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Submitted by: Karin

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This is a very interesting problem! This is a problem that can be solved using a system of equations. In order to write a system of equations, you must have two variables, so ask yourself, what two things do I need to know? You need to know how many dogs and how many chickens. So, we're going to let x= the number of dogs and y = the number of chickens. Now we need to write two equations. We know information about two things: the number of legs and the number of heads. Let's write an equation about the number of legs. How many legs does a dog have? 4. How many legs does a chicken have? 1. 4x + 2y = 148 (4 legs times number of dogs & 2 legs times number of chickens) Now lets write an equation about the number of heads. How many heads does a dog have? 1. How many heads does a chicken have? 1. x + y = 60. (Since each only has one head, we do not need to write the coefficient, 1). Now you need to solve your the system of equations: 4x + 2y = 148 x + y = 60. I would use the substitution method. x + y = 60 can be rewritten as y = -x + 60 Substitute this into the first equation for y. 4x + 2(-x + 60) = 148 4x -2x + 120 = 148 2x + 120 = 148 2x +120 -120 = 148 - 120 2x = 28 2x/2 = 28/2 x = 14 Then substitute your value for x into the second equation and solve. 14 + y = 60 14 -14 + y = 60 - 14 y = 46 There were 14 dogs and 46 chickens. Hope this helps. Visit <a href="http://www.algebra-class.com/system-of-equations.html"> System of Equations </a>for more help.

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