# Algebra - Applications

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Applications of algebra are found everywhere. The principles of algebra are applied in all branches of mathematics, for instance, **calculus**, **geometry**, and **topology**. They are applied every day by men and women working in all types of business. As a typical example of applying algebraic methods, consider the following problem. A painter is given the job of whitewashing three billboards along the highway. The owner of the billboards has told the painter that each is a **rectangle**, and all three are the same size, but he does not remember their exact dimensions. He does have two old drawings, one indicating the height of each billboard is two feet less than half its width, and the other indicating each has a perimeter of 68 feet. The painter is interested in determining how many gallons of paint he will need to complete the job, if a gallon of paint covers 400 square feet. To solve this problem three basic steps must be completed. First, carefully list the available information, identifying any unknown quantities. Second, translate the information into symbolic notation by assigning variables to unknown quantities and writing equations. Third, solve the equation, or equations, for the unknown quantities.

Step one, list available information: (a) three billboards of equal size and shape, (b) shape is rectangular, (c) height is 2 feet less than 1/2 the width, (d) perimeter equals 2 times sum of height plus width equals 68 feet, (e) total area, height times width times 3, is unknown, (f) height and width are unknown, (g) paint covers 400 sq ft per gallon, (h) total area divided by 400 equals required gallons of paint.

Step two, translate. Assign variables and write equations.

Let: A = area; h = height; w = width; g = number of gallons of paint needed.

Then: (1) h = 1/2w−2 (from [c] in step 1); (2) 2(h+w) = 68 (from [d] in step 1); (3) A = 3hw (from [e] in step 1); (4) g = A/400 (from [h] in step 1). Step three, solve the equations. The right hand side of equation (1) can be substituted into equation (2) for h giving 2(1/2w−2+w) = 68. By the **commutative property**, the quantity in parentheses is equal to (1/2w+w−2), which is equal to (3/2w−2). Thus, the equation 2(3/2w−2)=68 is exactly equivalent to the original. Applying the **distributive property** to the left hand side of this new equation results in another equivalent expression, 3w−4 = 68. To isolate w on one side of the equation, add 4 to both sides giving 3w−4+4 = 68+4 or 3w = 72. Finally, divide the expressions on each side of this last expression by 3 to isolate w. The result is w = 24 ft. Next, put the value 24 into equation (1) wherever w appears, h = (1/2(24)−2), and do the arithmetic to find h = (12−2) = 10ft. Then, put the values of h and w into equation (3) to find the area, A = 3×10×24 = 720 sq ft. Finally, substitute the value of A into equation (4) to find g = 720/400 = 1.8 gallons of paint.

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