Question

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

Answer 1

Let the tens digit of the original number be represented by x.  Let the ones digit be y.

The original number is then 10x + y, and the reversed number is 10y + x.

"Five times the sum of the digits is 13 less than the original number" becomes:

$5(x+y)=10x+y-13$

"Four times the sum of its digits is 21 less than the reversed number" becomes:

$4(x+y)=10y+x-21$

Work with the first equation until it is simpler:

$5x+5y=10x+y-13 \\ 5x-4y=13$

Do the same thing with the second equation:

$4x+4y=10y+x-21 \\ -3x+6y=21$

Now you have a system of two equations to solve.

$5x-4y=13 \\ -3x+6y=21$

To eliminate x, multiply the first equation by 3 and multiply the second equation by 5

$15x=12y=39 \\ -15x +30y=105$

Add the two equations.

$18y=144 \\ y=8$

Aha!  The one digit of the original number is 8.  To find y, put x = 8 into any equation containing both x and y.

$5x-4(8)=13 \\ 5x-32=13 \\ 5x=45 \\ x=9$

Sweet!  The tens digit of the original number is 9.  The original number is 98 and the reversed number is 89.

Their difference is 9.