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| Math Central | Quandaries & Queries |
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Question from Shawna, a student: You are told that you are working with three different numbers. When the first number is added to twice the other two numbers, the result s 64( i.e x+2y+2z= 64). When the second number is added to twice the other two numbers the result is 62 (y+2x+2z=62). Finally, when the third number is added to twice the other two numbers, the result is 59 (z+2x+2y=59) Determine the three numbers? |
Hi Shawna,
You have
x + 2y + 2z = 64 (1)
2x + y + 2z = 62 (2)
2x + 2y + z = 59 (3)
The first thing that I notice about these three equations is that if you add all three together then on the left side you get
5(x + y + z)
and on the right side
64 + 62 + 59 = 185
thus
x + y + z = 37 (4)
Now subtract equation (2) from equation (1) to get -x + y = 2 or
x = y - 2 (5)
Then subtract equation (3) from equation (2) to get -y + z = 3 or
z = y + 3 (6)
Substitute equations (5) and (6) into equation (4) and solve for y.
Penny
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