use of algebraic formulas and equations - skilled trades …：使用代数公式和方程-熟练工种…课件

Copyright 2013 - Michael Bush,1,Session 11 Complex Equations,Industrial Application of Basic Algebraic Principles,Industrial Application of Basic Algebraic Principles Session 10 In-Class,151 Equations Consisting of Combined Operations 383 152 Solving for the Unknown in Formulas 386 153 Substituting Values Directly in Given Formulas 387 154 Rearranging Formulas 390 UNIT EXERCISE AND PROBLEM REVIEW 394,Complex Equations,2,Copyright 2013 - Michael Bush,Equations Consisting of Combined Operations Page 382,Industrial Application of Basic Algebraic Principles Session 11 In-Class,3,Copyright 2013 - Michael Bush,Typically, in actual occupational applications, formulas are used and equations are developed that are complex. Each occupation typically has a select set of formulas that apply to their trade. The rules of algebra apply to all.,Complex equations require the use of two or more principles of equality for their solutions,Equations Consisting of Combined Operations Page 383,Industrial Application of Basic Algebraic Principles Session 11 In-Class,4,Copyright 2013 - Michael Bush,These equations, or job-specific formulas, require the use of two or more principles of equality for their solutions.,Remember Formulas are simply equations with a purpose.,Equations Consisting of Combined Operations Page 383,Industrial Application of Basic Algebraic Principles Session 11 In-Class,5,Copyright 2013 - Michael Bush,Use of proper procedure results in the unknown standing alone on one side of the equation with its value on the other.,For example, the equation 0.13x - 4.73(x + 6.35) = 5.06x - 2.87 requires a definite step-by-step procedure for determining the value of x.,Remove parentheses Combine like terms on each side of equation Get all unknown terms on one side and all known terms on the other using addition and subtraction principle Combine like terms Apply multiplication and division principles of equality Apply power and root principles of quality,Industrial Application of Basic Algebraic Principles Session 11 In-Class,6,Copyright 2013 - Michael Bush,Procedure for solving equations consisting of combined operations Page 383,Industrial Application of Basic Algebraic Principles Session 11 In-Class,7,Copyright 2013 - Michael Bush,Procedure for solving equations consisting of combined operations Page 383,Always solve for a positive unknown. A positive unknown may equal a negative value, but a negative unknown is not a solution. For example, x = -10 is correct, but -x = 10 is incorrect.,When solving equations where the unknown remains a negative value, multiply both sides of the equation by 1. Multiplying a negative unknown by 1 results in a positive unknown.,Get the unknown term all by itself on one side of the equation and all other terms on the other side of the equation It must be a whole amount (no fractions or decimals) Positive value (no negative sign),Industrial Application of Basic Algebraic Principles Session 11 In-Class,8,Copyright 2013 - Michael Bush,General rule for solving equations,1. Solve for x: 8x + 12 = 5 First subtract 12 from both sides,Industrial Application of Basic Algebraic Principles Session 11 In-Class,9,Copyright 2013 - Michael Bush,Example Page 383,x = 5,Next divide both sides by 8,2. Solve for D: -18D + 4D = 3D - 5D + 19 + 5,Industrial Application of Basic Algebraic Principles Session 11 In-Class,10,Copyright 2013 - Michael Bush,Example Page 384,D = -2,Divide both sides by -12,Combine like terms,-14D = - 2D + 24,Add 2D to both sides,+2D +2D,-12D = 24,3. Solve for y: 14y 6(y 3) = 22,14y 6y + 18 = 22,8y + 18 = 22, 18 18,Industrial Application of Basic Algebraic Principles Session 11 In-Class,11,Copyright 2013 - Michael Bush,Example Page 384,Remove parentheses,Combine like terms,Subtract 18 from both sides,8y = 4,Divide both sides by 8,Add 36.5 to both sides,Industrial Application of Basic Algebraic Principles Session 11 In-Class,12,Copyright 2013 - Michael Bush,Example Page 385,4. Solve for x:,Multiply both sides by 6,Extract square root of both sides,5. Solve for R:, 80 80,Industrial Application of Basic Algebraic Principles Session 11 In-Class,13,Copyright 2013 - Michael Bush,Example Page 385,Multiply by Denominator,Subtract 80 from both sides,4R = 50,Divide both sides by 4,Industrial Application of Basic Algebraic Principles Session 11 In-Class,14,Copyright 2013 - Michael Bush,Exercise 15-1 Page 386,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,2. 10M + 5 + 4M = 89,14M + 5 = 89,14M = 84,M = 6,Combine like terms,Subtract 5 from both sides,Divide both sides by 14,Industrial Application of Basic Algebraic Principles Session 11 In-Class,15,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,4. 4B - 7 = B + 14,4B = B + 21,3B = 21,B = 7,Add 7 to both sides,Divide both sides by 3,Subtract B from both sides,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,16,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,6. 6N + 4 = 84 + N,6N = 80 + N,5N = 80,N = 16,Subtract 4 from both sides,Divide both sides by 5,Subtract N from both sides,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,17,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,8. 