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Question 33419This question is from textbook algebra for college students
: This word problem is stumping me. I can't figure out how to write the equations needed for it. please help.
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than 5 times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then what is the number?
I have come up with x+y+z = 11
I'm lost on the second equation.
and i'm pretty sure the third equation is
x+2y = z
I hope I'm at least close. I would appreciate any help.
This question is from textbook algebra for college students
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! Let the number be xyz. We are told that x+y+z=11
original number is xyz
--> xyz really means 100x+10y+z
Reversing gives zyx
--> zyx really means 100z+10y+x
zyx is 46 more than 5 times the old number:
So, (100z+10y+x) = 5(100x+10y+z) + 46
--> 100z+10y+x = 500x+50y+5z + 46
--> 95z-40y-499x = 46
Also, hundreds digit (x) + twice tens (y) equals units (z)
--> x + 2y = z
So, we have 3 unknowns, xy and z and we have 3 equations relating them:
x+y+z=11
95z-40y-499x = 46
x+2y=z
You try to solve these 3 :-)
answer is 137 for the number.
jon.
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