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# word problems and turning them into algebraic equation to slove

Yet another farm problem to set up and solve.....
Tina looks out into the barnyard and sees pigs, chickens and ducks. She counts 38 heads and 86 feet. Knowing that that the sum of the number of pigs and chickens is 2 more than the number of ducks. (all animals are normal-each has 1 head, etc)  set up the using algebra equations and solve.

Let p = the number of pigs, c = number of chickens, and d = number of ducks.

p+c = d+2 ......(1)
P+c+d = 38 ......(2), balance by the total number of heads
4p+2c+2d = 86, balance by the total number of legs
/2, 2p+c+d = 43 ......(3)
(3)-(2): p = 5
(2)-(1): d = 36-d => d = 18
c = 38-p-d = 38-18-5 = 15

Answer: There are 5 pigs, 15 chickens and 18 ducks.

Hi Nancy;
Tina looks out into the barnyard and sees pigs, chickens and ducks. She counts 38 heads and 86 feet. Knowing that that the sum of the number of pigs and chickens is 2 more than the number of ducks. (all animals are normal-each has 1 head, etc) set up the using algebra equations and solve.

FIRST EQUATION...P+C+D=38

4P+2C+2D=86
SECOND EQUATION...2P+C+D=43

P+C=D+2
THIRD EQUATION...P+C-D=2
------------
SECOND EQUATION...2P+C+D=43
FIRST EQUATION...P+C+D=38
Let's subtract the first equation from the second...
P=5

SECOND EQUATION...2(5)+C+D=43
SECOND EQUATION...10+C+D=43
THIRD EQUATION...P+C-D=2
THIRD EQUATION...5+C-D=2
Let's add the second and third equations...
15+2C+0=45
Let's subtract 15 from both sides..
2C=30
Let's divide both sides by 2...
C=15

FIRST EQUATION...P+C+D=38
5+15+D=38
20+D=38
D=18

SECOND EQUATION...2P+C+D=43
2(5)+15+18=43
10+15+18=43
25+18=43
43=43

THIRD EQUATION..P+C-D=2
5+15-18=2
20-18=2
2=2