The sum of the digits in a two digit number is 9 .if 9 is added to the number by reversing the digits then the end result is thrice the original number . Find the original number.
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Let the digit in units place be x and digit in ten's place be y
x+y = 9 ; x = 9-y
Original number = 10y +x
New number = 10x+y+9
10x+y+9=3(10y+x)
10x+y+9= 30y+3x
10x-3x+9= 30y-y
7x+9=29y
7x+9-29y = 0
7(9-y) +9-29y = 0
63-7y+9-29y = 0
72-36y = 0
72=36y
y= 72/36
y=2
x+y = 9
x+2 = 9
x=9-2=7
Original number = 10y+x =10(2) +7= 27
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x+y = 9 ; x = 9-y
Original number = 10y +x
New number = 10x+y+9
10x+y+9=3(10y+x)
10x+y+9= 30y+3x
10x-3x+9= 30y-y
7x+9=29y
7x+9-29y = 0
7(9-y) +9-29y = 0
63-7y+9-29y = 0
72-36y = 0
72=36y
y= 72/36
y=2
x+y = 9
x+2 = 9
x=9-2=7
Original number = 10y+x =10(2) +7= 27
hope it shops u mate...
pls mark me brainliest
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