Question: ACTIVITY 3: FORM ME PROPORTIONALLY Direction: Solve the following problems involving proportion. 1. If 2 . find BY 2. Find the value of m if=

ACTIVITY 3: FORM ME PROPORTIONALLY Direction:ACTIVITY 3: FORM ME PROPORTIONALLY Direction:ACTIVITY 3: FORM ME PROPORTIONALLY Direction:ACTIVITY 3: FORM ME PROPORTIONALLY Direction:ACTIVITY 3: FORM ME PROPORTIONALLY Direction:
ACTIVITY 3: FORM ME PROPORTIONALLY Direction: Solve the following problems involving proportion. 1. If 2 . find BY 2. Find the value of m if= = Se-61-20 2 3.The Plantitas Mom buy "Money plant" for P 200.00. A week later, she is no longer interested in the plant and she posted in her FB page for online selling. If the ratio of a cost price to the selling price is 2:3, what is the selling price to be posted on her FB page?TOPIC: SOLVING PROBLEMS INVOLVING PROPORTIONS I. INTRODUCTION Knowing the rules or properties will guide us on how to arrive at the best solution or answer. As you go through this lesson, think of how the fundamental theorems of proportionality can be applied in order to solve problems and issues involving proportions. Learn and have fun accomplishing all the activities in this module. II. LEARNING OUTCOMES In this lesson, you will be able to: . apply the fundamental theorems of proportionality to solve problems involving proportions. III. LEARNING EXPERIENCES AND SELF-ASSESSMENT ACTIVITIES (SAA) FUNDAMENTAL RULES OF PROPORTION If w: x = y: z, then * * provided that x # 0; z # 0. PROPERTIES OF PROPORTION Cross-multiplication If = = 4. then wz = ry;x # 0,z # 0. Property Alternation Property If = = ! then= = !;x # 0,y # 0,z # 0. Inverse Property If= = 2. then! = =:w # 0,x # 0,y # 0,z # 0. Addition Property If = = 2. then"* = 2+4; x # 0,z # 0. Subtraction Property If = = 2. then -1= 2-; ;x # 0,z # 0. If = = = = 4 then= ="= = Sum Property Where k is a constant at proportionality and v # 0,x # 0,z # 0. Study the following examples: Example 1: Solve 12 using the fundamental rule of proportion.Example 1: Solve using the fundamental rule of proportion. Solution: Using Cross-Multiplication Using Alteration Using Inverse Property 12 12 12 3 12 3 12 12n or 24 120 = (8)(3) 24 = 128 120 = 24 Using Addition Property Using Subtraction Property Using Sum Property B 3 12 12 Ti 1+3 8 + 12 -3 8- 12 12 12 12 12 (m + 3) = 3(8 + 12) 12 (n - 3) -3(8 -12) 12n + 36 = 3(20) 12n -36 =3(-4) 12n + 36 = 60 12n - 36 = -12 12 3+ 12 12n = 24 121 = 24 3k 3k # 12k 1 =2 n =2 3+ 12 When k is considered in the sum property of the original proportion, the following proportions can be formed: n / 3 = k -+ n = 3k and 8 / 12 = k -+ 8 = 12k. When we substitute the 3k to n and 12k to 8 in the original proportion, all ratios in the proportion are equal to k, representing the equality of the ratios in the proportion. Example 2: If min = 4:3, find 3m - 2n 1 3m + n. Solution: "X4 3m =4mm= Using m = 10 Im- Im Therefore. 3m - 2n :3m + n = 2:5. Example 3: Page 3 / 5Using Addition Property Using Subtraction Property Using Sum Property B 8 12 12 12 1+3 8 + 12 8 - 12 12 3 12 3 12 12 (n + 3) = 3(8 + 12) 12 (n - 3) = 3(8 - 12) =k -1 = 3k, B 12 k - 8 = 12k 12n + 36 = 3(20) 12n - 36 = 3(-4) 8 n+ 8 = k 12n + 36 = 60 12n - 36 = -12 3 12 3+ 12 12n = 24 12n = 24 3k 12k 3k + 12k = k n = 2 n = 2 3 12 3 + 12 K When k is considered in the sum property of the original proportion, the following proportions can be formed: n / 3 = k - n = 3k and 8 / 12 = k - 8 = 12k. When we substitute the 3/ to n and 12k to 8 in the original proportion, all ratios in the proportion are equal to k, representing the equality of the ratios in the proportion. Example 2: If min = 4:3, find 3m - 2n : 3m + n. Solution: "X- + 3m = 4n - m= Using m = in An-20 20 3m+ 3+n in+n Therefore. 3m - 2n : 3m + n = 2:5. Example 3: 5q-or-42 find x. x Solution: Let ! = [=4= $9-6r-75 = k. Then. q = 2k,r = 3k.s = 4k, and 5q - 6r - 7s = kx. 5(2k) - 6(3k) - 7(4k) = kx 10k - 18k - 28k = kx -36k = kx -36 = xExample 3: 5q-or- Solution: = = k. Then, q = 2k,r = 3k.s = 4k, and 5q - 6r - 7s = kx. 5(2k) - 6(3k) - 7(4k) = kx 10k - 18k - 28k = kx -36k = kx -36 =X Example 4: If 22 doses of vaccines have been administered for every 100 persons. How many doses can be ordered for 5000 frontliners? Solution: Let x be the number of doses of vaccines. Based from the problem, the number of doses of vaccines needed to the number of persons are in the ratio 22:100. Then, 22 100 5000 22(5000) = 100x 110,000 = 100x 1100 = x Therefore, 1100 doses are needed to be ordered. IV. LEARNING RESOURCES For additional knowledge about this topic, you may watch the video in this link: (https://youtube/7gvvzCcnYfk) References: 1. Mathematics Learner's Material (Grade 9), pp. 358-361 2. E-math 9 Geometry, pp. 264-267

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