Machinery's Handbook, 31st Edition
ALGEBRA 25 Solving an Equation for an Unknown.— Solving an equation for an unknown (say, x ) requires isolating x from the other terms. This is accomplished by applying techniques of inverse operations , combining like terms , and factoring , as explained throughout ALGE- BRA , GEOMETRY , and TRIGONOMETRY . The simplest equations contain only one or two variables, such as linear equations and other polynomials in two dimensions. Others may contain many variables, such as for- mulas in physics and engineering. A formula is an equation that determines a physical quantity based on other known quantities. An example is area of a rectangle, A = lw , where l is length and w is width in like units. Another is horsepower transmitted by belting, P = SVW /33,000, where S , V , and W are, respectively, working stress of the belt, belt velocity, and belt width, all in appropriate units. The following examples give an overview of equation solving with general terms A, B, C , and D . The same processes apply for solving polynomial and other algebraic equations for an unknown. Solving by Adding or Subtracting : Given: B – C = A – D To solve for A : B – C + D = A, that is, A = B – C + D To solve for B : B = A – D + C To solve for C : B = A – D + C → B – A + D = C, that is, C = B – A + D To solve for D : B – C + D = A → D = A – B + C
In the last two, C and D are moved first so they do not have a minus sign before them in the answer. But keeping in mind that subtracting a term is the same as adding its negative (see Integers on page 3), another way to solve for C or D is given below. The sub- traction “minus C ” is treated as “negative C .” Then, multiplying through by – 1, all signs change but equality is maintained, leaving C : Solve for C : B – C = A – D → –C = A – D – B → (–1)(–C ) = (–1)( A – D – B ) → C = –A + D + B Solve for D : B – C = A – D → B – C – A = –D → (–1)( B – C – A ) = (–1)( –D ) → –B + C + A = D , or D = A – B + C Changing the sign of each term by multiplying through by –1 is a common technique. Solving by Multiplying or Dividing: A variable that is a factor or a divisor within a term is isolated using division ( multiplying by the reciprocal ) or multiplication ( cross- multiplying ). Example : In the equation AB C D = , each variable is isolated using multiplication. To isolate A , multiply each side by the reciprocal of C B AB C C B D A C B D CD B A CD B = = = 1 , so × × × = → B C : C B AB C C B D A C B D CD B A CD B = = = 1 , so × × × = → B C : To isolate B , multiply each side by the reciprocal of D B → B = = = , so × × × = :
C A
AB C
C A
C A
D CD A 1
CD A
A C
C A
AB C
C A
C A
D CD A 1
CD A
A C
D B →
:
B
=
=
=
, so
×
×
× =
AB C
To isolate C , multiply each side by C : C = × C D Then divide by D (multiply by the reciprocal of D ): AB D CD D AB D × 1 1 1 = × = ×
AB CD → =
AB D
1 → Equations with more complexity require more steps to solve for an unknown. = C C , s o
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