Download 265 10–1 Systems of Two Linear Equations ERGOGRIP INC

CANADIAN BIOGRAPHY
ERGOGRIP INC—SARAH AND ALEXANDRA LEVY
Success happens when you set out to help others. As the Canadian population ages, products
and services for elderly people become very important, allowing them to live in dignity
with independence. Just ask Sarah Levy and her daughter Alexandra. These two Montreal
entrepreneurs were associated with a company that provided institutional meals to nursing
homes. Working with seniors, they saw a need: They learned that normal utensils and dishware
used by people with symptoms characteristic of diseases such as arthritis and Parkinson’s
were not appropriate. They came to believe that the fundamental needs for a pleasant dining
experience, independence, and dignity were often denied to the residents.
Together, Sarah and Alexandra obtained financing and designed, manufactured, and marketed
a line of products used in homes, hospitals, and nursing homes. The products include mugs
and utensils with easy-grip handles and special thermal dishware to keep food warm longer. The
Levys’ company, Ergogrip Inc., has successfully penetrated the North American nursing home
market well beyond the borders of Quebec.
10–1 Systems of Two Linear Equations
Linear Equations
We have previously defined a linear equation as one of first degree.
◆◆◆
Example 1: The equation
3x 5 20
is a linear equation in one unknown. We learned how to solve this kind of equation in
◆◆◆
Chapter 3.
◆◆◆
Example 2: The equation
2x y 3
is a linear equation in two unknowns. A linear equation with the terms written in this order,
Ax By C 0, is said to be in general form. If we graph this equation, we get a straight
line, as shown in Fig. 10–2. Glance back at Sec. 5–3 if you’ve forgotten how to make such a
graph.
y
2x
−
y=
3
2
1
−2
−1
0
1
2
3
x
−1
slope = 2
−2
−3
FIGURE 10–2
y intercept
◆◆◆
265
266
Chapter 10
◆
Systems of Linear Equations
If a linear equation in two unknowns is solved for y, we get an equation in slope-intercept
form, as explained in Sec. 5–3.
Example 3: If we solve the equation from Example 2 for y, we get the same equation in
slope-intercept form.
◆◆◆
y 2x 3
Recall from Sec. 5–3 that the coefficient of the x term is the slope and the constant term is the
y intercept. Thus in our example,
slope 2
y intercept 3
The graph is, of course, the same as in Example 2.
◆◆◆
A linear equation can have any number of unknowns.
◆◆◆
Example 4:
(a) x 3y 2z 5 is a linear equation in three unknowns.
(b) 3x 2y 5z w 6 is a linear equation in four unknowns.
Systems of equations are also
called simultaneous equations.
Systems of Equations
A set of two or more equations that simultaneously impose conditions on all of the variables is
called a system of equations.
◆◆◆
Note that some variables may
have zero coefficients and may
not appear in every equation.
◆◆◆
Example 5:
(a) 3x 2y 5
x 4y 1
is a system of two linear equations in two unknowns.
(b) x 2y 3z 4
3x y 2z 1
2x 3y z 3
is a system of three linear equations in three unknowns.
(c) 2x y 5
x 2z 3
3y z 1
is also a system of three linear equations in three unknowns.
(d) y 2x2 3
y2 x 5
is a system of two quadratic equations in two unknowns.
◆◆◆
Solution to a System of Equations
The solution to a system of equations is a set of values of the unknowns that will satisfy every
equation in the system.
◆◆◆
Example 6: The system of equations
xy5
xy3
Remember: It is a good idea to
number your equations, as in
this example, to help keep track
of your work.
(1)
(2)
is satisfied only by the values x 4, y 1, and by no other set of values. Thus the pair (4, 1) is
◆◆◆
the solution to the system, and the equations are said to be independent.
To get a numerical solution for all of the unknowns in a system of linear equations, if one
exists, there must be as many independent equations as there are unknowns. We first solve
two equations in two unknowns; then later, three equations in three unknowns; and then larger
◆
267
Systems of Two Linear Equations
systems. But for a solution to be possible, the number of equations must always equal,
or exceed, the number of unknowns.
y
Approximate Graphical Solution to a Pair of Equations
Since any point on a curve satisfies the equation of that curve, the coordinates of
the points of intersection of two curves will satisfy the equations of both curves. Thus
we merely have to plot the two curves and find their point or points of intersection;
these will be the solution to the pair of equations.
