Mr. McCoy spent 35 cents at the hardware store buying nails and screws. Nails cost 4 cents, and screws cost 3 cents. If Mr. McCoy bought exactly 10 items, how many of each did he buy?
4 n + 3 s = 35
n + s = 10 so s = (10-n)
4 n + 3 (10-n) = 35
n + 30 = 35
n = 5
so s =5
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check
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4*5 +3*5 = 20 + 15 = 35 whew we did it
Are you a student?
Of course, also a retired professor.
To solve this problem, we can use a system of equations. Let's assume Mr. McCoy bought x nails and y screws.
We know that the total number of items he bought is 10, so we can write the equation: x + y = 10.
We also know that nails cost 4 cents and screws cost 3 cents, and he spent a total of 35 cents. So we can write the equation: 4x + 3y = 35.
Now, we have a system of equations:
Equation 1: x + y = 10
Equation 2: 4x + 3y = 35
To solve this system, we can use the method of substitution or elimination. Here, let's use the method of substitution.
First, solve Equation 1 for x in terms of y: x = 10 - y.
Substitute this value of x in Equation 2: 4(10 - y) + 3y = 35.
Simplify the equation: 40 - 4y + 3y = 35.
Combine like terms: -y = 35 - 40.
Simplify further: -y = -5.
Multiply both sides by -1 to get y: y = 5.
Now substitute the value of y back into Equation 1 to find x: x + 5 = 10.
Subtract 5 from both sides: x = 10 - 5.
Simplify: x = 5.
So, Mr. McCoy bought 5 nails and 5 screws.