1. The Game of Algebra or The Other Side of Arithmetic
Lesson 21 Part 1 by
Herbert I. Gross & Richard A. Medeiros
© Herbert I. Gross
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2. Systems of Linear Equations
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© Herbert I. Gross

3. Introduction
Suppose we are buying items at a cost of $3 each. In terms of function notation, if we let x denote the number of items and C the cost of the items in dollars, we see that…
C = f(x) = 3x
As soon as we know the number of items we want to buy, their cost is determined directly from the formula C = f(x) = 3x.
By knowing the value of one quantity (in this case, x), we can compute the value of the other quantity (in this case, C).
© 2007 Herbert I. Gross
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4. However in many “real life” situations we do not have this luxury.
More often than not, the quantity
we want to measure depends on our knowing the value of more than one other quantity.
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5. Stated another way, to determine
For example, suppose you once again are ordering items from a catalog. This time, however, you want to buy quantities of two items, one of which costs $3 each and the other of which costs $2 each.
Knowing how many $3 items you bought gives you no information about the number of $2 items you bought.
Stated another way, to determine
the total cost we would have to know both how many $3 items we bought and how many $2 items we bought.
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6. Again in terms of the language of functions, if we let x denote the
number of items we buy at $3 and let y denote the number of items we buy at $2, the total cost in dollars of buying this number of objects is given by…
C = g(x,y) = 3x + 2y
In this case, we would have to know the values of both x and y in order to determine the value of C.
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7. The point is that one important aspect of mathematics involves the situation in which we are dealing with several variables (the number of variables is called the dimension of the problem) and certain constraints are forced upon us.
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8. Definition D
The number of variables that are necessary to determine the output of a function is called the dimension of the domain of the function.
Any restriction that limits our choice for an input is called a constraint.
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9. In this Lesson we shall concentrate primarily on the case in which the
dimension of the domain is 2.
That is: we will be dealing with two independent variables, which, unless otherwise specified, will be denoted by x and y. We shall look in particular at the situation in which there are two linear constraints; that is: constraints of the form…
Ax + By = C
…where A, B, and C are constants.
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10. (which is true if C = 0; and false otherwise).
A Bit of Subtlety
In the constraint Ax + By = C, we assume that not both A and B are 0. For if they were the expression would become 0 = C; which is not a constraint but rather a statement
(which is true if C = 0; and false otherwise).
Note that in order for the product AB to equal 0 either A or B must equal 0. Hence, the mathematical way of saying that A and B cannot both be 0 is to write AB ≠ 0.
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11. Namely, we may rewrite y = mx + b
Notes
The form Ax + By = C (where AB ≠ 0) is an extension of the y = mx + b form.
That is, any equation of the form y = mx + b can be rewritten in the form Ax + By = C.
Namely, we may rewrite y = mx + b
as follows…
y = mx + b → y – mx = b → -mx + y = b; which is in the form Ax + By = C where,
A = -m, B = 1 and C = b.
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12. Notes On the other hand, if we let A = 1 and B = 0, the equation
Ax + By = C becomes x = C. Recall that
x = C is the equation of the vertical line that contains every point whose x-coordinate
is C. On the other hand, no vertical line can have the y = mx + b form. Namely, any line that can be written in that form intersects the y-axis at (0,b).
Notes
However, any vertical line, other than the
y-axis, doesn’t intersect the y-axis
anywhere; and if it is the y-axis, all of its points are y intercepts.
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13. Caution
Do not confuse x = C (a vertical line) with y = C ( a horizontal line). y = C has one and only one y-intercept, namely (0,C).
Caution
Even though the equation involves the letter x, the line x = C is parallel to the y-axis.
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14. Notes In summary, every straight line in the
xy-plane can be represented in the form
Ax + By = C. However, only non-vertical straight lines can be represented in the form y = mx + b.
Stated a bit differently, there is a
one-to-one correspondence between equations of the form Ax + By = C (where AB ≠ 0) and lines (including vertical lines) in the xy-plane.
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15. Note on Eliminating Fractions
Given an equation such as y = 1/2 x + 2/3, we may multiply both sides by 6 to get the equivalent equation 6y = 3x + 4.
This, in turn, may be rewritten in the form
-3x + 6y = 4; which has the form
Ax + By = C (where A = -3, B = 6 and C = 4).