3
x−1
1
◆◆◆
Example 7: Graphically find the approximate solution to the pair of linear equations
3x y 1
xy3
(1)
(2)
(1, 2)
2
0
y=
y=3
Section 10–1
1
2
3
−
x
3
Solution: We plot the lines as in Chapter 5 and as shown in Fig. 10–3. Note that the lines FIGURE 10–3. Graphical
solution to a pair of equations.
◆◆◆
intersect at the point (1, 2). Our solution is then x 1, y 2.
This is also a good way to get
A linear equation in two unknowns plots as a straight line. Thus two linear equations an approximate solution to a
plot as two straight lines. If the lines intersect in a single point, the coordinates of that point pair of nonlinear equations.
will satisfy both equations, and hence that point is a solution to the set of equations. If the
lines are parallel, there is no point whose coordinates satisfy both equations, and the set
of equations is said to be inconsistent. If the two lines coincide at every point, the coordinates of any point on one line will satisfy both equations. Such a set of equations is called
dependent.
Solving a Pair of Linear Equations by the Addition-Subtraction Method
The method of addition-subtraction, and the method of substitution that follows, both have the
object of eliminating one of the unknowns.
In the addition-subtraction method, we eliminate one of the unknowns by first (if necessary)
multiplying each equation by such numbers that will make the coefficients of one unknown in
both equations equal in absolute value. The two equations are then added or subtracted so as to
eliminate that variable.
Let us first use this method for the same system that we solved graphically in Example 7.
◆◆◆
Example 8: Solve by the addition-subtraction method:
3x y 1
xy3
(1)
(2)
Solution: Simply adding the two equations causes y to drop out.
Add :
3x y 1
xy3
4x
4
x1
Substituting into the second given equation yields
1y3
y2
Our solution is then x 1, y 2, as found graphically in Example 7.
◆◆◆
In the next example we must multiply one equation by a constant before adding.
◆◆◆
Example 9: Solve by the addition-subtraction method:
2x 3y 4
x y 3
(1)
(2)
x
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Chapter 10
◆
Systems of Linear Equations
Solution: Multiply the second equation by 3.
2x 3y 4
3x 3y 9
5x
5
Add:
We have thus reduced our two original equations to a single equation in one unknown. Solving
for x gives
x1
Substituting into the second original equation, we have
1y3
y2
So the solution is x 1, y 2.
Check: Substitute into the first original equation.
?
2(1) 3(2) 4
2 6 4 (checks)
Also substitute into the second original equation.
123
(checks)
◆◆◆
Often it is necessary to multiply both given equations by suitable factors, as shown in the
following example.
◆◆◆
Example 10: Solve by addition or subtraction:
5x 3y 19
7x 4y 2
(1)
(2)
Solution: Multiply the first equation by 4 and the second by 3.
20x 12y 76
21x 12y 6
41x
82
x 2
Add:
Substituting x 2 into the first equation gives
5(2) 3y 19
3y 9
y 3
So the solution is x 2, y 3.
These values check when substituted into each of the original equations (work not
shown). Notice that we could have eliminated the x terms by multiplying the first equation
by 7 and the second by 5, and adding. The results, of course, would have been the same:
◆◆◆
x 2, y 3.
The coefficients in the preceding examples were integers, but in applications they will
usually be approximate numbers. If so, we must retain the proper number of significant digits,
as in the following example.
◆◆◆
Example 11: Solve for x and y:
2.64x 8.47y 3.72
1.93x 2.61y 8.25
(1)
(2)
Section 10–1
◆
269
Systems of Two Linear Equations
Solution: Let us eliminate y. We multiply the first equation by 2.61 and the second equation by
8.47. We should carry at least two decimal places in our calculation, and round our answer to
three digits at the end.
