Find a two-digit number such that three times the tens digit is 2 less than twice the units digit and twice the number is 20 greater than the number obtained by reversing the digits
Let the tens digits of the number be x and the unit digit be y
3x = 2y - 2
3x - 2y = -2.......(i)
If the digits are interchanged, the tens digit becomes y, the unit digit becomes x. Hence 2(10x + y) = 10y + x + 20
(20x + 2y) - (10y + x) = 20
19x - 8y = 20.....(ii)
Multiply eqn.(i) by 8 and eqn.(ii) by 2
24x - 16y = -16......(iii)
38x - 16y = 40........(iv)
eqn(iv) - eqn(iii)
14x = 56
x = 4
Sub. for x = 4 in eqn(i)
3(4) - 2y = -2
14 = 2y
y = 7
So the original number is 10(4) + 7
i.e. 10x + y
= 47
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