More generally, if m and/or b are common fractions, the equation y = mx + b can be rewritten in the form Ax + By = C, where A, B and C are integers.
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16. A constraint will usually reduce the number of choices we can make in choosing values for the variables.
That is: with each constraint we usually lose a certain amount of freedom in our choice of values for the variables. If, for example, we start with five independent variables in the domain of a function (that is: the domain is 5 dimensional) and after a certain number of constraints, we are only free to choose two of the variables at random, we say that we have two degrees of freedom.
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17. Definitions D
By the (number of) degrees of freedom in the domain of a function,
we mean the number of variables that can be assigned values at random.
The special cases in which the constraints are either too limiting or not limiting enough are discussed in Lesson 22.
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18. Algebraic Translation
To see what this means in more specific terms, recall that we have already defined mathematical expressions in terms of “computer programs”.
For example, to visualize the mathematical expression 3x + 5, we may look at the “program”…
Program 1
Command
Algebraic Translation
Step 1
Pick a number x
Step 2
Multiply by 3 3x
Step 3
Add 5
3x + 5
Step 4
Write the output
3x + 5
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19. Notes
In the language of functions we usually abbreviate “Program 1” by writing
f(x) = 3x + 5.
In this example, the dimension of the domain (the set of inputs) is 1 because there is only one variable, namely: x.
And since there are no restrictions on what the input can be, we also have
1 degree of freedom.
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20. Notes Notice that “1-dimensional” tells us that there is only
1 variable (x) that is needed to determine the output. It doesn’t tell us anything about the number of steps in the program.
Notice also the connection between “dimension” as we use it in geometry and dimension” as defined in this lesson. Namely, the equation f(x) = 3x + 5 has
a 1-dimensional domain and the domain
in this case is the x-axis which is a line
(1 dimensional).
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21. Recall that the graph of f is the set of all ordered pairs (x,f(x)).
Hence, the graph of f is
2-dimensional.
In the language of sets, the graph of
f(x) = 3x + 5 consists of all ordered pairs
(x, 3x + 5)
(1,8)
(0,5)
(-1,2)
(-2,-1)
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22. Pictorially, the graph is the straight line
f(x) = 3x + 5 3
+ 5
Pictorially, the graph is the straight line
(1,8)
(0,5)
(0,5)
whose
slope is 3,
(-1,2)
and whose
y-intercept is (0,5).
(-2,-1)
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23. If we now impose a constraint on f(x) = 3x + 5, say, 3x + 5 = 26
(that is, the output in Program 1 has to be 26), the domain of f is now
0-dimensional.
That is: we no longer have any
degrees of freedom because in order to obey the constraint
the value of x must be 7.
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24. line whose y-coordinate is 26. This point is represented by the
In terms of sets {x: 3x + 5 = 26} consists of the single number 7.
In other words…
{x: 3x + 5 = 26} = {7}
In terms of the graph of f(x) = 3x + 5, (7,26) is the only point on the
line whose y-coordinate is 26.
This point is represented by the
intersection of the line y = 3x + 5
and the line y = 26.
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25. Algebraic Translation
Viewing the constraint 3x + 5 = 26 in terms of “Program 1”, we now have the following situation…
Program 1
Command
Algebraic Translation
Step 1
Pick a number x
Step 2
Multiply by 3 3x
Step 3
Add 5
3x + 5
Constraint
Output must be 26
3x + 5 = 26
In other words, up through Step 3, our input can be any number we want.
However, once the constraint 3x + 5 = 26 is imposed, the only input we can choose is 7.
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26. …where A, B and C represent constants.
Although the language may have changed, what we have done so far in this lesson is to review what we were already doing in Lessons 1 and 2. In this Lesson we shall be doing the same kind of thing but only in cases where the dimension of the domain of f is greater than 1.
In fact, we shall concentrate on linear equations in 2 unknowns. The most general form of a linear equation in 2 unknowns is…
Ax + By = C
…where A, B and C represent constants.
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27. number of elements in the solution set.
Important Aside
One of the ways in which linear equations differ from non-linear equations is in terms of the relationship between degrees of freedom and the
number of elements in the solution set.
For example, if a linear equation in x has no degrees of freedom, then the solution set cannot have more than a single member.
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28. For example, the solution set of the equation 3x + 5 = 26 has only the
one member, x = 7
However, look at an equation such as
x2 = 49. This equation is not linear (because the x term is squared) and it has no degrees of freedom. That is, in order to solve the equation, we are not free to choose values of x at random.