12 - (-x + 8) = 18,x = 14,Subtract 4 from both sides,Remove parenthesis,12 + x - 8 = 18,Combine like terms,4 + x = 18,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,18,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,10. 6 = -(2 + C) - (4 + 2C),12 = -3C,Add 6 to both sides,Remove parenthesis,6 = -2 - C - 4 - 2C,Combine like terms,6 = -6 3C,Divide both sides by -3,-4 = C,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,19,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,12. 0.29E = 9.39 - 0.01E,Add 0.01E to both sides,0.28E = 9.39,Divide both sides by 0.28,E = 31.3,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,20,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,14.,Subtract 8 from both sides,Multiply both sides by 7,P = -14.7,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,21,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,16.,Add 3 to both sides,Add 2D to both sides,Remove parenthesis,Combine like terms,Divide both sides by 8,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,22,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,18.,H2 = 4,Add 20 to both sides,Remove parenthesis,2H2 - 20 = H2 - 16,2H2 = H2 + 4,2H2 - 20 = (H + 4)(H - 4),Subtract H2 from both sides,Extract square root of both sides,H = 2,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,23,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,20.,4x + x2 = x2 - 15,Add x to both sides,Remove parenthesis,3x + x2 + 20 = x2 - x + 5,3x + x2 = x2 - x - 15,x(3 + x) + 20 = x2 - (x - 5),Subtract 20 from both sides,Divide both sides by 4,4x = - 15,Subtract x2 from both sides,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,24,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,22.,Combine like terms,Subtract 8F from both sides,Divide both sides by 3,Remove Parentesis,Subtract F2 from both sides,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,25,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,24.,Add 3 to both sides,Subtract B from both sides,Remove parenthesis,Extract square root from both sides,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,26,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,26.,Remove parenthesis,Square both sides,Combine,Divide both sides by 8,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,27,Copyright 2013 - Michael Bush,Solve for the unknown and check each of these combined operations equations. Round answers to 2 decimal places where necessary.,28.,Remove parenthesis,Square both sides,Combine,Divide both sides by 4.3,Exercise 15-1 Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,28,Copyright 2013 - Michael Bush,Round answers to 2 decimal places where necessary.,30.,Remove parenthesis,Extract cube root of both sides,Add 16.492 to both sides,Divide both sides by 8,Exercise 15-1 Page 386,Solving for the Unknown in Formulas Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,29,Copyright 2013 - Michael Bush,Occupational applications often require solving formulas in which all but one numerical value of the letter values are known. The unknown letter value can appear anywhere within the formula. There are two ways of solving for the unknown.,Solving for the Unknown in Formulas Page 386,Industrial Application of Basic Algebraic Principles Session 11 In-Class,30,Copyright 2013 - Michael Bush,Substituting known number values for letter values directly in the given formula and solving for the unknown. First rearranging the given formula so the unknown stands alone on one side of the formula, then substituting known number values for letter values, and solving for the unknown.,Solving for the Unknown in Formulas Page 387,Industrial Application of Basic Algebraic Principles Session 11 In-Class,31,Copyright 2013 - Michael Bush,You can do the Math then the Algebra You can do the Algebra then the Math,Substituting Values Directly in Given Formulas Page 387,Industrial Application of Basic Algebraic Principles Session 11 In-Class,32,Copyright 2013 - Michael Bush,To determine the numerical value of the unknown, write the original formula, substitute the known number values for their respective letter values, and simplify. Then follow the procedure given for solving equations consisting of combined operations.,Substituting into Formulas Page 387,To determine the value of the unknown Write the original formula Substitute the known number values for their respective letter values Simplify,Industrial Application of Basic Algebraic Principles Session 11 In-Class,33,Copyright 2013 - Michael Bush,An open belt pulley system is shown. The number of inches between the pulley centers is represented by x. The larger pulley diameter (D) is 6.25 in, and the smaller pulley diameter (d) is 4.25 in. The belt length (L) is 56.0 ins. The formula is: L = 3.14(0.5D + 0.5d) + 2x,Industrial Application of Basic Algebraic Principles Session 11 In-Class,34,Copyright 2013 - Michael Bush,Example Page 387,L = 3.14(0.5D + 0.5d) + 2x,56.0 = 3.140.5(6.25) + 0.5(4.25d) + 2x,56.0 = 3.14(5.25) + 2x,56.0 = 16.485 + 2x,39.515 = 2x,19.7575 = x,19.8 in. = x,A slot is cut in the circular piece shown. The piece has a radius (R) of 97.60 mm. The number of mm in the width is represented by W. Dimension A is 20.20 mm. Find W.,Industrial Application of Basic Algebraic Principles Session 11 