Simplify.
What are the excluded values of the original denominator?
1. (3 + 1/y) / (4 + 2/y)
2. [(5/x) - (3/x + 3)] / [(2/x^2 + 3) + (4/x)]
It looks hard, I know. I tried to show the work for the first one, but I cant figure out the answer. Help is greatly appreciated. The / sign means its a fraction, so there are fractions in the numerators and denominators. There is at least one small fraction within the overall fraction.
What are the excluded values of the original denominator?
1. (3 + 1/y) / (4 + 2/y)
2. [(5/x) - (3/x + 3)] / [(2/x^2 + 3) + (4/x)]
It looks hard, I know. I tried to show the work for the first one, but I cant figure out the answer. Help is greatly appreciated. The / sign means its a fraction, so there are fractions in the numerators and denominators. There is at least one small fraction within the overall fraction.
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I'm pretty sure it's this:
For number one: You have (3+1/y) in the numerator. So find a common denomitor for the "3" inorder to add it with the "1/y". (Thus you multiply 3 with "y/y" to get 3y/y) Now you have a common denominator for both. Add 3y with "+1" to get 3y+1 OVER y...Which will end up as (3y+1)/y for your numerator
For your denominator ---> do the same thing to find the common denomintor for the "4". If you repeat the steps you did in the numerator... you'll get (4y+2)/y as the result.
Next, divide (3y+1)/y with (4y+2)/y. (Note: you will end up multiplying (3y+1)/y with the reciprical of (4y+2)/y, which is y/(4y+2)) This will end up as (3y+1)/y "times" y/(4y+2))
the "y" 's cancel out and you're left with (3y+1)/(4y+2) as your answer. You can further simplify it to (3y+1)/(2(2y+1))
If by excluded value in the original denominator you mean, I'm guessing that it's the number that "y" can't be in the denominator because it will cause it to be a "zero". You can just set 4+2/y = 0 and solve for "y" to get -1/2. And another value can be "0" itself because you can't have a 0 in the denominator for a solution. Thus your excluded values in the original denominator will be -1/2 and 0
=D
For number one: You have (3+1/y) in the numerator. So find a common denomitor for the "3" inorder to add it with the "1/y". (Thus you multiply 3 with "y/y" to get 3y/y) Now you have a common denominator for both. Add 3y with "+1" to get 3y+1 OVER y...Which will end up as (3y+1)/y for your numerator
For your denominator ---> do the same thing to find the common denomintor for the "4". If you repeat the steps you did in the numerator... you'll get (4y+2)/y as the result.
Next, divide (3y+1)/y with (4y+2)/y. (Note: you will end up multiplying (3y+1)/y with the reciprical of (4y+2)/y, which is y/(4y+2)) This will end up as (3y+1)/y "times" y/(4y+2))
the "y" 's cancel out and you're left with (3y+1)/(4y+2) as your answer. You can further simplify it to (3y+1)/(2(2y+1))
If by excluded value in the original denominator you mean, I'm guessing that it's the number that "y" can't be in the denominator because it will cause it to be a "zero". You can just set 4+2/y = 0 and solve for "y" to get -1/2. And another value can be "0" itself because you can't have a 0 in the denominator for a solution. Thus your excluded values in the original denominator will be -1/2 and 0
=D