Tim and Judy mix two kinds of feed for pedigreed dogs. They wish to make 29 pounds of feed worth $0.47 per pound by mixing one kind worth $0.44 per pound with another worth $0.55 per pound.

How many pounds of the cheaper kind should they use in the mix?

-please help! so confused!

you have 29 total pounds. You want soo many pounds of the 44 cent and so many pounds of the 55 cent feed. So you can make the equation .44x + .55(29-x) = 29 x .47

Solve for x, that will tell you how many pounds of the cheaper kind they should use

To solve this problem, we can set up a system of equations.

Let:
x = pounds of the cheaper kind (worth $0.44 per pound)
y = pounds of the more expensive kind (worth $0.55 per pound)

We know that the total weight of the mixture is 29 pounds, so we have the equation:
x + y = 29

We also know that the value of the mixture is $0.47 per pound. Since the value is determined by the price per pound, we can set up another equation based on the values:
0.44x + 0.55y = 0.47(29)

To solve this system of equations, we can use substitution or elimination method. Let's solve it using the elimination method.

Multiply the first equation by 0.44 to match the coefficient of 'x' with that in the second equation:
0.44(x + y) = 0.44(29)
0.44x + 0.44y = 12.76

Now, we can subtract this equation from the second equation:
0.44x + 0.55y - (0.44x + 0.44y) = 12.76 - 12.76
0.55y - 0.44y = 0
0.11y = 0
y = 0

Substituting the value of y into the first equation:
x + 0 = 29
x = 29

According to the solution, they should use 29 pounds of the cheaper kind (worth $0.44 per pound) in the mix.