This week we will combine pairs of linear equations in two unknowns in order to determine a single value for each unknown. We will discuss the three basic methods, graphing, substitution, and addition. We will see that these methods can be extended to handle multiple variables. We will also touch on the Gaussian Elimination method.
One of the basic axioms in geometry is that any two separate lines are either parallel or have exactly one intersection. The same principle applies to the x-y coordinate system. Any two straight lines on the grid are either parallel (having the same slope so they never meet) or intersecting exactly once.
If there are two lines in the x-y coordinate system, representing equations, their intersection represents the solution to both equations. This fact means that we can solve a pair of equations simultaneously by plotting their graphs and finding the intersection.
Here
is an example of a problem that can be solved by graphing: Carol has
started saving for a home entertainment system. She calculates that
she can save the price of the new system she wants in 3 years. A
recent article projected that in the next three years the price of
such systems would gradually drop to half their present cost. Carol
thinks this is good news and plans to purchase the home entertainment
system as soon as she can do so from her savings. How long will this
take?
Carol's present savings are zero but will be 100% of the current price of the entertainment system in 3 years. The price of the entertainment system is now at 100% but it will be 50% in three years. This gives us two points for savings and two points for price. By connecting the lines, we find the intersection point at 2 years and 66.7%
Week Two of this class included the topic of solving one variable in an equation in terms of the others. The process of solving involves isolating the target variable on the left side of the equation.
If aX + bY =c, we can divide by b to get (a/b)X + Y = c/b - then subtract (a/b)X to get
Y = c/b - (a/b)X, which is the desired solution. Remember that a, b, and c are constants, so (c/b) and (a/b) will be numbers rather than expressions that require further calculation.
If we have two different two-variable equations in standard format and we solved the first one for Y, the next step is to substitute this solution for the "Y" in the second equation. The result will be a one-variable equation, which was covered in Week One.
For an example of this method, consider a recent school play for which 150 tickets were sold for a total of $650. If student tickets cost $3 and adult tickets cost $5 how many of each type were sold? Representing the number of student tickets as S and the number of adult tickets as A, we can write two equations in standard format:
A
+ S = 150
5A + 3S = 650
Solving the first of these equations for S gives S=150-A. We can substitute this solution, 150-A ,for the S in the second equation, giving
5A
+ 3( 150-A ) = 605
5A
+ 450-3A = 650 multiplying
2A + 450
= 650 combining terms
2A = 200 subtract 450 from both
sides
A = 100 divide by 2
So there were 100 adult tickets and 50 student tickets.
Another method of solving pairs of equations is to subtract one from the other (adding the arithmetic inverse) to eliminate one of the variables. Using the Tickets problem above (in reverse order)
5A
+ 3S = 650
A + S = 150
Let's eliminate the S in the first equation. Multiply the second equation by -3 and add it to the first
5A
+ 3S = 650
-3A -3S = -450
2A
= 200, so A=100
In general, the multiplier for the equation being added is the additive inverse of the coefficient for the variable being eliminated in the first equation (3 in this example) divided by its coefficient in the second equation (1 in this case). I reversed the order of the equations so that the one with the smaller coefficients came second.
Equations in one variable have inconsistent equations and identities giving no solution or an infinite number of solutions. In the same way, pairs of equations can be inconsistent, giving no solution or dependent, giving an infinite number of solutions.
In the School Play problem, if the tickets were all the same price,there would be no way to determine the number of adults attending from the revenue. If the tickets were all $4.00, we would have
A
+ S = 150
4A + 4S = 600
These two equations may look different; however, the second is the same as the first, multiplied by 4. When we graph dependent equations, the graphs overlap.
Suppose we had revenue from the school play of $650 with the tickets selling for $4.00 each. The equations would be
A
+ S = 150
4A + 4S = 650
If we multiply the first by -4 and add it to the second, we get
-4A
- 4S = 600
4A + 4S = 650
0
= 50
which is no solution at all. When we graph inconsistent equations, they are parallel.
In order to solve for a group of unknown variables, we need at least one linear equation for each variable. The methods for solution use the equations in combination to isolate each variable and match it to its value.