Yet the solution set, S, of this equation has two members, 7 and -7.
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29. For example, consider the equation…
In fact, an equation can have no degrees of freedom; yet its solution set can have many members.
For example, consider the equation…
(x – 1)(x – 2)(x – 3)(x – 4) = 0
The only way a product of numbers can equal 0 is if one of its factors is equal to 0.
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30. can be true only if either…
Hence, the equation
(x – 1)(x – 2)(x – 3)(x – 4) = 0
can be true only if either…
(x – 1) = 0; in which case, x = 1
(x – 2) = 0; in which case, x = 2
(x – 3) = 0; in which case, x = 3
(x – 4) = 0; in which case, x = 4
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31. In other words, the equation (x – 1)(x – 2)(x – 3)(x – 4) = 0
has no degrees of freedom (because we
are not free to choose a value of x at random and have it be a solution). Yet its solution set, S, contains four members.
That is…
S = {x: (x – 1)(x – 2)(x – 3)(x – 4) = 0}
= {1, 2, 3, 4}
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32. Let’s now suppose that instead of a 1-dimensional domain we had a
2-dimensional domain; that is, a function of the type f(x,y) = 3x + 2y.
In terms of a program, the situation might look something like this…
Program 2
Command
Algebraic Translation
Step 1
Pick a number (Input). x
Step 2
Multiply by 3. 3x
Step 3
Pick another number. y
Step 4
Multiply it by 2. 2y
Step 5
Add the number in Step 4
to the number in Step 2.
3x +2y
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33. Algebraic Translation
Command
Algebraic Translation
Program 2
Step 1
Pick a number (Input). x
Step 2
Multiply by 3. 3x
Step 3
Pick another number. y
Step 4
Multiply it by 2. 2y
Step 5
Add the number in Step 4
to the number in Step 2.
3x +2y
“Program 2” is 2-dimensional because there are two independent inputs (x and y), and
it also has 2 degrees of freedom because there are as yet no constraints on the choices for the two variables.
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34. Notes Another way to describe “Program 2” is by saying
that it has two independent variables. When we say that x and y are independent, we mean that in choosing the value of one of the variables, we do not, in any way, restrict the value of the other variable.
For example, in Step 3 when the “program” says “Pick another number”, we are free to choose any number (including the same
number we picked in Step 1) regardless of what number we chose in Step 1.
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35. Notes If we call the numbers x and y, the two numbers we
are determining must come from the set of all ordered number pairs (x,y).
Since there are two variables (or geometrically speaking, since the graph of this set is the xy-plane), we refer to the set as being 2-dimensional.
That is, the algebraic definition of the
xy-plane is…
{(x,y): x and y are any numbers}
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36. Notes Since we now have a 2-dimensional problem, we can expect to need
two constraints in order to determine what the two numbers are.
In general (but as we shall see in the next Lesson there are exceptions), each constraint reduces the number of degrees of freedom in our equation by 1.
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37. For example, suppose we add the single constraint in “Program 2” that
the sum in Step 5 has to be 18. That is…
Command
Algebraic Translation
Program 2
Step 1
Pick a number (Input). x
Step 2
Multiply by 3. 3x
Step 3
Pick another number. y
Step 4
Multiply it by 2. 2y
Step 5
Add the number in Step 4
to the number in Step 2.
3x +2y
Constraint
The sum is 18.
3x + 2y = 18
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38. Constraint
The sum is 18.
3x + 2y = 18
In other words, through Step 5, we are free to choose x and y in any way that we wish. However, the constraint 3x + 2y = 18 means that we can no longer choose
x and y at random.
That is: x and y are no longer independent variables because once we choose a value for either x or y, the value of the other is determined.
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39. we obtain the 1 dimensional equation…
For example, suppose we elect to let x = 2. If we replace x by 2 in the 2-dimensional equation…
3x + 2y = 18;
3(2) + 2y = 18;
we obtain the 1 dimensional equation…
6 + 2y = 18,
for which the solution is y = 6. In other words, while we aren’t forced to let x = 2, once we make this choice, y must be 6.
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40. More generally, we can solve the equation 3x + 2y = 18 for y to obtain that 2y = 18 – 3x, or, y = 1/2(18 – 3x) = 9 – 3/2x.
