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Question 351111: pls. help me solve this problem... thank you...
1. Three times the tens digit of a certain two digit number is two more than four times the units digit. The difference between the given number and the number obtained by reversing the digits is two less than twice the sum of the digits. Find the number.
2. Find the two digit number such that four times the units digit is six less than twice the tens digit, and the number is nine less than the three times the number obtained by reversing the digits.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Hi,
Let t represent the tens digit and d the ones digit of that certain number
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Question states
Write as You read
3*t =4*d + 2
t = (4/3)*d + (2/3)
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Note the value of the 2-digit number is
10*t + d
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Question states:
(10*t +d)-(10*d +t) = 2*(t + d) -2
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simplify
9*t - 9*d = 2t + 2d -2
7*t = 11*d -2
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substitute for t (in terms of d)
7*((4/3)*d + (2/3)) = 11*d - 2
(28/3)*d + 14/3 = 11*d - 2
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solve for d
20/3 = (5/3)d
d = 4 the units digit
Find the tens digit
t = (4/3)*d + (2/3) = 16/3 + 2/3 = 18/3 = 6
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Number is 64
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check your answer
3*t =4*d + 2
18 = 16 + 2
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