Question 166920: Algebra
You are working with three different numbers. When the first number is added to twice the other two numbers, the result is 64 (x + 2y + 2z= 64). When the second number is added to twice the other two numbers the result is 62. Finally, when the third number is added to twice the other two numbers, the result is 59. what are the three numbers??
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! You are working with three different numbers.
When the first number is added to twice the other two numbers, the result is 64
x + 2y + 2z = 64
;
When the second number is added to twice the other two numbers the result is 62.
2x + y + 2z = 62
:
Finally, when the third number is added to twice the other two numbers, the
result is 59.
2x + 2y + z = 59
:
Using the 1st two equations
x + 2y + 2z = 64
2x + y + 2z = 62
-------------------subtraction eliminates z
-x + y = 2
y = x+2
:
Multiply the 3rd equation by 2, and subtract the 1st equation
4x + 4y + 2z = 118
x + 2y + 2z = 64
-------------------subtraction eliminates z again
3x + 2y = 54
Substitute(x+2) for y
3x + 2(x+2) = 54
3x + 2x + 4 = 54
5x = 54 - 4
5x = 50
x = 10 is the 1st number
then
y = 10+2
y = 12 is the 2nd number
:
Find z using the 3rd equation
2(10) + 2(12) + z = 59
20 + 24 + z = 59
z = 59 - 44
z = 15 is the 3rd number
:
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Check solution in the 2nd equation
2(10) + 12 + 2(15) = 62
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what are the three numbers?? 10, 12, 15
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