2.07  Literal Equations

Intermediate Algebra: One Step at a Time.  Page 207 - 209:   #8, Extra Problem, 11, 12, 15, 17, 19

Dr. Robert J. Rapalje

Seminole State College of Florida

Sanford, FL  32773

  8.       Given    ,  solve for .

Solution:  Since there is a denominator of , multiply both sides by  to clear the fraction!

Next, remember that you are solving for , and the   has been multiplied by and .  In order to “undo” the multiplication, you must divide both sides by and :

In order to “undo” the square, you must take the square root of both sides:

There is no need for a in this case, since  is a radius, and it cannot be negative. 

Extra Problem (from Chris).     

Solve for x:          .

Solution:  First, remove parentheses by the distributive property.

Next, get all the x terms on the left side by subtracting   from each side.  At the same time, add  to each side to get all the non-x terms on the right side of the equation

Now, factor the common factor of x:

Finally, since the x has been multiplied by , you must divide both sides of the equation by .

                                           

11.    Given:    ,  solve for C.

Solution:  

There are at least two ways to solve for C in this problem.  Both are equally correct, but one is much easier that the other.  The easy way to solve this is to notice that the C has had two operations performed on it.  First, C is multiplied by the fraction , and then  was added.  To solve for C, you must UNDO these two operations in reverse order.  So, first undo the , by subtracting  from each side:

Now, undo the multiplication by  by multiplying both sides of the equation by the reciprocal of which is : 

 The other method involves multiplying both sides of the equation by the denominator which is  :

Subtract from each side:

To solve for   just divide both sides by :

This is a slightly different form of the answer obtained in the first method, if you factor out the 5, the answer will be the same as above:

              .

12.  Notice that this is the same problem as #11, but in reverse!  Begin with:          

          Given:      ,  solve for F.

Solution:  Begin by “undoing” the fraction   by multiplying both sides of the equation by  

Next, “undo” the  by adding a  to each side of the equation.

In conclusion, notice that the problem for #11 is the answer for #12, and vice-versa.     

15.                ,  solve for .

 There are two ways to begin this problem, both of which result in the same first step.  You might want to “undo” the fraction  by multiplying both sides of the equation by , which looks like this:

The result is EXACTLY the same:

Now, to solve for , you must get all the  terms on one side, and the “non-” terms on the other side.  To do this subtract  from each side.

The last step is to divide both sides by ,

The  on the left side divides out, but the  on the right side does NOT divide out!  Why???  (See answer below!)

ANSWER:  (Because they are FACTORS on the left side, but TERMS on the right side!!)

17. 

[Notice the similarity between this exercise and #15.  At first it looks like exactly the same problem.  The difference is that in #15 you were solving for , whereas in #17 you are solving for .  The problems begin the same, but they are NOT the same! ]

              ,  solve for .

 There are two ways to begin this problem, both of which result in the same first step.  You might want to “undo” the fraction  by multiplying both sides of the equation by , which looks like this:

The result is EXACTLY the same:

Now, to solve for , you must get all the   terms on one side, and the “non-” terms on the other side.  To do this subtract  from each side, exactly as in #15.

Now, to solve for , you must factor the common factor of , so as to get the  in one place. 

Divide out the factors of .

19.    Given    ,    solve for S.

First, find the LCD, which is FSU (to all the Florida Gator and Miami Hurricane fans,  GO  FLORIDA STATE!)                         

In the first position, the F divides out, leaving SU. 

In the second position the S divides out, leaving UF. 

In the third position, the U divides out, leaving FS.

Now, in order to solve for S, you have to get all the S terms on one side of the equation.  You can do that by subtracting FS from each side of the equation.

Now, to solve for S, you have to factor out the S on the left side of the equation:

and divide both sides by :  

IMPORTANT NOTE:  This problem is very much like my own career, in that I started (and graduated!) at FSU and then ended up (and graduated also!) at UF—except that I did NOT change colors!!

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