Topic 17 Changing The Subject of a Formula

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Bertram Horton
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1 Topic 17 Changing The Subject of a Formula Definition: When you write a formula like: 1. = Lb is called the subject of the formula.. = r h is called the subject of the formula.. E = mc 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: 1. L = where l is the new subject of the formula. b. h where r is the new subject of the formula. E. m = c 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.. Formula with add and subtract Example 1: Change the subject of the following formula to the desired variable. (a) T = m + n to m. (b) P = a b to a. (c) P = a b to b. Solution (a): T = m + n T n = m Subtract n to both sides m = T n Solution (b): P = a b P + b = a dd b to both sides a = P + b Solution (c): P = a b P a = b Subtract a to both sides P + a = b Multiply both sides by 1 b = a P Page 1

2 B. Formula with Multiply and Divide Example : Change the subject of the following formula to the desired variable. (a) C = D to D (b) F = x + c to c. (c) F = x + c to x (d) G = ½x to x. (e) H = ¾ y to y (f) R = I to (g) R = I to I Solution (a): C = D C = D Divide both sides by π D = C Solution (b): F = x + c F x = c Subtract x to both sides c = F x Solution (c): F = x + c F c = x Subtract x to both sides F c = x Divide both sides by x = F c Solution (d): G = ½ x G = x Multiply both sides by x = G Solution (e): H = ¾ y 4H = y Multiply both sides by 4 = y Divide both sides by y = Solution (f): R = I IR = Multiply both sides by I = IR Solution (g): R = I IR = Multiply both sides by I I = Divide both sides by R R Page

3 C. Formula with Squares and Square Roots Example : Change the subject of the following formula to the desired variable. (a) = r to r. (b) C = ab to b (c) B = g to g (d) = c d to d. Solution (a): = r = r Divide both sides by π = r Take the square root of both sides. Solution (b): C = ab C = ab Square both sides. C = b Divide both sides by a a b = Solution (c): B = g C a B = g Divide both sides by B = g Take the square root of both sides. g = B Solution (d): = c d = c + d Square both sides. c = d Subtract c from both sides c = d Divide both sides by c d = = d Take the square root of both sides. c Page

4 D. pplications of Changing the Subject of a Formula. Example 4: The olume of a cone of radius r and height h is given by the formula 1 = r h (a) If the radius is inches and the height is 10 inches what is he volume of the cone? (b) It the volume of the cone is 45 in and its height is 8 in what is the radius of the cone? 1 Solution (a): = r h 1 = (.14)() (10) = 94. in The cone would have a volume of 94. in 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 r h h = r Divide both sides by h = r Take the square root of both sides h We can now solve the problem So the radius of the cone will be. inches. h (45).14(8) inches Page 4

5 Example 5: The Energy that can be obtained from a mass of m kg is given by the formula E = mc where c is the speed of light. What is the formula for calculating the speed of light? Solution: E = mc E = c m Divide both sides by m E = c Take the square root of both sides m c = E m The formula for finding the speed of light c is c = Example 6: The formula for calculating the surface area of a sphere of radius r is = 4 r (a) What is the surface area of a sphere of radius.5 cm? (b) If a sphere has a surface area of 100 cm what is its radius? Solution (a): = 4 r = 4(.14)(.5) = The surface area of this sphere will be cm E m 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 r = r Divide both sides by 4 4 = r Take the square root of both sides 4 4 We can now solve the problem So the radius of this sphere will be 0.90 cm (.14) cm Page 5

6 Example 7: The formula for converting a temperature in Fahrenheit F into Centigrade C is Solution (a): C = C = (a) If the temperature is 75 o F what will it be in Centigrade? (b) If the temperature is 75 0 C what will it be in Fahrenheit? C = C = C =.9 o The temperature will be.9 o 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 = 9C = 5(F ) Multiply both sides by 9 C = F Divide both sides by 5 + = F dd to both sides F = + We can now solve the problem F = + F = + F = The temperature will be 167 o Fahrenheit.. 0 Page 6

7 Exercise 1 1. Change the subject of the given formula to the desired variable. (a) C = a + b to b. (b) W = x y to x. (c) Y = p t to t. (d) C = r to r. (e) y = mx + b to b. (f) y = mx + b to m. (g) G = 5x to x. (h) M = ¾ T to T (i) S = to D (j) S = to T (k) ax + by = c to y (l) y y 1 = m(x x 1 ) to m. Change the subject of the given formula to the desired variable. (a) x + y = r to x. (b) C = ab to b (c) C = a( x + 5) to x (d) E = ¾ + ½B to B (e) R = a b c to c (f) P = 1 k x to x. The period of a pendulum T in seconds is given by the formula T = 6.8 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 1.56 seconds how long is L the length of the pendulum chain? 4. The weight of an object in pounds is given by the formula W = where L is the length 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? 5. The surface area of a cylinder is = (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 00 m and its radius is m what is its height? Page 7

8 Solutions 1.(a) b = C a (b) x = W + y (c) t = p Y (d) 1.(e) b = y mx (f) m = (g) x = (h) T = 1.(i) D = ST (j) T = (k) y = (l) m =.(a) x =.(d) B = r y (b) b = 4E (e) c = C (c) x = a a b R (f) x = C 5 a P k.(a) T = Period of pendulum 8.88 sec.(b) L = length of pendulum =9. 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 94 m. 5.(b) he height is 1.9 m. Page 8

Topic 16 - Radicals. 1. Definition of a Radicals

Topic 16 - Radicals. 1. Definition of a Radicals Topic 16 - Radicals 1. Definition of a Radicals Definition: In Mathematics an efficient way of representing repeated multiplication is by using the notation a n, such terms are called exponentials where