Question 269458
In doing sums of digits problems, you have to keep track of the values as well as the counts.
To illustrate, the number 23 actually is 2 tens plus 3 ones.
The more general representation of the number 'xy' has the value 10x + y.
Similarly, the number depicted as 'abc' would have value 100a + 10b + c.
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It is critical that you do not get confused and think 'xy' means x time y!
It is easy to fall into this trap, especially with a long problem.  So beware!
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We are told there is a two-digit number.
We can call it 'xy'.
Remember, 'x' is just standing next to 'y'.  They are not being multiplied.
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We are told x+y = 12.
That means x = 12-y and y = 12-x.
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We are told that if you add 15 to the value of the number (i.e., 10x + y), the result is 6y.
10x + y + 15 = 6y
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Substituting:
10(12-y) +y + 15= 6y
120 -10y + y + 15 = 6y
135 + 9y = 6y
135 = 15y
15y = 135
y = 9
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x = 12-y = 12-9 = 3
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xy = 39
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Check by doing substituting back into the word problem.
39 can be viewed as xy, which means 3+9 = 12.  OK
39 + 15 = 54, which does = 6*9 = 6y. OK
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So, the number is:  39.
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BUT you teacher may want you to solve the problem using simultaneous equations.
In that case, we have two equations and two unknowns, so we can do it.
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Eq. 1:  x + y = 12
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Eq. 2:  10x + y + 15 = 6y
Subtracting 6y from both sides
10x -5y + 15 = 0
Subtracting 15 from both sides
10x - 5y = -15
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That gives us:
x + y = 12
10x - 5y = -15
Multiply the first equation by 10
10x + 10y = 120
10x - 5y = -15
Subtracting the second equation from the first
15y = 135
Dividing by 15
y = 9
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And therefore x=3.
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This means the number is 39.  
We checked it above, so there's no need to check it again.
Done.