The sum of the digits of a two digit number is 12.When 18 is added to the number its digits are reversed find the number
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the original number?
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Robert Nichols
Answered Feb 27, 2017
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Original Number 57, New number 75.
Let x represent the “tens” digit
let y represent the “ones” digit
So the original number is 10x + y
the reversed number is 10y + x
10x + y + 18 = 10y + x This is the new number is 18 more than the original
x + y =12 This is the sum of the digits is 12
Isolate x in x + y = 12 subtract y from both sides. x = 12 - y
Substitute 12-y for every x in: 10x + y + 18 = 10y + x =>
10(12-y) + y + 18 = 10y + (12-y) Distribute the 10
120 -10y + y + 18 = 10y + 12 - y Combine like terms
138 - 9y = 9y + 12 Add 9y on both sides
138 = 18y + 12 Subtract 12 on both sides
126 = 18 y divide both sides by 18
7 = y
x = 12 - y => x = 12 - 7 => x = 5
Original Number 57, new number 75. Check 75 -57 = 18
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Srinivasan, former Maths B.T.Asst Teacher (Retired) at P.S.G Sarvajana Hr.Sec School (1999-2010)
Answered Feb 27, 2017
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16 ANSWERS

Robert Nichols
Answered Feb 27, 2017
Continue Reading
Original Number 57, New number 75.
Let x represent the “tens” digit
let y represent the “ones” digit
So the original number is 10x + y
the reversed number is 10y + x
10x + y + 18 = 10y + x This is the new number is 18 more than the original
x + y =12 This is the sum of the digits is 12
Isolate x in x + y = 12 subtract y from both sides. x = 12 - y
Substitute 12-y for every x in: 10x + y + 18 = 10y + x =>
10(12-y) + y + 18 = 10y + (12-y) Distribute the 10
120 -10y + y + 18 = 10y + 12 - y Combine like terms
138 - 9y = 9y + 12 Add 9y on both sides
138 = 18y + 12 Subtract 12 on both sides
126 = 18 y divide both sides by 18
7 = y
x = 12 - y => x = 12 - 7 => x = 5
Original Number 57, new number 75. Check 75 -57 = 18
721 Views · View Upvoters · Answer requested by Aaron Woods
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Srinivasan, former Maths B.T.Asst Teacher (Retired) at P.S.G Sarvajana Hr.Sec School (1999-2010)
Answered Feb 27, 2017
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refer the pic.
hope it'll help
pls mark it as brainlist..
hope it'll help
pls mark it as brainlist..
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