Someone bought 7 ears of corn and 6 oranges for $3.66. At the same time a second person bought 10 ears of corn and 3 oranges for $3.39. Find the price of each.
Answer: oranges would be .78 cents and corn would be .41 cents each
You could have easily checked your answer by replacing them in the original question.
here is what I did:
7c + 6o = 366 , I will work in cents
10c + 3o = 339
double the 2nd:
20c + 6o = 678
7c + 6o = 366
subtract them:
13c = 312
c = 24
back into 1st:
7c + 6o = 366
7(24) + 6o = 366
168 + 6o = 366
6o = 198
o = 33
oranges cost 33 cents and corn costs 24 cents
check"
7(24) + 6(33) = 366 , check!
10(24) + 3(33) = 339, check!
do your previous questions the same way
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To find the price of each item, we can set up a system of equations based on the given information.
Let's assume the price of one ear of corn is "c" and the price of one orange is "o".
From the first person's purchase, we know that they bought 7 ears of corn and 6 oranges for $3.66. This can be expressed as the equation:
7c + 6o = 3.66
From the second person's purchase, we know that they bought 10 ears of corn and 3 oranges for $3.39. This can be expressed as the equation:
10c + 3o = 3.39
Now we have a system of two equations:
Equation 1: 7c + 6o = 3.66
Equation 2: 10c + 3o = 3.39
To solve this system of equations, we can use the substitution or elimination method. Let's use the elimination method:
Multiply Equation 1 by 10 and Equation 2 by 7 to get rid of the coefficients of "c":
70c + 60o = 36.6 (Equation 3)
70c + 21o = 23.73 (Equation 4)
Now, subtract equation 4 from equation 3:
(70c + 60o) - (70c + 21o) = 36.6 - 23.73
49o = 12.87
Divide both sides of the equation by 49:
o = 12.87 / 49
o ≈ 0.2627
Now substitute the value of "o" back into Equation 1:
7c + 6(0.2627) = 3.66
7c + 1.5762 = 3.66
7c = 3.66 - 1.5762
7c = 2.0838
c ≈ 2.0838 / 7
c ≈ 0.298
Therefore, the price of one ear of corn (c) is approximately $0.298 or approximately 0.41 cents each, and the price of one orange (o) is approximately $0.2627 or approximately 0.78 cents each.