An algebra problem by Palash Som

Digit product = 14 only one possibility 2 × 7 or 7 × 2 but after adding 45, 72 + 45 is 3 digit number as digits are reversed, sum is 2 digit number so answer is 27

Let the number be $\color{#20A900}{10x+y}$ .Then the equations are: $\color{#3D99F6}{xy=14\rightarrow (1)\\ 10x+y+45=10y+x\rightarrow (After\;Simplification)\;y-x=5\rightarrow (2)}$ Substituting $\color{#D61F06}{y=x+5}$ in $(1)$ we get: $\color{#69047E}{x(x+5)=14\\ x^2+5x-14=0 \\ (x-2)(x+7)=0\\ x=2\;or\;x=-7}$ As digits are positive so $x=2$ .Substituting this in $(2)$ ,we get $\color{#20A900}{y=7}$ .So the number is $\boxed{27}$

let the digit. at the 10's place be x . and the digit at one's place be y. then xy = 14. required no. = (10x+y) no. obtained by reversing its digit = (10y+x) therefore (10x+y)+45 = (10y+x) i.e.- 9(y-x)= 45 so y-x = 5 now (y+x)^2-(y-x)^2= 4xy so solving we get y+x = 9 solving the 2 equations i.e. y-x =5 and y+x = 9 we get y = 7 then x = 2 therefore the no. is 27