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Mastering Formula Transformations | Learn to Change the Subject Skillfully

Understand Changing the Subject of a Formula through algebraic rearrangements step by step. Examples with add, subtract, multiply, divide, squares, and square roots. Practice and master the skill.

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Mastering Formula Transformations | Learn to Change the Subject Skillfully

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  1. Olympic College Topic 17 Changing The Subject Of A Formula Topic 17 Changing The Subject of a Formula Definition: When you write a formula like:– 1. 2. 3. A V E = = = Lb  r2h mc2 A is called the subject of the formula. V is called the subject of the formula. E is called the subject of the formula. For each of this formula it is possible by using algebra to rearrange them so that they are written with a new subject , so for example: A b V h E c 2 1. 2. 3. L r m = = = where l is the new subject of the formula. where r is the new subject of the formula. where m is the new subject of the formula. To be able to do this we need to use algebra. The basic idea is to rearrange the formula so that the new subject of the formula is on its own . The one basic rule you use is that you do the same process to both sides of the equal sign . The main method to changing the subject is to try and get the new variable on its own , by slowly removing one of the other variables from it by doing some calculation to both sides such as add a to both sides or taking the square root . It is important that the order that you do this is correct. A. Formula with add and subtract Example 1: Change the subject of the following formula to the desired variable. (a) (b) (c) T P P = = = T m+n a – b a – b = to m. to a. to b. m+n Solution (a): T – n m = = m T – n Subtract n to both sides Switch the order a – b Solution (b): P = P+b a = = a P+b Add b to both sides Switch the order a – b Solution (c): P = P – a –P + a b – b b a – P = = = Subtract a to both sides Multiply both sides by – 1 Switch the order Page | 1

  2. Olympic College Topic 17 Changing The Subject Of A Formula B. Formula with Multiply and Divide Example 2: Change the subject of the following formula to the desired variable. (a) C =  D to D (b) (c) (d) (e) (f) (g) F F G H R R = = = = = = C C  D F 3x + c 3x + c ½x ¾y V I V I = = = = to c. to x to x. to y to V to I  D D C  3x + c Solution (a): Solution (b): Divide both sides by π Switch the order F – 3x = c Subtract 3x to both sides F – 3x c = Switch the order Solution (c): F = 3x + c F – c F  c 3 x = = = 3x x F  c 3 Subtract 3x to both sides Divide both sides by 3 Switch the order Solution (d): G = ½ x 2G = x Multiply both sides by 2 x = 2G Switch the order Solution (e): H = ¾y 4H = = 3y y Multiply both sides by 4 Divide both sides by 3 y = Switch the order V I Solution (f): R = IR = V Multiply both sides by I V = IR Switch the order V I Solution (g): R = IR = V Multiply both sides by I V R I = Divide both sides by R Page | 2

  3. Olympic College Topic 17 Changing The Subject Of A Formula C. Formula with Squares and Square Roots Example 3: Change the subject of the following formula to the desired variable.  r2 ab 2g2 c  2d 2  r2 = r2 = = r A =  ab = (a) (b) (c) (d) A C B V A A  A  r C = = = = to r. to b to g to d. Solution (a): Solution (b): Divide both sides by π Take the square root of both sides. Switch the order C2 b = = = ab b C 2 a Square both sides. Divide both sides by a Switch the order 2 a C 2g2 Solution (c): B = B 2 B 2 g g2 g = = = Divide both sides by 2 Take the square root of both sides. Switch the order B 2 c  2d 2 Solution (d): V = V2 V2 – c V 2  c 2 2 2 d c + 2d2 2d2 d2 d V 2  c 2 = = = = = Square both sides. Subtract c from both sides Divide both sides by 2 Take the square root of both sides. Switch the order V  c Page | 3

  4. Olympic College Topic 17 Changing The Subject Of A Formula D. Applications of Changing the Subject of a Formula. Example 4: The Volume of a cone of radius r and height h is given by the formula 1 3  r 2 h V = (a) If the radius is 3 inches and the height is 10 inches what is he volume of the cone? (b) It the volume of the cone is 45 in3 and its height is 8 in what is the radius of the cone? 1 3 1 3  r 2 h (3.14)(3) 2 (10) Solution (a): V V = = 94.2 in3 V = The cone would have a volume of 94.2 in3 Solution (b): To solve his problem we change the subject of the formula to r and then we can use this new formula to solve the given problem. 1 3 r2  r 2 h V 3V h = = Divide both sides by 3V h = r 3V h Take the square root of both sides Switch the order r = 3V h 3(45) 3.14(8) 135 25 . 12 5.374 2.32 inches We can now solve the problem r r r r r = = = = = So the radius of the cone will be 2.32 inches. Page | 4

