Question 1033321: The sum of three numbers is 10. The sum of twice the first number, 3 times the second number, and 4 times the third number is 31. The difference between 3 times the first number and the second number is -4. Find the three numbers.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Three numbers, a, b, c; write an equation for each statement
:
Th sum of three numbers is 10
a + b + c = 10
The sum of twice the first number, 3 times the second number, and 4 times the third number is 31.
2a + 3b + 4c = 31
The difference between 3 times the first number and the second number is -4.
3a - b = -4
:
Multiply the 3rd equation by -1. Add all three equations. a is eliminated
a + b + c = 10
2a+ 3b+ 4c = 31
-3a + b + 0 = 4
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0 + 5b + 5c = 45
simplify, divide by 5
b + c = 9
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In the 1st equation, replace a + b with 9
a + 9 = 10
a = 1
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Use the 3rd equation to find b, replace a with 1
3(1) - b = -4
-b = -4 - 3
-b = -7
therefore
b = 7
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I'll let you find c, check your solutions in all three equations
Find the three numbers.
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