In-Class,35,Copyright 2013 - Michael Bush,Example Page 388,5,990.76 = 9,525.76 - 0.2500W2,-3,535.00 = - 0.2500W2,-14,140 = W2,118.912= W,118.9 mm = W,Copyright 2013 - Michael Bush,36,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Exercise 15-3 Page 389,The formulas for this set of problems have been taken from various technical fields. Substitute the given numerical values for letter values and solve for the unknown.,Copyright 2013 - Michael Bush,37,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Exercise 15-3 Page 389,2. The inductive reactance is found from this formula.,where XL = inductive reactance, f = frequency, and L = inductance,Solve for L when f = 65 Hz and XL = 82 ohms. Round the answer to 2 significant digits.,Copyright 2013 - Michael Bush,38,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Exercise 15-3 Page 389,4. This formula is used to express a temperature in degrees Celsius as a temperature in degrees Fahrenheit.,Solve for C when the temperature is 28F. Round the answer to 2 significant digits.,Copyright 2013 - Michael Bush,39,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Exercise 15-3 Page 390,6. The accumulated amount in a savings account is found from this formula. Round the answer to 3 significant digits.,A = p + prt,where A = amount at end of period, p = principal invested, r = rate, and t = time in years.,$3312 = $1800 + ($1800)(0.07)t,A = p + prt,Solve for t when p = $1,800, r = 7%, and A = $3,312.,$3312 = $1800 + $126t,$1512 = $126t,t = 12 yr,12= t,Copyright 2013 - Michael Bush,40,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Exercise 15-3 Page 390,8. There are 4 cells connected in a series circuit. The internal resistance of each cell is 5.0 . The external circuit resistance is 8 . The current in the circuit is 4.0 A. Find the voltage of each cell.,where I = current in amperes, n = number of cells, E = voltage of one cell, r = internal resistance of one cell, and R = resistance of external circuit.,Copyright 2013 - Michael Bush,41,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Exercise 15-3 Page 390,10. This formula is used to find the impedance of a circuit.,where Z = impedance in ohms, R = resistance in ohms, and X = reactance in ohms.,Solve for X when R = 12 ohms and Z = 65 ohms. Round the answer to the nearest whole ohm.,652 = 122 + x2,4225 = 144 + x2,4081 = x2,63.882 = x,64 = x,Rearranging Formulas Page 390,Sometimes it is necessary to rearrange a formula to solve for another value The formula must be rearranged so that the unknown term is on one side of the equation and all other values are on the other side The formula is rearranged by using the same procedure that is used for solving equations consisting of combined operations,Industrial Application of Basic Algebraic Principles Session 11 In-Class,42,Copyright 2013 - Michael Bush,1. A screw thread is checked with a micrometer and 3 wires as shown. The measurement is checked by using this formula: M = D 1.5155p + 3w Solve the formula for w,Industrial Application of Basic Algebraic Principles Session 11 In-Class,43,Copyright 2013 - Michael Bush,Example Page 391,where M = measurement over the wires, D = major diameter, p = pitch, and w = wire size,M = D 1.5155p + 3w,Subtract D from both sides,M - D = -1.5155p + 3w,Add 1.5155p to both sides,M - D + 1.5155p = 3w,Divide both sides by 3.,Copyright 2013 - Michael Bush,44,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Example Page 391,2. The area of a ring shown is computed from this formula.,A = (R2 - r2),where A = area, = pi, R = outside radius, and r = inside radius. Solve for R.,Remove parentheses.,Add r2 to both sides,Divide both sides by .,Take the square root of both sides,3. A slot is cut in the circular piece shown. The piece has a radius (R) of 97.60 mm. The number of mm in the width is represented by W. Dimension A is 20.20 mm. Find W.,Industrial Application of Basic Algebraic Principles Session 11 In-Class,45,Copyright 2013 - Michael Bush,Example Page 392,Copyright 2013 - Michael Bush,46,Exercise 15-4 Page 392,Industrial Application of Basic Algebraic Principles Session 11 In-Class,The formulas for this set of problems are found in various occupational handbooks and manuals. Rearrange each to solve for the designated letter.,2. Solve for L,P = 2L + 2W,P 2W = 2L,Subtract 2W from both sides,Divide both sides by 2,Copyright 2013 - Michael Bush,47,Exercise 15-4 Page 393,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Rearrange the formula and solve for the designated letter.,4. Solve for w,Copyright 2013 - Michael Bush,48,Exercise 15-4 Page 393,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Rearrange the formula and solve for the designated letter.,6. Solve for M,Copyright 2013 - Michael Bush,49,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Rearrange the formula and solve for the designated letter.,8. Solve for r,Exercise 15-4 Page 393,Copyright 2013 - Michael Bush,50,Industrial Application of Basic Algebraic Principles Session 11 In-Class,Rearrange the formula and solve for the designated letter.,8. Solve for r,Exercise 15-4 Page 393,Copyright 2013 - Michael Bush,51,Industrial Application of Basic Algebraic Principles Session 11 In-Class,/,/,Exercise 15-4 Page 393,Copyright 2013 - Mich