6.89x 22.11y 9.71
16.35x 22.11y 69.88
Add:
9.46x
60.17
x
6.36
Substituting into the first equation yields
2.64(6.36) 8.47y 3.72
8.47y 3.72 16.79 13.07
y 1.54
◆◆◆
We are less likely to make a
mistake adding rather than
subtracting, so it is safer to
multiply by a negative number
and add the resulting equations,
as we do here.
Another approach is to divide
each equation by the coefficient
of its x term, thus making each
x coefficient equal to 1. Then
subtract one equation from the
other.
Substitution Method
To use the substitution method to solve a pair of linear equations, first solve either original equation for one unknown in terms of the other unknown. Then substitute this expression into the
other equation, thereby eliminating one unknown.
◆◆◆
Example 12: Solve by substitution:
7x 9y 1
y 5x 17
(1)
(2)
Solution: We substitute (5x 17) for y in the first equation and get
If, as in this example, one or
both of the given equations are
already in explicit form, then the
substitution method is probably
the easiest. Otherwise, many
students find the additionsubtraction method easier.
7x 9(5x 17) 1
7x 45x 153 1
38x 152
x4
Substituting x 4 into Equation (2) gives
y 5(4) 17
y3
So our solution is x 4, y 3.
◆◆◆
Systems Having No Solution
Certain systems of equations have no unique solution. If you try to solve either of these types,
both variables will vanish.
◆◆◆
Example 13: Solve the system
2x 3y 5
6x 9y 2
(1)
(2)
Solution: Multiply the first equation by 3.
Subtract:
6x 9y 15
6x 9y 2
0 13 (no solution)
◆◆◆
If both variables vanish and an inequality results, as in Example 13, the system is called
inconsistent. The equations would plot as two parallel lines. There is no point of intersection
and hence no solution.
If both variables vanish and an equality results (such as 4 4), the system is called dependent. The two equations would plot as a single line, indicating that there are infinitely many
solutions.
It does not matter much whether
a system is inconsistent or
dependent; in either case we
get no useful solution. But the
practical problems that we solve
here will always have numerical
solutions, so if your variables
vanish, go back and check your
work.
270
Chapter 10
◆
Systems of Linear Equations
A Computer Technique: The Gauss-Seidel Method
The Gauss-Seidel method is a simple computer technique for solving systems of equations.
Suppose that we wish to solve the set of equations
x 2y 2 0
3x 2y 6 0
(1)
(2)
We first solve the first equation for x in terms of y, and the second equation for y in terms of x.
This type of repetitive process is
called iteration, or the method of
successive approximations.
x 2y 2
(3)
3x 6
y (4)
2
We then guess at the value of y, substitute this value into Equation (3), and obtain a value for x.
This x is substituted into (4), and a value of y is obtained. These values for x and y will not be
the correct values; they are only our first approximation to the true values.
The latest value for y is then put into (3), producing a new x; which is then put into (4),
producing a new y; which is then put into (3); and so on. We repeat the computation until the
values no longer change (we say that they converge on the true values). If the values of x and
y get very large (diverge instead of converge), we solve the first equation for y and the second
equation for x, and the computation will converge.
Example 14: Calculate x and y for the equations above using the Gauss-Seidel method.
Take y 0 for the first guess.
◆◆◆
Solution: From Equation (3) we get
x 2(0) 2 2
Substituting x 2 into (4) yields
3(2) 6
y 6
2
With y equal to 6, we then use (3) to find a new x, and so on. We get the following values:
x
y
2
14
50
158
482
1 454
4 370
13 118
39 362
·
·
·
6
24
78
240
726
2 184
6 558
19 680
59 046
·
·
·
Note that the values are diverging. We try again, this time solving Equation (1) for y,
x2
y 2
and solving (2) for x,
2y 6
x 3
Section 10–1
◆
271
Systems of Two Linear Equations
Repeating the computation, starting with x 0, gives the following values:
x
y
2.666 667
3.555 556
3.851 852
3.950 617
3.983 539
3.994 513
3.998 171
3.999 39
3.999 797
3.999 932
3.999 977
3.999 992
3.999 998
3.999 999
4
4
·
·
·
1
2.333 334
2.777 778
2.925 926
2.975 309
2.991 769
2.997 256
2.999 086
2.999 695
2.999 899
2.999 966
2.999 989
2.999 996
2.999 999
3
3
·
·
·
which converge on x 4 and y 3.
