Download Math 121. Linear Systems of Equations (2 variables)

Math 121.
Linear Systems of Equations (2 variables)
1. (Graph a Rational Function from Chapter 3.5) Let f (x) =
Fall 2016
x2 − 4x + 4
.
x2 − 5x + 6
(a) Simplify f and find its domain.
(b) Find equations for the vertical asymptote(s) for the graph of f .
(c) Find the x- and y-intercepts of the graph of f .
(d) For each vertical asymptote found in part (b), determine the behavior of f just to the right and just to the
left of the vertical asymptote.
(e) Find all values of c for which there is a “hole” in the graph of f above x = c.
(f) Find all horizontal asymptote of f .
(g) Use the information above and plot additional points a necessary to graph f .
Solution: (a) The domain of f is {x : x 6= 3, x 6= 2} (see the simplified form of f below):
f (x) =
x−2
(x − 2)(x − 2)
=
(x − 3)(x − 2)
x−3
x 6= 3, x 6= 2
(b) The zeros in the denominator of the simplified rational function provide the locations of the vertical
asymptotes. Therefore, the graph of f has vertical asymptote x = 3, there will be a hole in the graph at
x = 2.
(c) The x-intercept is (2, 0) and the y-intercept is (0, 32 ).
(d) When x is close to, but larger than 3 we have
f (x) =
(x − 2)
1
≈
= large positive number.
(x − 3)
small positive number
When x is close to, but smaller than 3 we have
f (x) =
(x − 2)
1
=
= large negative number.
(x − 3)
small negative number
Thus the graph increases without bound as x approaches 3 from the right and decreases without bound as
x approaches 3 from the left.
(e) There would be a hole in the graph above x = 2 (since 2 is not in the domain of f , but there is no
vertical asymptote at x = 2).
(f) The degrees of the numerator and denominator are the same, so the horizontal asymptote is found by
setting y equal to the ratio of the leading coefficients, that is y = 1
(g) A graph of f along with its asymptotes is as follows
10 y
8
6
4
2
−10 −8 −6 −4 −2
−2
x
2
4
6
8
10
−4
−6
−8
−10
2. A metallurgist made two purchases. The first purchase, which cost $554, included 18 kilograms of an iron
alloy and 14 kilograms of a lead alloy. The second purchase, at the same prices, cost $937 and included 36
kilograms of the iron alloy and 19 kilograms of the lead alloy. Solve a system of linear equations to determine
the price per kilogram of each alloy.
Solution: Let x be the price per kilogram of the iron alloy, and let y be the price per kilogram of the lead
alloy. Then
18x + 14y = 554
and
36x + 19y = 937
Multiply the first equation by −2 and add it to the second to obtain
−36x − 28y
36x + 19y
−9y
= −1108
= 937
= −171
−171
= 19, and so 18x + (14)(19) = 554 which implies x = 16.
−9
Thus, the iron alloy is $16 per kilogram, and the lead alloy is $19 per kilogram.
Then y =
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3. Use the method of substitution to solve the following system of equations.
−2x + 5y = 13
x − 5y = −14
and
Solution: The second equation is easy to solve for x. Namely, x = −14 + 5y, and we substitute this into
the first equation to obtain
5y − 2(−14 + 5y)
=
13
⇒
5y + 28 − 10y
=
13
⇒
−5y
and thus y =
= −15
−15
= 3. Then x = −14 + 5(3) = 1. That is, the solution is (x, y) = (1, 3).
−5
4. Use the method of elimination to solve the following system of equations.
4x + 5y = 2
and
7x + 4y = 13
Solution: To eliminate y we can multiply the first equation by 4 and the second by −5 and then add
them together to obtain
+
16x + 20y
=
8
−35x − 20y
=
−65
−19x
= −57
Therefore, x = −57/ − 19 = 3. Then plugging x = 3 into the second equation yields
7(3) + 4y = 13
⇒
4y = 13 − 21
⇒
4y = −8
⇒
y = −8/4 = −2.
That is, the solution to the system is (x, y) = (3, −2).
5. Use the method of substitution to solve the following system of equations.
−3x + y = −8
7x − 2y = 20
and
Solution: The first equation is easy to solve for y. Namely, y = −8 + 3x, and we substitute this into the
second equation to obtain
and thus x =
7x − 2(−8 + 3x)
=
20
⇒
7x + 16 − 6x
=
20
⇒
1x
=
4
4
= 4. Then y = −8 + 3(4) = 4. That is, the solution is (x, y) = (4, 4).
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6. Solve each of the following systems of equations, or state that there is no solution. Then classify the system
as independent, dependent, or inconsistent.
(a) 2x − 7y = −5
and −4x + 14y = 10
(b) −2x + 14y = 6
and −x + 7y = 5
Solution: (a) Multiply the first equation by 2 and then add it to the second to obtain
4x − 14y
−4x + 14y
0
=
=
=
−10
10
0
This means 0x = 0, and so x = c where c is any real number, and then putting x = c in the first equation
2c − 7y = 5 ⇒
7y = 2c − 5
⇒
y=
2c − 5
7
The solution is c, 2c−5
where c is any real number. This system is dependent (if you were to graph both
7
equations, they would be the same line).
(b) Multiply the second equation by −2 and add it to the first to obtain
2x − 14y
=
−2x + 14y
=
−10
6
0
=
−4
It is never true that 0 = −4 and so this system is inconsistent, it has no solution (if your were to graph
both equations, they would be parallel nonintersecting lines).
7. Flying with the wind, a plane traveled a 1200 miles in 5 hours. Against the wind, the return trip covering
the same distance took 6 hours. Find the rate of the plane in calm air and find the rate of the wind.
Solution: Use the uniform motion formula d = rt. Let p be the rate of the plane, and w be the rate of
the wind. This leads to equations 5(p + w) = 1200 and 6(p − w) = 1200. These equations simplify to
p + w = 240 and p − w = 200. Adding these two equations together yield 2p = 440 and so p = 220. Then
w = 240 − 220 = 20. Thus the rate of the plane is 220 miles per hour and the rate of the wind is 20 miles
per hour.
8. A goldsmith has two gold alloys. The first alloy is 36% gold; the second alloy is 78% gold. Set-up and solve
a system of two linear equations to determine how many grams of each alloy should be mixed to produce 105
grams of an alloy that is 64% gold?
Solution: Let x be the number of grams of 36% gold, and let y be the number of grams of 78% gold.
Then
x + y = 105
and
0.36x + 0.78y = (0.64)(105)
Then x = 105 − y and substituting this into the second equation implies
0.36(105 − y) + 0.78y = 67.20
Then 0.42y = 67.20 − 37.80 and so y =
29.40
0.42
and so
37.80 + 0.42y = 67.20
= 70 and x = 105 − 70 = 35.
That is, the goldsmith should mix 35 grams of 36% gold with 70 grams of 78% gold.
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