# Algebra

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Tara’s website, Garden Edibles, specializes
in the sale of herbs and flowers for colorful meals and garnishes. Tara
sells packets of nasturtium seeds for \$0.95 each and packets of Johnny-jumpup
seeds for \$1.43 each. She decides to offer a 16-packet spring-garden combination,
combining packets of both types of seeds at \$1.10 per packet. How
many packets of each type of seed should be put in her garden mix?
1. Familiarize. To familiarize ourselves with the problem situation, we
make a guess and do some calculations. The total number of packets of
seed is 16. Let’s try 12 packets of nasturtiums and 4 packets of Johnnyjump-
ups.
The sum of the number of packets is , or 16.
The value of these seed packets is found by multiplying the cost per
packet by the number of packets and adding:
, or \$17.12.
The desired cost is \$1.10 per packet. If we multiply \$1.10 by 16, we get
, or \$17.60. This does not agree with \$17.12, but these calculations
give us a basis for understanding how to translate.
We let the number of packets of nasturtium seeds and
the number of packets of Johnny-jump-up seeds. Next, we organize the
information in a table, as follows.
a  b 
16\$1.10
\$0.9512  \$1.434
12  4
\$ 0.95
\$ 1.10
\$ 1.43
575
8.4 Solving Applied Problems:
Two Equations
a  b  16
 17.60
0.95a  1.43b
NASTURTIUM
JOHNNYJUMP-
UP SPRING
Number
of Packets a b 16
Price
per Packet \$0.95 \$1.43 \$1.10
Value
of Packets 0.95a 1.43b or 17.60
16  1.10,
ISBN:0-536-47742-6
2. Translate. The total number of packets is 16, so we have one equation:
.
The value of the nasturtium seeds is 0.95a and the value of the Johnnyjump-
up seeds is 1.43b. These amounts are in dollars. Since the total
value is to be , or \$17.60, we have
.
We can multiply by 100 on both sides of this equation in order to clear the
decimals. Thus we have the translation, a system of equations:
, (1)
. (2)
3. Solve. We decide to use substitution, although elimination could be
used as we did in Example 1. When equation (1) is solved for b, we
get . Substituting for b in equation (2) and solving
gives us
Substituting
Using the distributive law
Subtracting 2288 and collecting
like terms
.
We have . Substituting this value in the equation , we
obtain , or 5.
4. Check. We check in a manner similar to our guess in the Familiarize
step. The total number of packets is , or 16. The value of the packet
mixture is
, or \$17.60.
Thus the numbers of packets check.
5. State. The spring garden mixture can be made by combining 11 packets
of nasturtium seeds with 5 packets of Johnny-jump-up seeds.
Do Exercise 2.
EXAMPLE 3 Student Loans. Jed’s student loans totaled \$16,200. Part was
a Perkins loan made at 5% interest and the rest was a Stafford loan made at 4%
interest. After one year, Jed’s loans accumulated \$715 in interest. What was the
amount of each loan?
1. Familiarize. Listing the given information in a table will help. The
columns in the table come from the formula for simple interest: .
We let the number of dollars in the Perkins loan and the number
of dollars in the Stafford loan.
x  y 
I  Prt
\$0.9511  \$1.435
11  5
b  16  11
a  11 b  16  a
a  11
48a  528
95a  2288  143a  1760
95a  14316  a  1760
b  16  a 16  a
95a  143b  1760
a  b  16
0.95a  1.43b  17.60
16\$1.10
a  b  16
2. Blending Coffees. The Coffee
Counter charges \$9.00 per
pound for Kenyan French Roast
coffee and \$8.00 per pound for
Sumatran coffee. How much of
each type should be used to
make a 20-lb blend that sells
for \$8.40 per pound?
Sumatran
Coffee
\$8.00 lb
Kenyan French
Roast
\$9.00 lb
HOUSE
BLEND
\$8.40 lb
576
CHAPTER 8: Systems of Equations
x  y  16,200
0.05x  0.04y  715
PERKINS
LOAN
STAFFORD
LOAN TOTAL
Principal x y \$16,200
Rate of
Interest 5% 4%
Time 1 yr 1 yr
Interest 0.05x 0.04y \$715
ISBN:0-536-47742-6
2. Translate. The total of the amounts of the loans is found in the first row
of the table. This gives us one equation:
.
Look at the last row of the table. The interest totals \$715. This gives us a
second equation:
, or .
After we multiply on both sides to clear the decimals, we have
.
3. Solve. Using either elimination or substitution, we solve the resulting
system:
,
.
We find that and .
4. Check. The sum is , or \$16,200. The interest from \$6700
at 5% for one year is , or \$335. The interest from \$9500 at 4% for
one year is , or \$380. The total interest is , or \$715.
The numbers check in the problem.
5. State. The Perkins loan was for \$6700 and the Stafford loan was for
\$9500.
Do Exercise 3.
EXAMPLE 4 Mixing Fertilizers. Yardbird Gardening carries two kinds of
fertilizer containing nitrogen and water. “Gently Green” is 5% nitrogen and
“Sun Saver” is 15% nitrogen. Yardbird Gardening needs to combine the two
types of solution to make 90 L of a solution that is 12% nitrogen. How much
of each brand should be used?
1. Familiarize. We first make a drawing and a guess to become familiar
with the problem.
We choose two numbers that total 90 L—say, 40 L of Gently Green