In other words, to belong to the solution set of 3x + 2y = 18, the ordered number pair must have the form (x,9 – 3/2x)
In this form, we see that the solution set is now 1-dimensional because its members are determined by a single variable (x).
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41. Notice that while there are infinitely many number pairs (x,y) that will satisfy the equation 3x + 2y = 18, the likelihood that an ordered pair of numbers (x,y) chosen at random will satisfy the equation is
very small.
In terms of a graph, this is the set of all points (x,y) that are on the line whose equation is 3x + 2y = 18; or, in the mx + b form, y = -3/2x + 9 (that is, the line whose slope is -3/2 and whose y-intercept is (0,9)).
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42. Thus, even with the constraint we still have one degree of freedom.
So even though a point (x,y) chosen at random is very likely not to be on this line there are, nevertheless, infinitely many points (x,y) that are on the line.
Thus, even with the constraint
3x + 2y = 18,
we still have one degree of freedom.
What we’ll be doing in this Lesson and the next is to see what happens to a linear expression in two variables when we impose two linear constraints.
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43. For example, let’s add another step
And a second constraint to Program #2.
Command
Algebraic Translation
Program 2
Step 1
Pick a number (Input). x
Step 2
Multiply by 3. 3x
Step 3
Pick another number. y
Step 4
Multiply it by 2. 2y
Step 5
Add the number in Step 4
to the number in Step 2.
3x +2y
Constraint
The sum is 18.
3x + 2y = 18
Step 6
Subtract the number in
Step 4 from the number
in Step 2.
3x – 2y
Constraint
The difference is 6.
3x – 2y = 6
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44. To indicate that both equations have to be satisfied simultaneously (that is, by the same ordered number pair), we usually connect the equations by means of a brace. That is, rather than write, for
example, that…
3x + 2y = 18
3x – 2y = 6
and at the same time…
we write instead…
3x + 2y = 18
3x – 2y = 6
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45. In the language of sets, the system of simultaneous linear equations defined by
3x + 2y = 18
3x – 2y = 6
means…
{(x,y): 3x + 2y = 18 and 3x – 2y = 6}.
If we pick values for x and y at random the chances are that the ordered pair (x,y) will satisfy neither the equation 3x + 2y = 18 nor the equation 3x – 2y = 6.
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46. As we shall soon prove, the ordered pair must be (4,3).
Important Point
There will be some pairs that will satisfy one equation but not the other.
However, there will be only one ordered pair that satisfies both of the equations
3x + 2y = 18 and 3x – 2y = 6.
As we shall soon prove, the ordered pair must be (4,3).
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47. Geometric Note The use of graphs gives us an easy way
to visualize how the combination of both constraints determines a unique pair of numbers, x and y , that satisfy the system of linear equations…
Geometric Note
3x + 2y = 18
3x – 2y = 6
Namely, suppose we let L1 denote the line represented by the equation 3x + 2y = 18 and we let L2 denote the line represented by equation 3x – 2y = 6.
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48. Then any pair of numbers, (x,y), that satisfiesL1
(0,9)
Then any pair of numbers, (x,y), that satisfies
3x + 2y = 18 can be represented by a point on L1.
(2,6)
(4,3)
(6,0)
(8,-3)
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49. Similarly, any pair of numbers, (x,y)
(8,9)
Similarly, any pair of numbers, (x,y)
that satisfies 3x – 2y = 6 can be represented by a point on L2.
(6,6)
(4,3)
(2,0)
(0,-3)
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50. Therefore, for a pair of numbers, (x,y), to satisfy both equations, the pair of numbers must correspond to a point that is
on both lines.
But since the two lines have different slopes (i.e., they are not parallel), they can meet at one and only one point; and it is this point
that represents the unique pair of numbers that satisfies each of the two constraints.
If we were to plot the lines L1 and L2 we would find that the one point at which they meet is (4,3).
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51. L1 L2
(0,9)
(8,9)
If we now plot the lines L1 and L2 we would find that the one point at which they meet is (4,3).
(2,6)
(6,6)
(4,3)
(4,3)
(2,0)
(6,0)
(0,-3)
(8,-3)
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52. Notice that if we let x = 4 and y = 3, we see that this number pair is a solution of the simultaneous equations defined by…
3x + 2y = 18
3x – 2y = 6
Namely…
3(4) + 2(3) = 18
3(4) – 2(3) = 6
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53. How would we graph this 5-dimensional equation?
