Does a two-digit number exist such that the digits sum to 9 and when the digits are reversed the resulting number is 9 greater t
2 answers:
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The given fraction is:
We cannot take anything common from the numerator but from the denominator we can take out 2 as common because 2 is a factor of both the terms i.e. 2y and 10.
2y + 10 can be written as 2y +2(5) which can further be written as 2(y+5)
So, our fraction now becomes:
y+5 occurs in both the numerator and denominator and thus be cancelled out leaving behind:
Therefore, the given fraction in lowest term is equal to 1/2. So option B is the correct answer
We cannot take anything common from the numerator but from the denominator we can take out 2 as common because 2 is a factor of both the terms i.e. 2y and 10.
2y + 10 can be written as 2y +2(5) which can further be written as 2(y+5)
So, our fraction now becomes:
y+5 occurs in both the numerator and denominator and thus be cancelled out leaving behind:
Therefore, the given fraction in lowest term is equal to 1/2. So option B is the correct answer
