A number consists of two digits . The digit in tens place exceeds the digit in unit place by 4 . The sum of the digits is 1/7 of the number. Find the number
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Establish the variables
x = tens place
y = ones place
...
Given
x = y+3
x+y = (10x+y)
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Substitute x into the equation
y+3 + y = (10(y+3) + y)
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Distribute the 10
2y + 3 = (10y + 30 + y)
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Combine like terms
2y + 3 = (11y + 30)
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Multiply both sides by 7 to be rid of the fraction
14y + 21 = 11y + 30
Isolate y
14y - 11y = 30 - 21
3y = 9
y =9
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x = y+3 = 3+3
x=6
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Then number is 63
...
Check
x = tens place
y = ones place
...
Given
x = y+3
x+y = (10x+y)
...
Substitute x into the equation
y+3 + y = (10(y+3) + y)
...
Distribute the 10
2y + 3 = (10y + 30 + y)
...
Combine like terms
2y + 3 = (11y + 30)
...
Multiply both sides by 7 to be rid of the fraction
14y + 21 = 11y + 30
Isolate y
14y - 11y = 30 - 21
3y = 9
y =9
...
x = y+3 = 3+3
x=6
...
Then number is 63
...
Check
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