While the graphing method is nice, it is very limited. Among other things, suppose we had a linear equation in, say, five unknowns.
How would we graph this 5-dimensional equation?
Therefore what we’d like to do is find an algebraic way that does the same thing as the graphing method. The reason for this is that we can usually find ways of extending an algebraic technique to higher
dimensional equations which cannot be solved graphically.
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54. Other Graphing Problems
Even in the 2 dimensional case, there can be a problem finding the exact point of intersection of the two lines.
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55. If the two lines are almost parallel…
Example 1
If the two lines are almost parallel…
Where would the exact point of intersection be?
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56. If the two lines do not intersect on the grid lines…
Example 2
If the two lines do not intersect on the grid lines…
Where would the exact point of intersection be?
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57. Still Other Graphing Problems
Even in a case where the lines seem to meet at the point of intersection of two grid lines; because of the thickness of the lines, how would we distinguish between, say, (4,3) and (4.02,2.93), etc.?
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58. notice that by the “equals added to equals”
One strategy for solving systems of simultaneous equations makes use of the fact that the sum of a number and its opposite is 0. For example, 2y + -2y = 0; or in more informal language, 2y – 2y = 0.
To apply this strategy to the system of simultaneous linear equations defined by…
3x + 2y = 18
3x – 2y = 6
notice that by the “equals added to equals”
principle, we can add the two left hand sides of the two equations and equate this to the sum of the two right hand sides.
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59. This illustrates the power of logical thinking.
That is…
3x + 2y = 18 +
3x – 2y = 6
6x (+0) = 24
x = 4
This illustrates the power of logical thinking.
Simply by adding both equations and then dividing by 6, we were able to prove that if the system of equations defined by the above is to be satisfied, then x must equal 4.
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60. equation 3(4) + 2y = 18; from which it follows that y = 3.
From a different point of view, we have proven that if x ≠ 4, then it is impossible to find values for x and y for which the equations 3x + 2y = 18 and 3x – 2y = 6
are both satisfied.
Once we know that x = 4, we may replace x by 4 in either equation to find the value of y.
For example, if we replace x by 4 in the equation 3x + 2y = 18, we obtain the
equation 3(4) + 2y = 18; from which it follows that y = 3.
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61. As a check, we may replace x by 4 and
y by 3 in the equation 3x – 2y = 6 to obtain the true statement…
3x – 2y = 6
3(4) – 2(3) = 6
12 – 6 = 6
6 = 6
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62. That is, we obtained y = 3 using only the equation 3x + 2y = 18
That is, we obtained y = 3 using only the equation 3x + 2y = 18. If we had made a mistake, our answer would not have worked in the equation 3x – 2y = 6.
As usual, notice that there is more than one way to solve a system of equations.
For example, we could have used the equation 3x + 2y = 18 to express y in terms of x. Namely, if we subtract 3x from both sides of the equation, we get the equivalent equation 2y = 18 – 3x from which it follows that…
y = 18 – 3x 2
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63. in equation 3x – 2y = 6 to get the equation
We may then replace y by
y = 18 – 3x 2
in equation 3x – 2y = 6 to get the equation
3x – – 3x = 6 2
( )
which is a linear equation in x.
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64. One way we can solve this equation is by the following sequence of steps…
3x – – 3x = 6 2
( )
3x – (18 – 3x) = 6
3x + -1( x) = 6
3x x = 6
6x – 18 = 6
6x = 24
x = 4
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65. The method described in the above note is often referred to as the method of substitution. While this method works very nicely for a two dimensional linear system of equations, it becomes quite
cumbersome when we are involved with linear systems that have a greater dimension. On the other hand, as we shall soon see, our first method generalizes very nicely to the case of the general system of linear equations in more than two unknowns.
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66. of either x or y are equal also eliminates one of the variables.
Adding in the case in which
the multipliers of either x or y
(these multipliers are usually called coefficients) are opposites eliminates one of the variables. –
In a similar way, subtracting
in the case in which the coefficients
of either x or y are equal also eliminates one of the variables.
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67. With respect to this example, if we subtract 3x – 2y = 6 from 3x + 2y = 18, the x terms will cancel. That is, recalling that to subtract, we add the opposite, we obtain…
3x + 2y = 18 –
3x – 2y = 6 – –
0 + 4y = 12
y = 3
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68. Caution Be careful when you subtract!
