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The sum of three numbers is 11 . The sum of twice the first​ number, 5 times the second​ number, and 6 times the third number is 44. The difference between 4 times the first number and the second number is 10. Find the three numbers.

Sagot :

Answer:

4, 6, 1

Step-by-step explanation:

We can solve this problem using a system of equations:

1) a + b + c = 11

2) 2a + 5b + 6c = 44

3) 4a - b = 10

First, we can substitute the value of b from equation #3 into equation #1:

b = 4a - 10

a + (4a - 10) + c = 11

5a - 10 + c = 11

5a + c = 21

c = 21 - 5a

Now, we can plug the values of b and c into equation #2, as b and c are represented in terms of a:

2a + 5(4a - 10) + 6(21 - 5a) = 44

2a + 20a - 50 + 126 - 30a = 44

-8a + 76 = 44

-8a = -32

a = 4

b = 4a - 10 = 4(4) - 10 = 6

c = 21 - 5a = 21 - 5(4) = 1

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