could someone help me with math? (fractions/like terms)

Question

Perform the following additions and/or subtractions:

(1/8v^3 −1/3v^2 −2v +2) − (9/4v^3 −1v^2+1/4v−2/3)

 

please help its too confusing

Answer 1

(1/8v³ - 1/3v² - 2v + 2) - (9/4v³ - 1/v² + 1/4v - 2/3)

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Are these terms written as:   1/(8v³)  or  (1/8)v³ ?
Your grouping is not clear.  I believe Marlene has worked

it the way you intended it... If not then, as written in the

way you posted it, below is the way it would be done:

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First off realize that since we have the negative sign outside

the parenthesis of the 2nd set of terms we are subtracting

each term inside that set of parent esis...

 

1/8v³ - 1/3v² - 2v + 2 - 9/4v³ + 1/v² + 1/4v - 2/3

 

Let's put the exponents in descending order with all constants

at the end

 

1/8v³ - 9/4v³ - 1/3v² + 1/v² - 2v + 1/4v + 2 - 2/3

 

= (1/8)v^-3 - (9/4)v^-3 -(1/3)v^-2 + v^-2 -2v + (1/4)v^-1 -4/3

 

Now we can see all our similar terms and combine like terms more easily

 

Look at the v^-3 terms:  (1/8)v^-3 - (9/4)v^-3 = (-17/8)v^-3

 

Look at the v^-2 terms: (-1/3)v^-2 + v^-2 = (2/3)v^-2

 

There is only one v term...   -2v

 

There is only one v^-1 term ... (1/4)v^-1

 

The constant is -4/3

 

Putting this back together:

 

(-17/8)v^-3 + (2/3)v^-2 - 2v + (1/4)v^-1 - 4/3

 

If you need to get rid of the negative exponents rewrite this as:

 

-17/8v³ + 2/3v² - 2v + 1/4v - 4/3

 

Should you need to combine this to a single term you need

a common denominator.  Our current denominators are

8v³, 3v², 1, 4v, 3 

The least common multiple of the constants 8, 3, and 1 is 24.

The least common multiple of v³, v², v  is  v³

So our common denominator is 24v³

 

First term:  -17/8v³ = -51/24v³

2nd term:  2/3v² = 16v/24v³

3rd term:  -2v = -48v³/24v³

4th term:  1/4v = 6v²/24v³

5th term: -4/3 = -32v³/24v³

 

Combining these together we have:

 

(-51 + 16v - 48v³ + 6v² -32v³)/24v³

 

= (-80v³ + 6v² + 16v - 51)/24v³

 

 

 

 

Answer 2

Hi Jared,

 

The most important thing to remember about problems like this is we need common denominators just as we would when adding/subtracting fractions like

 

1/2 + 1/4 = 2/4 + 1/4 = 3/4

 

So

   1            1                                 9                      1         2

(--- v³  -  --- v²  -  2v  +  2)   -  ( --- v³  -  v²  +  --- v  -  --- )  

   8           3                                  4                      4         3

 

Let's look at all the fractional coefficients.

 

The coefficients are 1/8, 1/3, 9/4, 1/4 and 1/3.  This would suggest a common denominator of 24.

 

   1     3         1     8               24          24         9     6         24          1     6       2     8
(--- •--- v³ - --- • --- v² - 2v • ---- + 2 • ----) - ( --- • --- v³ - ---- v² + --- • --- v - --- • ---)
   8     3        3      8               24          24        4      6         24          4     6       3     8

 

This gives us

   3           8          48       48        54        24          6        16
(---- v³ - ---- v² - ---- v + ----) - ( ---- v³ - ----- v² + ---- v - ---- )
   24        24         24       24        24        24          24       24

 

 

With all the denominators equaling 24, we can rewrite the above as,

 

(3v³ - 8v² - 48v + 48) - (54v³ - 24v² + 6v - 16)

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                                 24

 

 

and

 

3v³ - 8v² - 48v + 48 - 54v³ + 24v² - 6v + 16)
-------------------------------------------------------------
                          24

 

 

Now we can combine like terms

 

-51v³ + 16v² - 42v + 64

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                   24