For example, when we subtract the equation
3x – 2y = 6 from the equation 3x + 2y = 18,
we are subtracting -2y from 2y.
That is, by the “add the opposite” rule
2y – -2y = 2y + 2y.
Don’t make the mistake of calling the answer 0.
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69. convert a subtraction problem into an addition problem.
A Helpful Strategy…
Because it is less confusing to add than to subtract, it may be helpful to you if you multiply both sides of either one of the
equations by -1.
Since equals multiplied by equals are equal, this will not change the solution set of the system, but it manages to
convert a subtraction problem into an addition problem.
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70. is equivalent to the system…
For example, if we multiply both sides of the equation 3x – 2y = 6 by -1 (and if we recall that multiplying by -1 changes the sign of each term), we see that the system…
3x + 2y = 18
3x – 2y = 6
is equivalent to the system…
3x + 2y = 18
-3x + 2y = -6
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71. -4y = -12 and the solution is still y = 3.
Now if we add these two equations…
3x + 2y = 18
-3x + 2y = -6
we obtain 4y = 12 or y = 3.
4y = 12
If you had wanted to you could have multiplied both sides of 3x + 2y = 18 by -1 and obtained -3x + -2y = -18.
Now when you add,
-3x + -2y = -18
3x + -2y = 6
the sum becomes
-4y = -12 and the solution is still y = 3.
-4y = -12
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72. However, since most people prefer to work with positive rather than negative coefficients, it is often helpful to make our choice based on which one allows us to work with positive coefficients.
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73. For what values of x and y is it true that…
To help ensure that you understand our discussion, try to do the following example. If you think you know how to solve the problem do it before you read our solution. Otherwise you may read our solution first.
For what values of x and y is it true that…
2x + 3y = 16
4x – 3y = 14
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© 2007 Herbert I. Gross 73

74. Once we know that x = 5, we may replace
By adding equals to equals (i.e., the sum of the two left sides must equal the sum of the two right sides), we see that…
Solut
ion
2x + 3y = 16 +
4x – 3y = 14
6x (+0) = 30
x = 5
Once we know that x = 5, we may replace
x by 5 in either of the two equations and then solve for y.
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75. 1-dimensional equation…
For example, if we replace x by 5 in the equation 2x + 3y = 16, we obtain the
1-dimensional equation…
2x + 3y = 16
2(5) + 3y = 16
10 + 3y = 16
3y = 6
y = 2
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© 2007 Herbert I. Gross 75

76. As a check we can now replace x by 5 and
y by 2 in the equation 2x + 3y = 16 and see whether we get a true statement.
2x + 3y = 16
2(5) + 3(2) = 16
= 16
16 = 16
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77. …which is a true statement.
So we replace x by 5 and y by 2 in the equation 4x – 3y = 14 to obtain…
4x – 3y = 14
4(5) – 3(2) = 14
20 – 6 = 14
14 = 14
…which is a true statement.
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78. Notes Notice that the check is part of the solution.
Namely, once we knew the value of x, we found the value of y from the equation
2x + 3y = 16.
Notes
To make sure we didn’t make a numerical error we check the result in the equation that we didn’t use to find the value of y.
For example, once we knew that x = 5, we could have solved for y using the equation 4x – 3y = 14. In this case we would have
checked the result by using the equation 2x + 3y = 16.
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79. Notes The big question that remains, however, is
how likely is it that the coefficients of y in each of the two equations will be either the same or opposites?
Notes
The answer is that while it is very unlikely that in a pair of simultaneous equations chosen at random that the coefficients of
either x or y will be opposites, we can always replace the given system by an equivalent system in which the coefficients will be opposites.
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80. Thus, the equations 2x = 6 and 10x = 30 are equivalent.
Definition D
By two equations being equivalent we mean that they have the same solution set.
For example, if we multiply both sides of the equation 2x = 6 by 5, we obtain the equation 10x = 30. The solution for each of these equations is x = 3.
Thus, the equations 2x = 6 and 10x = 30 are equivalent.
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81. Let’s practice this technique by solving the system of equations…
3x + 4y = 25
2x + 3y = 18
We have a choice of solving the system for either x or y. If we elect to solve for x it means that we want to eliminate y.
