Question 21508: Here is my problem:
A father and his two sons bought a drug store for $320,000. If the father
invested twice as much as the older son, and the older son invested
$45,000 less than twice the younger son, how much did the father and
his sons each invest?
This needs to be solved as a linear equation.
The answer is: younger son, $65,000; older son, $85,000; father $170,000
I started with:
Father = A, older son = B and younger son = C
A + B + C = 320,000
2B = A
2C - 45,000 = B
But I don't know where to go from there.
Any help would be appreciated, Thanks!
Sandy
Found 2 solutions by venugopalramana, josmiceli:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE MY COMMENTS BELOW
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Here is my problem:
A father and his two sons bought a drug store for $320,000. If the father
invested twice as much as the older son, and the older son invested
$45,000 less than twice the younger son, how much did the father and
his sons each invest?
This needs to be solved as a linear equation.
The answer is: younger son, $65,000; older son, $85,000; father $170,000
I started with:
Father = A, older son = B and younger son = C..........GOOD...BUT THERE IS A BETTER OPTION...THE LESS THE NUMBER OF UNKNOWNS THE BETTER..I SHALL SHOW IT SEPERATELY...FIRST LET US GO BY YOUR THINKING
A + B + C = 320,000....VERY GOOD....BETTER NAME IT AS EQN.I
2B = A........VERY GOOD ......BETTER NAME IT AS EQN.II
2C - 45,000 = B .....EXCELLENT.....BETTER NAME IT AS EQN.III
But I don't know where to go from there.....AS USUAL YOU ARE AT THE END OF THE TUNNEL ,BUT NOT ABLE TO SEE THE LIGHT...THINK IT OUT THIS WAY..WE HAVE 3 UNKNOWNS..A,B,C....SO WE NEED 3 INDEPENDENT RELATIONS TO FIND THEM...AND WE HAVE 3 EQNS.NUMBERED ABOVE...WE TRY TO MAKE 3 TO 2 AND FROM 2 TO 1 UNKNOWNS.
EQN.I HAS ALL 3 UNKNOWNS..SO LET US KEEP IT OUT FOR SOME TIME...EQNS.II AND III HAVE ONLY 2 UNKNOWNS...SO LET US TAKE THEM...FROM EQN.II,WE HAVE A=2B............................EQN.IV
SO LET US TRY TO FIND C ALSO INTERMS OF B FROM EQN.III.
2C-45000=B
2C=(B+45000)
C=(B+45000)/2.................EQN.V
NOW SUSTITUTE EQNS.IV AND V IN EQN.I
2B+B+(B+45000)/2=320000
3B+(B+45000)/2=320000
MULTIPLY THROUGH OUT WITH 2
6B+B+45000=640000
7B=640000-45000=595000
B=595000/7=85000
A=2B=170000
C=(B+45000)/2=65000
Any help would be appreciated, Thanks!.....
Sandy
NOW ON TO BETTER PRACTICE...TAKE ANY CONVENIENT UNKNOWN AS X ..HERE YOUNGER SON IS THE CONVENIENT CHOICE SO LET YOUNGER SON'S INVESTMENT=X
TWICE THIS =2X
LESS 45OOO IS =2X-45000=OLDER SON'S INVESTMENT
TWICE THIS =2(2X-45000)=4X-90000=FATHER'S INVESTMENT
BUT TOTAL INVESTMENT IS 3200000
SO X+2X-45000+4X-90000=320000
7X=320000+90000+45000=455000
X=65000...ETC.....
Answer by josmiceli(19441)  (Show Source):
You can put this solution on YOUR website! The solution that you present as given must be incorrect. The problem states
that the older son invested $45,000 less than the younger son. Your solution
as stated has the older son investing 20,000 more than the younger son.
Check the information again.
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