#### crappiefisher26

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It says Adding Rational Expressions with a Common Denominator

Express the following sum as a single fraction.

> the 6z is also negative in the 1st equation-(8z-4y)/6z - (7z+y)/6z

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It says Adding Rational Expressions with a Common Denominator

Express the following sum as a single fraction.

> the 6z is also negative in the 1st equation-(8z-4y)/6z - (7z+y)/6z

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if the denominator is the same, then simplify the numerator.crappiefisher26 said:

It says Adding Rational Expressions with a Common Denominator

Express the following sum as a single fraction.

> the 6z is also negative in the 1st equation-(8z-4y)/6z - (7z+y)/6z

-8z+4y-7z-y. Solve that then put it over the denominator. Simplify more if you have to.

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well the denominators are -6 and 6 so which one would i use?

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I thought you said that the first 6z was negative also.crappiefisher26 said:well the denominators are -6 and 6 so which one would i use?

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the first 6z is negative the second is positive

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Has this not been covered in class?crappiefisher26 said:I have never done a question like this.

"Rational expressions" are just polynomial fractions, so you'd work with them in the same way you would numerical fractions.crappiefisher26 said:It says Adding Rational Expressions with a Common Denominator

Is the first "minus" a negative sign, or a divider between your comment and the first expression? If the denominators are negative, shouldn't there be "minus" signs on them? Such as:crappiefisher26 said:the 6z is also negative in the 1st [expression]-(8z-4y)/6z - (7z+y)/6z

. . . . .[(8z - 4y) / (-6z)] - [(7z + y) / (-6z)]

Does the variable "z" not then belong in the denominators?crappiefisher26 said:the denominators are -6 and 6

Thank you.

Eliz.

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heres the equation the right way.

[-(8z-4y)/(-6z)] - [(7z+y)/(6z)]

[-(8z-4y)/(-6z)] - [(7z+y)/(6z)]

Distribute the negative:\(\displaystyle \L \;\frac{\,-\,(8z\,-\,4y)}{\,-\,6z}\,\to\,\frac{\,-\,8z\,+\,4y}{\,-\,6z}\)

Get like denominators by multiplying the whole second term by an negative:\(\displaystyle \L \;\,-\,\frac{7z\,+\,y}{6z}\,\to\,-\,\,\frac{\,-\,7z\,-\,y}{\,-\,6z}\)

So now we have:\(\displaystyle \L \;\frac{\,-\,8z\,+\,4y}{\,-\,6z}\,-\,\,\frac{\,-\,7z\,-\,y}{\,-\,6z}\,=\,\frac{15z\,+\,5z}{-6z}\)

There you are, your one fraction!

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Thank you.crappiefisher26 said:heres the equation the right way.

[-(8z-4y)/(-6z)] - [(7z+y)/(6z)]

A good way to start would be to note that you can simplify the first fraction a bit:

. . . . .\(\displaystyle \L \frac{-(8z\,-\,4y)}{-6z}\,- \,\frac{7z\,+\,y}{6z}\)

. . . . .\(\displaystyle \L \frac{-1(8z\,-\,4y)}{-1(6z)}\,- \,\frac{7z\,+\,y}{6z}\)

Cancel off the common factor of -1 to get:

. . . . .\(\displaystyle \L \frac{8z\,-\,4y}{6z}\,- \,\frac{7z\,+\,y}{6z}\)

. . . . .\(\displaystyle \L \frac{(8z\,-\,4y)\,-\,(7z\,+\,y)}{6z}\)

. . . . .\(\displaystyle \L \frac{8z\,-\,4y\,-\,7z\,-\,y}{6z}\)

...and so forth.

Eliz.

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- (2x-u)/(6x) + (8x+9u)/(6x)

^

for example -1/4 the negative is in lign with the divided by sign .

help please

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Note: An "equation" contains an "equals" sign, and is "solved". You have posted an "expression", to be "simplified".crappiefisher26 said:ok guess i was wrong with the equation

If the "minus" sign is on the entire expression, then it can be moved onto the numerator (standard practice) or onto the denominator (non-standard), but not both.crappiefisher26 said:- (2x-u)/(6x) + (8x+9u)/(6x)

^

for example -1/4 the negative is in lign with the divided by sign

With which? You have been provided with most of the worked solution to the posted exercise. If you are asking for assistance with the new expression (which should then have been posted as a new thread), then please reply showing how far you have gotten. (You can model your work on the solutions posted earlier.)crappiefisher26 said:help please

Thank you.

Eliz.

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-(7z-9t/2z) - (7z+3t/2z)

Before i actually subtract the two fractions, i distributed the two negative signs:

-(7z+9t/2z) + -(7z-3t/2z)

Since each term in the sum has the same denominator, we add the numerators together and divide the result by the common denominator to obtain the single fraction

(-7z+9t-7z-3t)/2z

Regrouping like terms, the fraction simplifies to

(-14z+6t)/2z

The last step is to factor and reduce the fraction:

(-14z+6t)/2z=2(-7z+3t)/2(z)=-7z+3t/z thats the final right answer.

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heres how he showed it but im confused.

Well for example this is one that i got wrong.

-(7z-9t/2z) - (7z+3t/2z)

Before i actually subtract the two fractions, i distributed the two negative signs:

-(7z+9t/2z) + -(7z-3t/2z)

Since each term in the sum has the same denominator, we add the numerators together and divide the result by the common denominator to obtain the single fraction

(-7z+9t-7z-3t)/2z

Regrouping like terms, the fraction simplifies to

(-14z+6t)/2z

The last step is to factor and reduce the fraction:

(-14z+6t)/2z=2(-7z+3t)/2(z)=-7z+3t/z thats the final right answer.

Crappy, (-7z + 3t) / z IS CORRECT (with brackets added!!)crappiefisher26 said:(-14z + 6t) / 2z = 2(-7z + 3t) / 2(z) = -7z + 3t / z thats the final right answer.

So what the heck is your problem :shock: