The sum of the digits of a two-digit counting number is 10. If the digits are reversed, the new number is one less than twice the original number. What was the original number?

chilledz3non Apr 25, 2016

#2**+5 **

A + B=10

2[10A + B] - 1=10B + A

So the original number is=37

Solve the following system:

{A+B = 10 | (equation 1)

2 (10 A+B)-1 = A+10 B | (equation 2)

Express the system in standard form:

{A+B = 10 | (equation 1)

19 A-8 B = 1 | (equation 2)

Swap equation 1 with equation 2:

{19 A-8 B = 1 | (equation 1)

A+B = 10 | (equation 2)

Subtract 1/19 × (equation 1) from equation 2:

{19 A-8 B = 1 | (equation 1)

0 A+(27 B)/19 = 189/19 | (equation 2)

Multiply equation 2 by 19/27:

{19 A-8 B = 1 | (equation 1)

0 A+B = 7 | (equation 2)

Add 8 × (equation 2) to equation 1:

{19 A+0 B = 57 | (equation 1)

0 A+B = 7 | (equation 2)

Divide equation 1 by 19:

{A+0 B = 3 | (equation 1)

0 A+B = 7 | (equation 2)

Collect results:

**Answer: | A = 3 and B = 7**

Guest Apr 25, 2016