
David W. answered 03/20/16
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Algebra provides a way to put complicated descriptions into equations using variables and operations (like multiply, add, etc.). To “solve” a system of equations means to find the value(s) of the variables that make all the equations true at the same time.
The units of the equations are important:
( (2 legs)/(1 chicken) ) * (number of chickens) = number of legs for chickens
( ( 4 legs)/(1 doc) ) * (number of dogs) = number of dog legs
If we made a table, we could be here for a long time:
Number of chickens Number of dogs Total number of legs
------------------------- --------------------- ---------------------------
0 38 152
1 37 150
2 36 148
3 35 146
. . .
30 8 92
. . .
38 0 76
Let:
C= number of chickens
D = number of dogs
Translate:
“38 animals on a farm, all chickens and dogs” means C + D = 38 [eq1]
“There are 92 legs in all” means
(number of chicken legs) + (number of dog legs) = 92
2C + 4D = 92 [eq2]
“How many dogs are on the farm?” means report D
The math:
We may either substitute or eliminate to solve for one variable. Let’s eliminate C and solve for D (which is what the question asks us to find):
2C + 2D = 76 [2*eq1]
2C + 4D = 92 [eq2 again]
----------------------------- [elimination; subtract equations]
-2D = -16
D = 8 [divide both sides by (-2)]
There are 8 dogs on the farm.
[note that this agrees with the table above]
The problem doesn’t asks for C, but to check our answer:
Is 30 + 8 = 38 ?
38 = 38 ?yes
Is 2(30) + 4(8) = 92 ?
60 + 32 = 92 ?
92 = 92 ?yes