◆◆◆
How do we know what value to take for our first guess? Choose whatever values seem
“reasonable.” If you have no idea, then choose any values, such as the zeros in this example.
Actually, it does not matter much what values we start with. If the computation is going to converge, it will converge regardless of the initial values chosen. However, the closer your guess,
the faster the convergence.
CASE STUDY—ELECTRICAL CURRENTS
As an electronics technologist, you are analyzing a robotic servo system. After some initial calculations
and analysis, you have reduced the system to the equivalent circuit shown below.
1.5 kΩ
6.8 kΩ
–
8V
+
4.8 kΩ
I1
I2
2.2 kΩ
+
12 V
–
1.8 kΩ
I3
We can write an equation for each loop based on the voltage drops and sources. Electronics technology students may be familiar with these equations:
Loop 1: 6.8(I1 I2) 1.5I1 8
Loop 2: 2.2(I2 I3) 4.8I2 6.8(I2 I1) 12
Loop 3: 2.2(I3 I2) 1.8I3 12
Now, the “k” in the values near the resistors (the squiggly lines) stands for kilo (or thousand), so
2.2k would be 2200. However, to keep this straightforward, the resistance is divided by 1000, so the
answers will be the currents in milliamperes. This is not a concern, but it helps this to makes sense for
both electronics and non-electronics students. To determine the currents, remove the parentheses and
collect like terms. Now you will have three equations with three unknowns. Solve for I1, I2, and I3. This
chapter will show you how.
272
Chapter 10
◆
Systems of Linear Equations
Exercise 1
◆
Systems of Two Linear Equations
Graphical Solution
Graphically find the approximate solution to each system of equations.
1. 2x y 5
x 3y 5
3. x 2y 3
3x y 5
5. 2x 5y 4
5x 2y 3
x 2y 7
5x y 9
4. 4x y 8
2x y 7
6. x 2y 2 0
3x 6y 2 0
2.
Algebraic Solution
Solve each system of equations by addition-subtraction, or by substitution.
No applications are given here;
they are located in Exercise 3.
7. 2x y 11
3x y 4
9. 4x 2y 3
4x y 6
11. 3x 2y 15
5x 6y 3
13. x 5y 11
3x 2y 7
15. x 11 4y
5x 2y 11
17. 7x 4y 81
5x 3y 57
19. 3x 2y 1
2x y 10
21. y 9 3x
x 8 2y
23. 29.1x 47.6y 42.8
11.5x 72.7y 25.8
25. 4n 18 3m
m 8 2n
27. 3w 13 5z
4w 7z 17 0
29. 3.62x 11.7 4.73y
4.95x 7.15y 12.8 0
31. 4.17w 14.7 3.72v
v 8.11 2.73w
8.
10.
12.
14.
16.
18.
20.
22.
24.
26.
28.
30.
32.
xy7
3x y 5
2x 3y 5
3x 3y 10
7x 6y 20
2x 5y 9
4x 5y 34
2x 3y 22
2x 3y 3
4x 5y 39
3x 4y 85
5x 4y 107
5x 2y 3
2x 3y 5
y 2x 3
x 19 3y
4.92x 8.27y 2.58
6.93x 2.84y 8.36
5p 4q 14 0
17p 31 3q
3u 5 2v
5v 2u 16
3.03a 5.16 2.11b
5.63b 2.26a 18.8
5.66p 4.17q 16.9 0
13.7p 32.2 3.61q
Computer
33. Use a spreadsheet for the Gauss-Seidel method described in Sec. 10–1. Use it to solve any
of the sets of equations in this exercise set.
10–2
Other Systems of Equations
Systems with Fractional Coefficients
When one or more of the equations in our system have fractional coefficients, simply multiply
the entire equation by the LCD, and proceed as before.