  5. Olympic College Topic 17 Changing The Subject Of A Formula Example 5: The Energy that can be obtained from a mass of m kg is given by the formula E = mc2 where c is the speed of light. What is the formula for calculating the speed of light? mc2 Solution: E = E m E m c c2 c = = = Divide both sides by m Take the square root of both sides Switch the order E m E m The formula for finding the speed of light c is c = 4  r2 Example 6: The formula for calculating the surface area of a sphere of radius r is A = (a) What is the surface area of a sphere of radius 3.5 cm? (b) If a sphere has a surface area of 100 cm3 what is its radius? 4  r2 4(3.14)(3.5)2 153.86 Solution (a): A A A = = = The surface area of this sphere will be 153.86 cm2 Solution (b): To solve his problem we change the subject of the formula to r and then we can use this new formula to solve the given problem. 4  r2 A = A 4 A 4 r r2 r = = = Divide both sides by 4 Take the square root of both sides Switch the order A 4 A 4 100 4(3.14) 7.96 0.90 cm We can now solve the problem r r r r = = = = So the radius of this sphere will be 0.90 cm Page | 5

  6. Olympic College Topic 17 Changing The Subject Of A Formula Example 7: The formula for converting a temperature in Fahrenheit F into Centigrade C is C = (a) If the temperature is 75oF what will it be in Centigrade? (b) If the temperature is 750C what will it be in Fahrenheit? Solution (a): C C C = = = 23.9o C = The temperature will be 23.9o Centigrade. Solution (b): To solve his problem we change the subject of the formula to F and then we can use this new formula to solve the given problem. C = 5(F – 32) F – 32 F 9C = C = + 32 = Multiply both sides by 9 Divide both sides by 5 Add 32 to both sides F = + 32 Switch the order We can now solve the problem F F F = = = + 32 + 32 0 The temperature will be 167o Fahrenheit.. Page | 6

  7. Olympic College Topic 17 Changing The Subject Of A Formula Exercise 1 1. Change the subject of the given formula to the desired variable. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) C W Y C y y G M S S = = = = = = = = = = a+b x – y p – t 2  r mx + b mx + b 5x ¾T to b. to x. to t. to r. to b. to m. to x. to T to D to T (k) ax + by = c to y y – y1 m(x – x1) (l) = to m 2. Change the subject of the given formula to the desired variable. x2 + y2 = r2 (a) to x. ab a( x + 5) ¾A + ½B a  b c 2 (b) (c) (d) (e) C C E R = = = = to b to x to B to c 1 2 k x 2 (f) P = to x 3. The period of a pendulum T in seconds is given by the formula T = 6.28 Where L is the length of the pendulum chain in feet. (a) If the length of the pendulum is 19.6 feet what will be the period of the pendulum? (b) If the period of a pendulum is 12.56 seconds how long is L the length of the pendulum chain? 4. 5. The weight of an object in pounds is given by the formula W = of the object in inches. (a) What is the weight of an object if its length is 10 inches? (b) If the weight of the object is 10 pounds what is its length? The surface area of a cylinder is A = where L is the length (a) What is the surface area of a cylinder of radius 10 cm and height 5 cm? (b) If the surface area of a cylinder is 200 m2 and its radius is 2 m what is its height? Page | 7

  8. Olympic College Topic 17 Changing The Subject Of A Formula Solutions 1.(a) b = C – a 1.(e) b = y – mx 1.(i) D = ST (c) t = p – Y (g) x = (k) y = (b) x = W + y (f) m = (j) T = (d) r = (h) T = (l) m = C 2 a a  b R C a = r 2  y 2 4 E  3 A 2 – 5 2 P k 2.(a) x = 2.(d) B = (b) b = (e) c = (c) x = (f) x 3.(a) T = Period of pendulum 8.88 sec 3.(b) L = length of pendulum =39.2 feet 4.(a) The weight of the object is 5 pounds. 4.(b) The length of the object is 47.5 in 5.(a) The surface area is 942 m2. 5.(b) he height is 13.9 m. Page | 8

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