“Systems of Equations”
Project
Serena Mohamed
Due Date: 3/8/13
Mrs. Castro/Math
Group D
Grade 7 AEP
Room 323
1.
Farmer Ben has only ducks and cows. He can’t remember how many of each he has.
Well he doesn’t need to remember because he knows he has 22 animals and that 22 is
also his age. He also knows that the animals have a total of 56 legs, because 56 is also
his father’s age. Assuming that each animal has all legs intact and no more, how
many of each animal does Farmer Ben have?
d=ducks
c=cows
1. d+c=22
2d+4c=56
2
=d+2c=28
3. 22-c=28-2c
+2c +2c
22+c=28
-22 -22
c=6
2. d=22-c
d=28-2c
4. d=22-6=16
d=28-2(6) =16
d=16
Farmer Ben has 6 cows and 16 ducks.
First, I wrote two equations: d+c=22 and 2d+4c=28. Next, I simplified 2d+4c=28
by dividing each side by 2. The simplified equation I got was d+2c=28. The third
step I did was I switched the equations a little bit to make it look like this: d=22-c
and d=28-2c. Then I set them equal to each other to make it look like this: 22c=28-2c. Finally I solved the equation and my answers were: c=6 and d=16.
2.
A group of 148 people is spending five days at a summer camp. The cook
ordered 12 pounds of food for each adult and 9 pounds of food for each child. A
total of 1,410 pounds of food was ordered. What is the total number of adults
and total number of children in the group?
a=adults
c=children
1. a+c=148
2. . a+c=148 x 12= 12a+12c=1776
12a+9c=1410
3. 12a+12c=1776
-12a+9c= 1410
3c=366
3
c=122
4. 148-122=26
1410-9(122) =312
312/12
=26
12(26) +9(122)
=1410
a=26
The total number of adults is 26 and the total number of children is 122
.
The first step I did was write the equation for the number of people and the
equation for the pounds of food. The next step I did was multiply a+c=148 by 12
in order to make one variable identical. Then I subtracted one equation
(12a+12c=1776) from another (12a+9c=1410). By doing that, I found the answer
to the number of children. So I substituted c with the answer and solved for a.
and the answers were: c=122 and a=26.
The owner of a movie theater was counting the money from one day’s
ticket sales. He knew that a total of 150 tickets were sold. Adult tickets
3.
cost $7.50 each and children’s ticket’s cost $4.75 each. If the total
receipts for the day were $891.25, how many of each kind of ticket were
sold?
a=adults
c=children
1. a+c=150
7.5a+4.75c=891.25
3. 7.5(150-c) +4.75c=891.25
7.5 x 150=1125-7.5c+4.75c
-7.5c
1125-2.75c=891.25
+4.75c
-2.75c
2. a=150-c
1125-891.25=2.75c
1125-891.25=233.75
233.75=2.75c
2.75 2.75
85=c
4. 150-85=65
a=65
The number of adult tickets sold was 65 and the number of children
tickets sold was 85.
First, I wrote the equation for the number of tickets which was a+c=150.
Next, I wrote the equation for the total receipts which was
7.5a+4.75c=891.25. Then I solved for the number of children and finally, I
solved for the number of adults and the answers were a=65 and c=85.