost per pound times the number of pounds of beans, 5.50g or 8p.



Notice that when we calculate how much 70 pounds of Guatemalan beans (385 dollars) and 30 pounds of Panamanian beans (240 dollars), the sum is 625 dollars.

How Do I Solve A System of Linear Equations By Graphing?

To solve a system of linear equations in two variables by graphing, you must first solve each equation for the dependent variable. Once this is done, we can use the equations to find an appropriate window for the graph. It is often useful to also solve the system algebraically…this helps us to establish the horizontal extent of the window.

Problem The annual number of cars produced t years after 2000 by a small car manufacturer is $latex y=6.032t+34.543$ thousand cars. A larger producer has annual production $latex y=-10.564t+100.340$ thousand cars. In what year will the annual production be equal? What will the production be then?

Solution Examine the two equations. The vertical intercepts are 34.543 and 100.340. From 34.543, the first line rises. From 100.340, the second line decreases. Based on this, we can deduce that a vertical window from 0 to 110 is appropriate. If we solve the system by substitution, we see that the point of intersection should be t = 3.965. This suggests a horizontal window of 0 to 5 or 0 to 10. In this window, we can find the point of intersection at (3.965, 58.460).



The value 3.965 corresponds to a time late in the year 2003. The number of cars produced at that time is 58.460 thousand cars or 58,460 cars. Be careful in rounding the t value. Although t = 3.965 rounds to 4 (the year 2004), this time is in the year 2003. Rounding to the nearest integer would put the point of intersection in the next year…a mistake when the problem asked in what year.