Question 70032: I need help with the equation.
The sum of three numbers is 47. Three times the smallest number is nine more than the largest number. Two times the middle number is one more than the sum of the other two numbers. Find the three numbers.
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let the three numbers be A, B, and C with A being the smallest and C being the largest. From the problem description, you can write:
1) A+B+C = 47 The sum of three numbers is 47.
2) 3A = C+9 Three times the smallest is 9 more than the largest.
3) 2B = A+C+1 Two times the middle number is 1 more than the sum of the other two.
So, now you have three equations with three unknowns.
Rewrite equation 2) as:
2a) C = 3A-9 and substitute into equation 3) and solve for B.
3a) 2B = A+(3A-9)+1 Simplify.
2B = 4A-8 Divide both sides by 2.
3b) B = 2A-4 Now substitute this B and the C from equation 2a) into equation 1) and solve for A.
A+(2A-4)+(3A-9) = 47 Simplify and solve for A.
6A-13 = 47 Add 13 to both sides.
6A = 60 Divide both sides by 6.
A = 10 Substitute this value of A into equation 3b) and solve for B and then into equation 2a) and solve for C.
B = 2(10)-4
B = 20-4
B = 16
C = 3(10)-9
C = 30-9
C = 21
The three numbers are:
10, 16, and 21
Check:
A+B+C = 10+16+21 = 47
3A = C+9
3(10) = 21+9
30 = 30
2B = A+C+1
2(16) = 10+21+1
32 = 32
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