Question 1041233: I shall write a two digit number sum of two digit number is 14 and if 29 is subtracted from the number two digits will be equal .Led us from the simultaneous equations by solving them let us see what will be the two digit number.
Found 3 solutions by ankor@dixie-net.com, Edwin McCravy, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! I shall write a two digit number sum of two digit number is 14 and if 29 is subtracted from the number two digits will be equal .
Led us from the simultaneous equations by solving them let us see what will be the two digit number.
:
No one has answered this question because there are no integer solutions using the given values. It can be done if you change 29 to 20 so it will read>
:
" I shall write a two digit number sum of two digit number is 14 and if 20 is subtracted from the number two digits will be equal ."
Let a = the 10's digit
let b = the units, this will also be the equal digits
Two equations
a + b = 14
and
10a + b - 20 = 10b + b
10a + b - 20 = 11b
10a + 1 - 11b = 20
10a - 10b = 20
simplify, divide by 10
a - b = 2
:
Use elimination on these two equations
a + b = 14
a - b = 2
------------ adding eliminates b, find a
2a = 16
a = 8 is the first digit
then
14 - 8 = 6 is the 2nd digit
:
86 is the two digit number
:
if you subtract 20 from 86 you get 66, the digits are equal
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website! The other tutor has made the error of assuming that the
number with the equal digits must have both its digits
as the same digit as the units digit of the desired number.
This is not a requirement.
Let a = the tens digit
Let b = the units digit
The number = 10a + b
Since the sum of the digits is 14,
a + b = 14, or b = 14 - a
Let c = the digit that is the same in the
answer when 29 is subtracted from the number.
So the answer when 29 is subtracted from the
number is 10c + c or 11c
10a + b - 29 = 11c
So the system of equations is
This is underdetermined because there are more
unknowns than equations.
We substitute 14 - a for b
10a + (14 - a) - 29 = 11c
10a + 14 - a - 29 = 11c
9a - 15 = 11c
9 is the smallest number in that equation in
absolute value. So write 15 and 11 in terms
of their nearest multiple of 9. So we write
15 as 18 - 3 and 11 as 9 + 2
9a - (18 - 3) = (9 + 2)c
9a - 18 + 3 = 9c + 2c
Divide through by 9
a - 2 + 3/9 = c + 2c/9
Get the fractions on one side,
other terms on the other side:
2c/9 - 3/9 = a - 2 - c
The right side is an integer, so the left
side is too. Let that integer be I.
2c/9 - 3/9 = I a - 2 - c = I;
2c - 3 = 9I
2 is the smallest number in that equation in
absolute value. So write 3 and 9 in terms
of their nearest multiple of 2. So we write
3 as 2+1 and 9 as 8+1
2c - 3 = 9I
2c - (2 + 1) = (8 + 1)I
2c - 2 - 1 = 8I + I
Divide through by 2
c - 1 - 1/2 = 4I + I/2
Get the fractions on one side,
other terms on the other side:
c - 1 - 4I = I/2 + 1/2
The left side is an integer, so the
right side is too. Let that integer be J.
c - 1 - 4I = J; I/2 + 1/2 = J
I+1 = 2J
I = 2J - 1
Substitute for I
c - 1 - 4(2J - 1) = J
c - 1 - 8J + 4 = J
c - 8J + 3 = J
c = 9J - 3
The only value of J that will permit c to be
a digit is J = 1
So c = 9J - 3 = 9(1) - 3 = 6
And since I = 2J - 1,
I = 2(1) - 1, = 2 - 1 = 1
Now the system of equations becomes:
Since a - 2 - c = I
a - 2 - 6 = 1
a - 8 = 1
a = 9
Since b = 14 - a
b = 14 - 9 = 5
So the number is 95
Checking:
95
-29
66
Edwin
Answer by MathTherapy(10552)  (Show Source):
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