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                      Equations may contain different symbols, each represent-  Since m/m is equal to 1, the a remains by itself on the right side.
                   ing some unknown quantity. In science, the phrase  solve the   For convenience, the whole equation may be flipped to move
                   equation means to perform certain operations with one sym-  the unknown to the left side.
                   bol (which represents some variable) by itself on one side of the                F _

                   equation. This single symbol is usually, but not necessarily, on             a =
                                                                                                    m
                   the left side and is not present on the other side. For example,
                   the equation F = ma has the symbol F on the left side. In sci-  Thus, a quantity that indicated a multiplication (ma) was re-
                   ence, you would say that this equation is solved for F. It could   moved from one side by an inverse operation of dividing by m.
                   also be solved for m or for a, which will be considered shortly.      Consider the following inverse operations to “cancel” a quan-
                   The equation F = ma is solved for F, and the indicated oper-  tity from one side of an equation, moving it to the other side:
                   ation is to multiply m by a because they are in the form ma,   If the Indicated Operation  Perform This Inverse
                   which means the same thing as m × a. This is the only indicated   of the Symbol You Wish   Operation on Both Sides
                     operation in this equation.                          to Remove Is:           of the Equation
                      A solved equation is a set of instructions that has an order
                                                                          multiplication          division
                   of indicated operations. For example, the equation for the rela-
                   tionship between a Fahrenheit and Celsius temperature, solved   division       multiplication
                   for °C, is C = 5/9(F – 32). A list of indicated operations in this   addition  subtraction
                   equation is as follows:                                subtraction             addition
                     1. Subtract 32° from the given Fahrenheit temperature.  square               square  root
                     2.  Multiply the result of (1) by 5.                 square root             square
                     3.  Divide the result of (2) by 9.
                   Why are the operations indicated in this order? Because the
                   bracket means 5/9 of the quantity (F – 32°). In its expanded form,   EXAMPLE A.1
                   you can see that 5/9(F – 32°) actually means 5/9(F) – 5/9(32°).
                                                                          The equation for finding the kinetic energy of a moving body is
                   Thus, you cannot multiply by 5 or divide by 9 until you have     2
                                                                          KE = 1/2 mv . You need to solve this equation for the velocity, v.
                   found the quantity (F – 32°). Once you have figured out the
                   order of operations, finding the answer to a problem  becomes
                   almost routine as you complete the needed operations on both   SOLUTION
                   the numbers and the units.                             The order of indicated operations in the equation is as follows:

                                                                            1. Square v.
                                                                                     2
                                                                            2. Multiply v  by m.
                    A.2 SOLVING EQUATIONS                                   3.  Divide the result of (2) by 2.
                                                                          To solve for v, this order is reversed as the “canceling operations” are
                   Sometimes it is necessary to rearrange an equation to move a
                                                                          used:
                   different symbol to one side by itself. This is known as solving
                   an equation for an unknown quantity. But you cannot simply   Step 1:   Multiply both sides by 2.
                   move a symbol to one side of an equation. Since an equa-                         1 _

                                                                                               KE =       mv 2
                   tion is a statement of equivalence, the right side has the same                  2
                   value as the left side. If you move a symbol, you must perform                   2 _

                                                                                              2KE =     mv 2

                   the operation in a way that the two sides remain equivalent.                     2
                   This is  accomplished by “canceling out” symbols until you                  2KE = mv 2
                   have the unknown on one side by itself. One key to under-
                                                                          Step 2:   Divide both sides by m.
                   standing the canceling operation is to remember that a frac-
                   tion with the same number (or unit) over itself is equal to 1.              _ _     2
                                                                                               2KE
                                                                                                     mv



                                                                                                        =
                   For example, consider the equation F = ma, which is solved                   m     m
                   for F. Suppose you are considering a problem in which F and                  _     2
                                                                                                2KE
                                                                                                          = v
                   m are given, and the unknown is  a. You need to solve the                     m
                   equation for a so it is on one side by itself. To eliminate the m,   Step 3:   Take the square root of both sides.
                   you do the  inverse of the indicated operation on m, dividing
                                                                                               _
                                                                                                2KE

                                                                                                         2
                                                                                                       v

                   both sides by m. Thus,                                                         √      m              =     √
                                                                                                 2KE

                                           F = ma                                                   √     _             = v
                                                                                                  m
                                         F _ _
                                             ma



                                             =                                                    or
                                         m    m
                                         F _
                                                                                                    _
                                                                                                    2KE



                                         m   = a                                               v =     √      m
                   624     APPENDIX A Mathematical Review                                                                 A-2
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