and 50 L of Sun Saver—for the amounts of each fertilizer. Will the
resulting mixture have the correct percentage of nitrogen? To find out,
we multiply as follows:
and .
Thus the total amount of nitrogen in the mixture is 2 L  7.5 L, or 9.5 L.
5%40 L  2 L of nitrogen 15%50 L  7.5 L of nitrogen
5
5% nitrogen
Gently
Green
Sun
Saver
g liters s liters 90 liters
1 15% nitrogen 12% nitrogen
4%\$9500 \$335  \$380
5%\$6700
\$6700  \$9500
x  6700 y  9500
5x  4y  71,500
x  y  16,200
5x  4y  71,500
5%x  4%y  715 0.05x  0.04y  715
x  y  16,200
3. Client Investments. Kaufman
Financial Corporation makes
investments for corporate
clients. It makes an investment
of \$3700 for one year at simple
interest, yielding \$297. Part of
the money is invested at 7%
and the rest at 9%. How much
was invested at each rate?
Do the Familiarize and
Translate steps by completing
the following table. Let the
number of dollars invested at
7% and the number of
dollars invested at 9%.
y 
x 
577
8.4 Solving Applied Problems:
Two Equations
FIRST
INVESTMENT
SECOND
INVESTMENT TOTAL
Principal, P x \$3700
Rate of
Interest, r 9%
Time, t 1 yr 1 yr
Interest, I 0.07x \$297
x     3700
0.07x   297
ISBN:0-536-47742-6
The final mixture of 90 L is supposed to be 12% nitrogen. Now
.
Since 9.5 L and 10.8 L are not the same, our guess is incorrect. But these
calculations help us to become familiar with the problem and to make
the translation.
We let the number of liters of Gently Green and the number
of liters of Sun Saver.
The information can be organized in a table, as follows.
2. Translate. If we add g and s in the first row, we get 90, and this gives us
one equation:
.
If we add the amounts of nitrogen listed in the third row, we get 10.8, and
this gives us another equation:
, or .
After clearing the decimals, we have the following system:
(1)
. (2)
3. Solve. We solve the system using elimination. We multiply equation (1)
by 5 and add the result to equation (2):
Multiplying equation (1) by 5
; Dividing by 10
Substituting in equation (1) of the system
g  27. Solving for g
g  63  90
s  63
10s  630
5g  15s  1080
5g  5s  450
5g  15s  1080
g  s  90,
5%g  15%s  10.8 0.05g  0.15s  10.8
g  s  90
g  s 
12%90 L  10.8 L
578
CHAPTER 8: Systems of Equations
g  s  90
0.05g  0.15s  10.8
GENTLY
GREEN
SUN
SAVER MIXTURE
Number
of Liters g s 90
Percent
of Nitrogen 5% 15% 12%
Amount
of Nitrogen 0.05g 0.15s or 10.8 liters
0.12  90,
ISBN:0-536-47742-6
4. Check. Remember that g is the number of liters of Gently Green, with
5% nitrogen, and s is the number of liters of Sun Saver, with 15% nitrogen.
Total number of liters of mixture:
Amount of nitrogen:
Percentage of nitrogen in mixture:
The numbers check in the original problem.
5. State. Yardbird Gardening should mix 27 L of Gently Green and 63 L of
Sun Saver.
Do Exercise 4.
Motion Problems
When a problem deals with speed, distance, and time, we can expect to use
the following motion formula.
THE MOTION FORMULA
TIPS FOR SOLVING MOTION PROBLEMS
1. Draw a diagram using an arrow or arrows to represent distance
and the direction of each object in motion.
2. Organize the information in a table or chart.
3. Look for as many things as you can that are the same, so you can
write equations.
EXAMPLE 5 Auto Travel. Your brother leaves on a trip, forgetting his suitcase.
You know that he normally drives at a speed of 55 mph. You do not
discover the suitcase until 1 hr after he has left. If you follow him at a speed
of 65 mph, how long will it take you to catch up with him?
1. Familiarize. We first make a drawing. From the drawing, we see that
when you catch up with your brother, the distances from home are the
same. We let the distance, in miles. If we let the time, in hours, for
you to catch your brother, then the time traveled by your brother
at a slower speed.
55 mph
t + 1 hours d miles
65 mph
t hours d miles
Cars
meet
here
t  1 
d  t 
d  rt
Distance  Rate or speed  Time
10.8
90
 0.12  12%
5%27  15%63  1.35  9.45  10.8 L
g  s  27  63  90
4. Mixing Cleaning Solutions.
King’s Service Station uses two
kinds of cleaning solution
containing acid and water.
“Attack” is 2% acid and “Blast”
is 6% acid. They want to mix
the two to get 60 qt of a
solution that is 5% acid. How
many quarts of each should
they use?
Do the Familiarize and
Translate steps by completing
the following table. Let the
number of quarts of Attack
and the number of quarts
of Blast.
Review examples 2, 3, and 4 in section 8.4 of the text. How does the author determine what the first equation should be? What about the second equation? How are these examples similar? How are they different? Find a problem in the text that is similar to examples 2, 3, and 4. Post the problem for your classmates to solve.

Consider responding to your classmates by asking clarifying questions or by expanding a classmate’s response. Also, help students solve the problem you posted by providing feedback or hints if necessary. You may also want to provide an explanation for your solution after a sufficient number of students have replied.

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