As we discussed last week, the preparation of three dimensional graphs is rather time consuming. While a two-dimensional graph can be drawn in seconds, a three dimensional graph can take hours. As our textbook points out, graphs are not feasible as a solution method beyond two variables.
Example 1 on page 7-24 of our textbook shows the addition method for solving a linear system with three variables. How would we solve it with substitution?
(1) x + y - z = -1 (2) 2x – 2y + 3z = 8 (3) 2x – y + 2z = 9
First, let's use equation (1) to solve for z
x + y - z = -1 x + y = -1 + z add z to both sides x + y + 1 = z add 1 to both sides z = x + y + 1 reverse sides
Now, we can substitute this formula for z into the other two equations.
(2) 2x – 2y + 3( x + y + 1 ) = 8
2x – 2y + ( 3x + 3y + 3 ) = 8 apply distribution
5x + y + 3 = 8 combine terms
5x + y = 5 subtract 3 from both sides
(3) 2x – y + 2( x + y + 1 ) = 9
2x – y + (2x + 2y + 2 ) = 9 apply distribution
4x + y + 2 = 9 combine terms
4x + y = 7 subtract 2 from both sidesThis leaves us with the new pair of equations
(2) 5x + y = 5
(3) 4x + y = 7
We can use equation (2) to give y in terms of x:
5x + y = 5
y = 5 - 5x subtract 5x from both sides
Now substitute this formula into equation 3
4x + (5 - 5x) = 7
-x + 5 = 7 combine terms
-x = 2 subtract 5 from both sides
x = -2
Use this value in the second equation (2)
5x + y = 5
-10 + y = 5 substitute
y = 15 add 10 to both sides
Use these values in the original solution for z
z = x + y + 1 = -2 +15 + 1 = 14
When you compare this method against the solution on page 7-24 of our textbook, you see a strong argument for using the addition method.
The Gaussian Elimination method is described in section 7.4 of the textbook. It is similar to the addition method, with a simplified notation. The Example 1 set of equations
(1)
x + y - z = -1
(2) 2x – 2y + 3z = 8
(3) 2x – y + 2z
= 9
can be written
/ \ | 1 +1 -1 | -1| | 2 –2 3 | 8| | 2 –1 2 | 9| \ / The “x”, “y”, and “z” are represented by position and the “=” is represented by a “|”. / \ | 1 +1 -1 | -1| | 5 1 0 | 5| 3R1 + R2 ->R2 | 2 –1 2 | 9| \ / The strange expression, “3R1 + R2 ->R2” means “the sum of rows one and 2 replace what was in row 2”. Just as with the addition method, the goal is to have zero coefficients for the variables except for one number in each row. / \ | 1 +1 -1 | -1| | 5 1 0 | 5| | 4 1 0 | 7| 2R1 + R3 -> R3 \ / / \ | 1 +1 -1 | -1| | 4 1 0 | 7| R2 <-> R3 | 5 1 0 | 5| \ / / \ | 1 +1 -1 | -1| | 4 1 0 | 7| | 1 0 0 | -2| R3 – R2 -> R3 \ / / \ | 0 +1 -1 | 1| R1 – R3 -> R1 | 0 1 0 | 15| | 1 0 0 | -2| \ / / \ | 0 0 -1 |-14| R1 – R2 -> R1 | 0 1 0 | 15| | 1 0 0 | -2| \ / / \ | 0 0 1 | 14| –R1 -> R1 | 0 1 0 | 15| | 1 0 0 | -2| \ /
Like addition, this method easily expands to a large number of variables.
How would you solve this problem (exercise
75 in section 7.2)?
Nickels
and dimes. Winborne has 35 coins consisting of dimes and
nickels. If the value of his coins is $3.30, then how many of each
type does he have?
How did you decide which of the
methods for solving pairs of equations (graphing, substitution,
addition, or reduction) to use in this particular situation?
In another showing of the school play, the 150 seat theater was again sold out. This time, the revenue from the tickets ($3 for students and $5 for adults) was $655. How many tickets of each type were sold?
Have you dealt with any real-life problems involving pairs of equations?