Variable Linear
2010
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Variable Linear
Linear Systems and the Substitution Method
When we have two lines, and we try to graph them on the same coordinate plane, three things can happen: one, the lines may never intersect, in which case they are parallel; two, the lines may coincide, in which case they are identical; or three, the lines can intersect in one point, the unique solution of the system. Here we are going to look at how to find that unique solution using the method known as substitution.
When we have a linear system, or pair of lines, whose solution we are looking to find, we usually write them in standard form. One of the advantages of standard form is that it allows us to solve the system more easily when we try to eliminate one of the variables. This method, known as elimination or linear combinations, is another method that we will look at in an upcoming article. For now, we are going to focus on the method of substitution, which
works particularly nicely when one of the equations is in slope-intercept form.
The method of substitution, as the name implies, allows us to substitute one of the variables by its value in terms of the other, and then solve for the variable. For example, take the following system: y = 2x + 1 and 3x - 2y = 5. The first equation tells us what y is in terms of the variable x. We use this in the second equation and replace y with 2x + 1. Doing this we obtain,
3x - 2(2x + 1) = 5. Now that we have an equation in x only, we can solve to get x = -7. We then plug this value in the first equation to get y = 2(-7) + 1 or -13. The unique solution of this system then is -7, -13, which is a point and the unique point where, if we were to graph these two lines on the same coordinate plane, the two lines intersect.
Although this method usually gives students trouble, there really is no difficulty with it if you keep your head on straight and your pencil sharpened. For example, take the following system:
3x - 5y = 4, and y = -3x + 10. By using the second equation in the first, we obtain
3x - 5(-3x + 10) = 4. Simplifying, we have 3x + 15x -50 = 4. Rearranging terms, we get
18x = 54 or x = 3. We use this value of x to get y = -3(3) + 10 or y = 1.
Lines and more specifically, linear systems, find important applications in the fields of telecommunications, signal processing, and automatic control, the last field of which deals with such interesting things as the programming, guidance, and control of ballistic missiles. In the first article in this series, we examined how to solve a linear system by the method of substitution. Here we will look at some basic problems which employ such linear systems.
Let us look at the following example which deals with museum admissions at, let us say, the Museum of Natural History in New York City. Suppose that in one day, this museum collected $1590 from 321 people admitted to see its splendors. The price of each adult admission is $6. People between the ages of 4-17 pay the child admission of $4. Let us calculate how many of each type, adult and child, were admitted to the museum.
Linear systems give us a neat way of solving this problem. To appreciate the power of linear systems and the method of substitution, which we are going to use to solve this problem,
try to hazard a guess as to how you would figure this out. You will quickly see that there is no convenient way to get the number of adults and the number of children that were admitted. Yet by creating some linear models in the form of a system, we can quickly arrive at the answer to this problem.
Let us start with a verbal model and then translate this into mathematics. This is a convenient and helpful strategy which will allow us to solve the problem more readily. We have from the information that the number of adults plus the number of children is equal to 321. We also have that the number of adults times the price of an adult admission plus the number of children times the price of a child's admission is equal to the total amount collected.
If we let x represent the number of adults and y represent the number of children, we can translate the verbal model into a linear system of equations. Since the price for adults is $6 and the price for children $4 and the total number of people attending 321, we have
x + y = 321 and 6x + 4y = 1590. Notice that both of these equations are in standard form. We can easily take the first equation and put into slope-intercept form by writing
y = -x + 321 (moving x to the other side). We now substitute into the second equation to get
6x + 4(-x + 321) = 1590. Simplifying, we have 2x = 306, or x = 153. Using this value of x to get y, we have y = 168.
Alfred Q Richardson
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What is the Linear Equations in One Variable?
A student has earned scores of 87, 81, and 88 on the first 3 of 4 tests. If the student wants an average (arithmetic mean) of exactly 87, what score must she earn on the fourth test?
A. 85
B. 86
C. 87
D. 92
E. 93
In order to have an average of exactly 87 for four tests, the student must have accumulated exactly
4 * 87 = 348 test points
In the first three tests, the student has earned [87 + 81 + 88 = 256] test points.
How much must she earn on the fourth test ? This is easy ! You can do it.
QED
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