To do this we will replace the two equations above by equivalent equations in which the coefficients of y are opposites.
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© 2007 Herbert I. Gross 81

82. 4 and 3 and that 12 is the least common multiple of 3 and 4.
We might start by observing that the coefficients of y in equation below are
4 and 3 and that 12 is the least common multiple of 3 and 4.
3x + 4y = 25
2x + 3y = 18
So we can multiply both sides of 3x + 4y = 25 by 3 to obtain the equivalent equation…
3( )
3x + 4y = 25
3( )
9x + 12y = 75
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83. And we can multiply both sides of
2x + 3y = 18 by -4 to obtain the equivalent equation…
-4( )
2x + 3y =
-4( )
-8x + -12y = -72
In other words, the system of equations defined by…
3x + 4y = 25
2x + 3y = 18
is equivalent to the system of equations defined by…
9x + 12y = 75
-8x y = -72
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84. ~ Notation We sometimes use the symbol “~” to indicate
that two systems are equivalent.
Notation
In terms of this notation…
3x + 4y = 25
2x + 3y = 18
9x + 12y = 75
-8x y = -72 ~
x = 3
The beauty of the latter system is that the coefficients of y are now opposites.
Hence, when we add the two equations in the latter system the terms involving y cancel, and we are left with x = 3.
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85. ~ Notes Once we know that x = 3, we may go to either of
the two equations in either of the systems…
3x + 4y = 25
2x + 3y = 18
9x + 12y = 75
-8x y = -72 ~
(It doesn’t matter which because the systems are equivalent) to determine the value of y.
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86. This time, however, we want to eliminate the terms that involve x.
Notes
However, we want to practice the method of
solving for y by first eliminating the terms involving x.
3x + 4y = 25
2x + 3y = 18
This time, however, we want to eliminate the terms that involve x.
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87. Notes Noticing that the coefficients of x are 2 and 3, we see that
6 is the least common multiple.
Hence, we can multiply both sides of the first equation by 2…
and both sides of the second equation by -3…
to get the
equivalent system of equations…
2( )
3x + 4y = 25
2( )
-3( )
2x + 3y =
-3( )
6x + 8y = 50
-6x + -9y = -54
6x + 8y = 50
-6x + -9y = -54
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88. If we now add these two equations…
Notes
6x + 8y = 50
-6x + -9y = -54
If we now add these two equations…
-y = -4
we obtain -y = -4 or y = 4.
As a check, if x = 3 and y = 4,
3x + 4y = 3(3) + 4(4) = 25.
and
2x + 3y = 2(3) + 3(4) = 18
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89. In solving the system of equations…
Notes
3x + 4y = 25
2x + 3y = 18
In solving the system of equations…
we chose to multiply the top equation by 2 and the bottom equation by -3.  It would have been just as logical to have multiplied the top equation by -2 and the bottom equation by 3 to obtain the equivalent system…
-6x + -8y = -50
6x + 9y = 54
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90. If we now add these two equations…
Notes
-6x + -8y = -50
6x + 9y = 54
If we now add these two equations…
y = 4
we obtain immediately
that y = 4 .
Either way we obtain the correct answer but most of us would rather work with positive coefficients than negative ones.  Thus, a good strategy would be to choose our signs in such a way that we wind up dealing with positive coefficients.
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91. One way to do this is to ignore the signs until
Notes
One way to do this is to ignore the signs until
the last step. 
For example, suppose we want to eliminate y from the system of equations…
3x + 5y = 30
4x + 2y = 40
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92. Notes We begin by multiplying both sides of the top equation by 2
and both sides of the bottom equation by 5
to obtain the equivalent system of equations…
2( )
3x y =
2( )
2( )
5( )
4x y =
5( )
5( )
6x + 10y = 60
8x + 10y = 200
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93. In this way, when we add we obtain the equation 2x = 140 or x = 70.
Notes
Since 8 is greater than 6, we can make sure when we add that the coefficient of x is positive if we multiply the top equation by -1 to obtain…
-6x + -10y = -60
8x + 10y = 200
In this way, when we add we obtain the equation 2x = 140 or x = 70.
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© 2007 Herbert I. Gross 93

94. This concludes our lesson.
In essence, Lesson 21, part 2 may be viewed as a continuation of
Lesson 21 part 1 in the sense that it introduces the cases in which the dimension of the linear system is greater than 2.
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