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Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two-digit number, four times the sum of its two digits is 21 less than the reversed two-digit number.

(Hint: You can use variables to represent the digits of a number. If a two-digit number has the digit x in tens place and y in one’s place, the number will be 10x + y. Reversing the order of the digits will change their place value and the reversed number will 10y + x.)

The difference of the original two-digit number and the number with reversed digits is
.

Respuesta :

ivycoveredwalls
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:

[tex]5(x+y)=10x+y-13[/tex]

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

[tex]4(x+y)=10y+x-21[/tex]

Work with the first equation until it is simpler:

[tex]5x+5y=10x+y-13 \\ 5x-4y=13[/tex]

Do the same thing with the second equation:

[tex]4x+4y=10y+x-21 \\ -3x+6y=21[/tex]

Now you have a system of two equations to solve.

[tex]5x-4y=13 \\ -3x+6y=21[/tex]

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

[tex]15x=12y=39 \\ -15x +30y=105[/tex]

Add the two equations.

[tex]18y=144 \\ y=8[/tex]

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

[tex]5x-4(8)=13 \\ 5x-32=13 \\ 5x=45 \\ x=9[/tex]

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.

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