Question 26175: Hello. I am having trouble finding a system of equations for this word problem. "For a certain 3 digit number, interchanging the first and last digits gives a three digit number 297 less than the original number. Doubling the first and last digits of the original number gives a three digit number that increases the sum of the digits to 15" I have figured out the answer to the problem (154) but i need it to be put into a system of equations. I am trying my best to figure out the solution so i can hand it in TOMORROW!(I know i need a solution quickly!) I am in eighth grade and am in algebra 1 and no problem has stumped me this much! Please Math Gods, set up a system of equations for this word problem. If you do, my day will be saved! Thank you and have a nice day!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! For a certain 3 digit number, interchanging the first and last digits gives a three digit number 297 less than the original number. Doubling the first and last digits of the original number gives a three digit number that increases the sum of the digits to 15" I have figured out the answer to the problem (154)
NO THE ANSWER IS NOT 154..BUT IT IS 451.
but i need it to be put into a system of equations.
LET THE 3 DIGIT NUMBER BE
XYZ....ITS VASLUE IS 100X+10Y+Z
INTERCHANGING FIRST AND LAST DIGITS GIVE US
ZYX...ITS VALUE IS 100Z+10Y+X..THIS IS 297 LESS THAN ORIGINAL NUMBER.HENCE
100x+10y+z-100z-10y-x=297
99x-99z=297
99(x-z)=297
x-z=297/99=3
x-z=3...........I
dobling the first and last digits of original number gives sum of the digits as
2x+y+2z which is equal to 15
2x+y+2z=15...........II
we have 3 unknowns and only 2 eqns.but we know x,y and z have to be integers between 0 and 9.so let us try as follows
2*eqn.I+eqn.II....gives
4x+y=21..............III
using eqn.I let us make a table of possible values of x and z and test it with the eqn.III
x-z=3.since minimum value of z is zero,x has to be atleast 3. hence let us try x=3...then...z=0..and from eqn.III...y=21-4x=9...now we see that if x is more than 5, y becomes negative from eqn.III.so the only possibilities are x=3,y=9,z=0........390....390-93=297......2*3+9+2*0=15
x=4,y=5,z=1........451....451-154=297.....2*4+5+2*1=15
x=5,y=1,z=2........512....512-215=297.....2*5+1+2*2=15
are the possible numbers.
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