Need Help Quick: Express the following sum as a single fract

crappiefisher26

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I have never done a question like this.

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
 

letsgetaway

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Re: Need Help Quick

crappiefisher26 said:
I have never done a question like this.

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
if the denominator is the same, then simplify the numerator.

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

crappiefisher26

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

letsgetaway

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

crappiefisher26

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the first 6z is negative the second is positive
 

stapel

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Re: Need Help Quick: Express the following sum as a single f

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

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

crappiefisher26 said:
the 6z is also negative in the 1st [expression]-(8z-4y)/6z - (7z+y)/6z
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:

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

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

Thank you.

Eliz.
 

crappiefisher26

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heres the equation the right way.

[-(8z-4y)/(-6z)] - [(7z+y)/(6z)]
 

jonboy

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If this is what you mean:\(\displaystyle \L \;\frac{\,-\,(8z\,-\,4y)}{\,-\,6z}\,-\,\frac{(7z+y)}{6z}\)


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! :)
 

stapel

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crappiefisher26 said:
heres the equation the right way.

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

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.
 

crappiefisher26

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ok guess i was wrong with the equation

- (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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stapel

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

crappiefisher26 said:
- (2x-u)/(6x) + (8x+9u)/(6x)
^
for example -1/4 the negative is in lign with the divided by sign
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:
help please
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.)

Thank you.

Eliz.
 

crappiefisher26

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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.
 

Denis

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

Crappy, you GOT to start being more careful;
the way you show that, 2z is NOT a common denominator;
you need more brackets:
-[(7z - 9t) / (2z)] - [(7z + 3t) / (2z)]
 

crappiefisher26

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Heres the original equation could somebody please help. Cause i cant figure this out. -[(7z - 9t) / (2z)] - [(7z + 3t) / (2z)]

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.
 

Denis

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crappiefisher26 said:
(-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!!)
So what the heck is your problem :shock:
 
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