History On The Use Of Variables

The use of letters to represent general numbers goes back to the Greeks. Aristotle frequently used capital letters of two letters for the designation of magnitude or number.

The use of z,y,x to represent unknowns is due to Rene Descartes in his La geometrie. He introduces the use of the first letters to signify known quantities while he used last letters to signify unknown quantities. He used the last letter z for the first unknown and proceeded backwards to y and x for the second and third, respectively.

The predominant use of the letter x to represent an unknown value came in an interesting way. During the printing of Descartes’ La Geometrie which introduced coordinate geometry, the printer reached a dilemma. While the text was being typeset, the printer began to run short of the last letters of the alphabet. The publisher asked Descartes if it mattered whether x, y, or z was used in each of the book’s many equations. He replied that it made no difference. The printer selected x for the most unknowns, since the letters y and z are used in the French language more frequently than is x.

Rene Descartes

Discussion with Examples

* What is a linear equation in two variables?

A linear equation in two variables is a first-degree equation that can be put in the form:

Ax + By = C Where A, B, and C are real numbers, A and B are not 0, and x and y are the two variables

e.g. x + y = 10, 2x + 5y = 15

* What is the solution of a linear equation in two variables?

A solution of linear equation in two variables requires two values, one for each variable, which will satisfy both equations. These two variables can be represented by an ordered pair (a,b).

* How do you decide whether an ordered pair satisfies an equation in two variables?

An ordered pair (a,b) can belongs to an equation in two variables if a true statement results when a and b are substituted for x and y.

e.g. (3,2); 2x + 3y = 12

To see whether (3,2) is a solution of the equation 2x + 3y = 12, we substitute 3 for x and 2 for y in the given equation..

2x + 3y = 12 2(3) + 3(2) = 12 ?

Let x = 3; let y = 2

6 + 6 = 12 ? 12 = 12 True

The result is true, so (3,2) satisfies 2x + 3y = 12

* What are the three kinds of the systems of equations?

To illustrate what the three kinds of the systems are, let us observe these graphs of a linear system:

1) x – y = 4, x + y = 2

2) 5x – 2y = 10, 5x – 2y = 6

3) x + y = 2, 2x + 2y = 41
1) In the first graph, the lines intersect at a single point. In this case, it has only one solution. This is a consistent system.

2) In the second graph, the lines are parallel to each other. Therefore, the lines do not intersect at any point and sp the system has no solution. This is an inconsistent system.

3) In the third graph, the lines are identical; thus, they overlap and end up being the same line. In this case, the system has an infinite number of solutions since all the solutions of one equation are also solutions of the other equation.

* How do you determine whether a system is dependent, consistent, or inconsistent?

In general, the system of equations Ax + By = C and ax + by = c is:

1) consistent if A/a is not equal to B/b and B/b is not equal to C/c e.g. Consider the consistent system (x – y = 4, x + y = 2) given in the sample graph. Substituting the coefficients as stated in the formula above, 1/1 is not equal to -1/1 and -1/1 is not equal to 4/2, proving that it is indeed a consistent system.

2) inconsistent if A/a is equal to B/b and B/b is not equal to C/c e.g. Consider the inconsistent system (5x – 2y = 10, 5x – 2y = 6) given in the sample graph. Substituting the coefficients as stated in the formula above, 5/5 is equal to -2/-2 and -2/-2 is not equal to 10/6, proving that it is indeed an inconsistent system

3) dependent if A/a is equal to B/b and B/b is equal to C/c e.g. Consider the inconsistent system (x + y = 2, 2x + 2y = 4) given in the sample graph.
Substituting the coefficients as stated in the formula above, 1/2 is equal to 1/2 and 1/2 is equal to 1/2, proving that it is indeed a dependent system

* How do you solve for the solution of systems of linear equations?

There are three ways to solve systems of linear equations in two variables:

1) graphing

2) elimination method

3) substitution method

* How do you use graphing to solve for the solution of systems on linear equations?

