The original number will be "63".
Let,
- The ten's digit be "x".
- One's digit be "y".
The equation will be:
→ [tex]x=y+3[/tex]...(equation 1)
Original number will be:
→ [tex]10x+y[/tex]
then,
→ [tex]2 (10x+y)-\frac{(10y+x)}{2} = 108[/tex]
→ [tex]\frac{4(10x+y)}{1} -(10y+x) = 216[/tex]
→ [tex]40x+4y-10y-x = 216[/tex]
→ [tex]39x-6y=216[/tex]...(equation 2)
By using (equation 1) and (equation 2), we get
→ [tex]39(y+3)-6y=216[/tex]
→ [tex]39y+117-6y=216[/tex]
→ [tex]33y=99[/tex]
→ [tex]y = \frac{99}{33}[/tex]
→ [tex]= 3[/tex]
Now,
[tex]x = y+3[/tex]
[tex]=3+3[/tex]
[tex]=6[/tex]
hence,
The original number will be:
= [tex]10x+y[/tex]
= [tex]10\times 6+3[/tex]
= [tex]60+3[/tex]
= [tex]63[/tex]
Thus the above approach is right.
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