A 2-digit number is increased by 36 when the digits are reversed. The sum of the digits is 10. What is the original number?

Work Shown:

T = tens digit of the original number

U = units digit of the original number

U+T = 10 since the digits add to 10. Solve for U to get U = 10-T

10T+U = original number

10U+T = new number where digits are reversed

new number = (old number) + 36

10U+T = (10T+U)+36

10U+T = 10T+U+36

10(10-T)+T = 10T+10-T+36 ... plug in U = 10-T

100-10T+T = 10T+10-T+36

-9T+100 = 9T+46

100-46 = 9T+9T

54 = 18T

18T = 54

T = 54/18

T = 3 is the tens digit of the original number

U = 10-T

U = 10-3

U = 7 is the units digit of the original number

10T+U = 10*3+7 = 30+7 = 37 is the original number

73 is the reversed number.

73-37 = 36 is the difference, showing that 37 increased by 36 leads to 73. The answer is confirmed.

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation:

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