To solve for the solution of systems on linear equations using graphing, find the x and y intercepts of each solution. The x-intercept is found by taking y = 0 and then solving for the value of x. The y-intercept is found by taking x = 0 and solving for the value of y. The x-intercept (a,0) and y-intercept (0,b) will serve as the two points to create the line.

e.g. Solve the system of equation by graphing:

In x + y = 3 and x- y = 1, you will get (respectively):

x - intercept - (3,0); y - intercept - (0,3)

x - intercept - (1,0); y - intercept - (0,-1)

Therefore, (3, 0) and (0, 3) are on the line x + y = 3 and (1, 0) and (0, -1) are on the line x – y = 1. Plotting these points and drawing the line that contains each pair we obtain this graph:

Conclusion: Since the lines intersect at (2,1), the solution is (2,1)

* How do you use the substitution method to solve for the solution of systems on linear equations?

To solve for the solution of systems on linear equations using the substitution method, follow these steps:

1) Simplify if needed. This involves removing parentheses and fractions

2) Solve one equation for either variable.

3) Substitute what you get for step 2 into the other equation.

4) Solve for the remaining variable.

e.g. Solve the system of graphing by the substitution method:

Step 1: It is already simplified

Step 2: Solving for the second equation

Step 3 and 4: Substitute the expression y + 1 for x into the first equation and solve for y:

Step 5: Plug in 3 for y into the equation in step 2 to find the value of x.

Conclusion: (4, 3) is the solution

* How do you use the elimination method to solve for the solution of systems on linear equations?

To solve for the solution of systems on linear equations using the elimination method, follow these steps:

1) Simplify and put both equations in the form Ax + By = C if needed.

2) Multiply one or both equations by a number that will create opposite coefficients for either x or y if needed.

3) Add equations.

4) Solve for second variable

5) Solve for remaining variable.

e.g. Solve the system of graphing by the elimination method:

Step 1: Multiplying each equation by it's respective LCD we get:

Step 2: Multiplying the first equation by 5 and the second equation by -2 we get:

Step 3:

Step 4:

Step 5: I choose to plug in 10 for x into the first simplified equation (found in step 1) to find the value of y. Conclusion: The solution is (10, 24)

Exercises

A. Determine if the given system of equation is consistent, dependent, or inconsistent

1. x + 3y = 5
3x + 9y = 45

2. 3x – 6y =5
2x – 4y = 11

3. 6x + 4y = 3
6x + 4y = 5


B. Solve the following using the method of elimination

1. 4x – y = 9
x – 3y = 16

2. 0.7x – 0.3y = 0.5
-0.4x + 0.7y = 1.3

C. Use substitution to solve these equations

1. 5x + y = 8
3x – 4y =14

2. 4x + 12y = 4
5x – y = -11

D. Use comparison method to solve these equations

1. x – 3y = 5
5x – y = -14

2. 3x – 8y = 11
x + 6y = 8

Answers

A

1. dependent
2. inconsistent
3. inconsistent

B

1. (-1 , -17/3)
2. (2 , 3)

C

1. (2 , -2)
2. (-2 , 1)

D

1. (-3 , -1)
2. (5 , ½)

Applications

The systems of linear equations can be used for a variety of problems such as coin, geometry, age, work, mixture, investment, motion, and digits. That is why we should exert effort in understanding them because they can be helpful to real life situations like other math concepts.

Age Word Problem

In January of the year 2000, my husband John was eleven times as old as my son William. In January of 2012, he will be three times as old as my son. How old was my son in January of 2000?

Let "J " stand for my husband John's age, and let "W " stand for William's age. Then J = 11W in the year 2000. In the year 2012, John and William will be twelve years older, so their ages will be J + 12 and W + 12. Also, John will be three times as old as William, so J + 12 = 3(W + 12). So now you've got two equations, each with two variables:

J = 11W J + 12 = 3W + 36

If you know how to solve systems of equations, you can proceed with those techniques. Otherwise, use the first equation to simplify the second; since J = 11W, plug in "11W " for "J " in the second equation:

J + 12 = 3W + 36 (11W) + 12 = 3W + 36 11W – 3W = 36 – 12 8W = 24 W = 3

Remember that the problem did not ask for the value of the variable W; it asked for the age of a person. So the answer is: William was three years old at the beginning of the year 2000.

