Three softball teams ordered equipment from the same catalog. The first team spent $285 on 5 shirts, 4 caps, and 8 bats. The second team spent $210 on 12 shirts and 6 caps. The third team spent $250 for 7 shirts, 10 caps, and 3 bats. What were the catalog prices for shirts, caps, and bats?
I have to use matrices for this one
so, your equations are
5s+4c+8b = 285
12s+6c = 210
7s+10c+3b = 250
Use your favorite method of solution, and see whether you arrive at this solution:
http://www.wolframalpha.com/input/?i=%7B%7B5,4,8%7D,%7B12,6,0%7D,%7B7,10,3%7D%7D*%7B%7Bs%7D,%7Bc%7D,%7Bb%7D%7D+%3D+%7B%7B285%7D,%7B210%7D,%7B250%7D%7D
To find the catalog prices for shirts, caps, and bats, we need to solve the system of equations. Let's assume the price of a shirt is S, the price of a cap is C, and the price of a bat is B.
From the first team's purchase:
5S + 4C + 8B = $285
From the second team's purchase:
12S + 6C = $210
From the third team's purchase:
7S + 10C + 3B = $250
Let's solve this system of equations step-by-step using the method of substitution:
From the second team's purchase, we can simplify the equation to:
2S + C = $35 (Dividing the equation by 6)
We can solve this equation for C:
C = $35 - 2S
Now, substitute this value of C in the first and third equations:
From the first team's purchase:
5S + 4($35 - 2S) + 8B = $285
5S + $140 - 8S + 8B = $285
-3S + 8B = $145 (Simplifying the equation)
From the third team's purchase:
7S + 10($35 - 2S) + 3B = $250
7S + $350 - 20S + 3B = $250
-13S + 3B = -$100 (Simplifying the equation)
Now, solve these two equations simultaneously:
Multiply the first equation by 13:
-39S + 104B = $1885
Multiply the second equation by 8:
-104S + 24B = -$800
Add these two equations:
-39S + 104B + (-104S + 24B) = $1885 + (-$800)
-39S - 104S + 104B + 24B = $1885 - $800
-143S + 128B = $1085
Now, we have the system of equations:
-143S + 128B = $1085 (Equation 1)
-13S + 3B = -$100 (Equation 2)
To eliminate S, multiply equation 2 by 11:
-143S + 33B = -$1100 (New Equation 2)
Now, add the new equation 2 to equation 1:
-143S + 128B + (-143S + 33B) = $1085 + (-$1100)
-143S - 143S + 128B + 33B = -$15
Combine like terms:
161B = -$15
Divide by 161 on both sides:
B = -$15/161
Therefore, the catalog price for a bat is approximately -$0.09.
Now, substitute the value of B in equation 2 to find the catalog price for a shirt:
-13S + 3(-$0.09) = -$100
-13S - $0.27 = -$100
-13S = -$100 + $0.27
-13S = -$99.73
Divide by -13 on both sides:
S = -$99.73 / -13
S ≈ $7.67
So, the catalog price for a shirt is approximately $7.67.
Finally, substitute the values of S and B in equation 2 to find the catalog price for a cap:
-13($7.67) + 3(-$0.09) = -$100
-99.71 - 0.27 = -$100
-$100 ≈ -$100
Therefore, the catalog price for a cap is approximate to $0.
To summarize:
The catalog prices for shirts, caps, and bats are approximately:
Shirt: $7.67
Cap: $0
Bat: -$0.09
To find the catalog prices for shirts, caps, and bats, we can set up a system of equations using the information given.
Let's denote the catalog price for a shirt as "s", the catalog price for a cap as "c", and the catalog price for a bat as "b".
For the first team, they spent $285 on 5 shirts, 4 caps, and 8 bats. We can set up the equation:
5s + 4c + 8b = 285
For the second team, they spent $210 on 12 shirts and 6 caps. We can set up the equation:
12s + 6c = 210
For the third team, they spent $250 on 7 shirts, 10 caps, and 3 bats. We can set up the equation:
7s + 10c + 3b = 250
Now we have a system of three equations:
5s + 4c + 8b = 285 (Equation 1)
12s + 6c = 210 (Equation 2)
7s + 10c + 3b = 250 (Equation 3)
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
From Equation 2, we can isolate "s" in terms of "c":
12s + 6c = 210
12s = 210 - 6c
s = (210 - 6c) / 12
s = (35 - c)/2 (Equation 4)
Substitute Equation 4 into Equations 1 and 3:
5((35 - c)/2) + 4c + 8b = 285 (Equation 5)
7((35 - c)/2) + 10c + 3b = 250 (Equation 6)
Now we have a system of two equations:
5((35 - c)/2) + 4c + 8b = 285 (Equation 5)
7((35 - c)/2) + 10c + 3b = 250 (Equation 6)
Simplify these equations:
(175 - 5c)/2 + 4c + 8b = 285 (Equation 5 simplified)
(245 - 7c)/2 + 10c + 3b = 250 (Equation 6 simplified)
To eliminate the fractions, multiply each equation by 2:
175 - 5c + 8c + 16b = 570 (Equation 5 multiplied by 2)
245 - 7c + 20c + 6b = 500 (Equation 6 multiplied by 2)
Now simplify these equations:
175 + 3c + 16b = 570 (Equation 5 simplified further)
245 + 13c + 6b = 500 (Equation 6 simplified further)
Rearrange the equations:
3c + 16b = 570 - 175
3c + 16b = 395 (Equation 7)
13c + 6b = 500 - 245
13c + 6b = 255 (Equation 8)
Let's multiply Equation 7 by 6 and Equation 8 by 16 to eliminate "b":
18c + 96b = 2370 (Equation 7 multiplied by 6)
208c + 96b = 4080 (Equation 8 multiplied by 16)
Subtract Equation 7 multiplied by 208 from Equation 8 multiplied by 18:
208c + 96b - (18c + 96b) = 4080 - 2370
190c = 1710
Divide both sides by 190:
c = 9
Substitute the value of "c" into Equation 7:
3c + 16b = 395
3(9) + 16b = 395
27 + 16b = 395
16b = 395 - 27
16b = 368
Divide both sides by 16:
b = 23
Now substitute the values of "b" and "c" into Equation 5 to find "s":
175 + 3c + 16b = 570
175 + 3(9) + 16(23) = 570
175 + 27 + 368 = 570
570 = 570
So the catalog prices for shirts, caps, and bats are $9, $23, and $35, respectively.