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Activity 3         UNDERSTANDING ALGEBRA MULTIPLICATION

                   Problem 3.1Multiplication of algebraic forms in

                   everyday life


               Pak Idris has a square apple orchard and Pak Tohir has a rectangular orange orchard. The
               length of Pak Tohir's orange orchard is 20 m more than the length of the side of Pak Idris'

               apple orchard. Meanwhile, the width is 15 m less than the length of the side of Pak Idris'

               apple orchard. If it is known that the area of Pak Idris and Pak Tohir's gardens is the same,
               then determine the area of Pak Idris' apple orchard?


               Solution:

               To solve this problem, you can assume the length of the side of Mr. Idris's apple orchard with

               a variable, for example the variable x. Mr. Tohir's orange orchard is 20 meters longer than the
               side of the apple orchard. It can be written as x + 20. The width is 15 meters less than the length

               of the side of Mr. Idris's apple orchard. It can be written as x 15. As we know that the area of

               a rectangle is length × width. But in the problem of determining the length of the side of the
               garden, we have a little difficulty because what is being multiplied is an algebraic form. In this

               problem, the area of Pak Tohir's garden is the product of x + 20 with x 15.

               Pak Tohir's garden area can be written in algebraic form


               Area = length × width = (x + 20) × (x 15)


                                     = x2 15x + 20x 300

                                     = x2 + 5x 300 unit area


               So, the area of Mr. Tohir's garden is x2 + 5x 300 units of area.

               Since it is known that the area of Mr. Idris's apple orchard is the same as the area of Mr. Tohir's

               orange orchard, then we get: The area of Mr. Idris's apple orchard  =  area  of  Mr.  Tohir's

               orange garden

                                               2
                                     (x)2   = x + 5x – 300

                                     x2     = x2 + 5x – 300


                                     x2- x2 = 5x – 300

                                     0      = 5x – 300


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