Coin Word Problem

Your uncle walks in, jingling the coins in his pocket. He grins at you and tells you that you can have the coins if you can figure out how many of each kind of coin he is carrying. You're not too interested until he tells you that he's been collecting those gold-tone dollar coins. The twenty-six coins in his pocket are dollars and quarters, and they add up to seventeen dollars. How many of each coin does he have? Here's how you figure it out:

There are twenty-six coins in total. Some of them are dollar coins; let "d" stand for the number of dollar coins. The rest of the coins are quarters; let "q" stand for the number of quarters. Then d + q = 26.
If your uncle has only one quarter, then 25×1 = 25 cents comes from quarters. If he has two quarters, then 25×2 = 50 cents comes from quarters. Since he has q quarters, then 25×q = 25q cents comes from quarters.
For the dollar part to work, you'll have to convert to cents. That is, one dollar is one hundred cents. Since he has d dollars, then he has 100d cents from the dollar coins.
He has seventeen dollars in total, or 1700 cents, part of which is from quarters and part of which is from dollars. To help keep things straight, I'll set up a table:

If you know how to solve systems of linear equations, you can see that you have the following system:
d + q = 26 100d + 25q = 1700
If you don't know about systems yet, don't worry: we're nearly done. Since we have two equations related to the same situation, I'll solve one of them and plug the result into the other one. For instance, starting with the two equations::
d + q = 26 100d + 25q = 1700
...I'll solve the first equation above to get:
q = 26 – d
(I could have solved the first equation for d instead of for q; the end result would have been the same, though the computations would have looked different in the middle. Choosing to solve for q in this case was purely a matter of taste.)
Then I'll plug this into the second equation in place of q to get:
100d + 25(26 – d) = 1700
Then I'll solve:
100d + 25(26 – d) = 1700 100d + 650 – 25d = 1700 75d + 650 = 1700 75d = 1050 d = 14
In other words, fourteen of the coins are dollar coins. Since the remainder of the twenty-six coins are quarters, there are 12 quarters.

The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

Now, though, you can use systems of equations, which means you can use two variables.
a + c = 2200 4a + 1.5c = 5050
Then solve the system for the number of adults and the number of children:
a = 2200 – c
4(2200 – c) + 1.5c = 5050

8800 – 4c + 1.5c = 5050

8800 – 2.5c = 5050

–2.5c = –3750

c = 1500
a = 2200 – (1500) = 700
There were 1500 children and 700 adults.

Distance Problem

A 555-mile, 5-hour plane trip was flown at two speeds. For the first part of thtrip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?

Let's set up a grid:

Using "d = rt", the first row gives us d = 105t and the second row gives us:
555 – d = 115(5 – t)
Since the two distances add up to 555, add the two distances:
555 = 105t + 115(5 – t)
Then we get:
555 = 105t + 575 – 115t 555 = 575 – 10t –20 = –10t 2 = t
According to our grid, "t" stands for the time spent on the first part of the trip, so the answer is "The plane flew for two hours at 105 mph and three hours at 115 mph."
You can add distances and you can add times, but you cannot add rates. Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph?

Digit Problem

The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number.

I'll use "t" for the "tens" digit of the original number and "u" for the "units" (or "ones") digit. Keeping in mind that the tens digit stands for "ten times this value", I then have:
t + u = 7 10u + t = 10t + u + 27
(The "10t + u" on the right-hand side of the second equation represents the original number.) First I'll simplify the second equation:
10u + t = 10t + u + 27 9u – 9t = 27 u – t = 3
This gives me:
u + t = 7 u – t = 3
Adding down, I get 2u = 10, so u = 5. Then t = 2. Checking, this means that the original number was 25 and the new number (gotten by switching the digits) is 52. Since 52 – 25 = 27, this solution checks out.

The number is 25.

Business Problem

A landscaping company placed two orders with a nursery. The first order was for 13 bushes and 4 trees, and totalled $487. The second order was for 6 bushes and 2 trees, and totalled $232. The bill does not list the per-item price. What is the cost of one bush and of one tree?

I could try to add the bushes and trees, to get 19 bushes and 6 trees, but this wouldn't get me anywhere, because I don't have subtotals for the bushes and trees. So I'll pick variables ("b" for the number of bushes and "t" for the number of trees) and set up a system of equations:
13b + 4t = 487 6b + 2t = 232
Multiplying the second row, I get:
13b + 4t = 487 –12b – 4t = –464
This says that b = 23. Back-solving, I get that t = 47.
Bushes cost $23 each; trees cost $47 each.