The sum of the digits of a $2$ digits no is $9$ . If the digits are reversed the new no decreased by $9$ is equal to $4$ times the original no, find the orginal numbers.

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Solution

Let ten's digit be $x$ and unit's be $y$
$∴$ original no. $= 10 x + y$
$∵ x + y = 9$
$\Rightarrow$ $x = 9 - y$ ___ $(1)$
New no. found by reversing the digits $= 10 y + x$
$(10 y + x - 9) = 4 (10 x + y)$
$\Rightarrow 10 y + x - 9 = 40 x + 4 y$
$\Rightarrow 39 x - 6 y = - 9$
$\Rightarrow 13 x - 2 y + 3 = 0$
$\Rightarrow 13 (9 - y) - 2 y + 3 = 0$
$\Rightarrow 117 + 3 - 15 y = 0$
$\Rightarrow y = 8$
$x = 9 - 8 = 1$
$∴$ The original no. is $= 10 \times 1 + 8$
$= 18$

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The sum of the digits of a 2 digits no is 9. If the digits are reversed the new no decreased by 9 is equal to 4 times the original no, find the orginal numbers.

Let ten's digit be x and unit's be y∴ original no.=10x+y∵x+y=9⇒ x=9−y___(1)New no. found by reversing the digits =10y+x(10y+x−9)=4(10x+y)⇒10y+x−9=40x+4y⇒39x−6y=−9⇒13x−2y+3=0⇒13(9−y)−2y+3=0⇒117+3−15y=0⇒y=8x=9−8=1∴ The original no. is =10×1+8=18

The sum of the digits of a $2$ digits no is $9$ . If the digits are reversed the new no decreased by $9$ is equal to $4$ times the original no, find the orginal numbers.