113 1.2 Applications and Modeling with Linear Equations Geometry Problems x x x + 3 x + 3 Original square Side is increased by 3. x and x + 3 are in centimeters. Figure 1 EXAMPLE 1 Finding the Dimensions of a Square If the length of each side of a square is increased by 3 cm, the perimeter of the new square is 40 cm more than twice the length of each side of the original square. Find the dimensions of the original square. SOLUTION Step 1 Read the problem. We must find the length of each side of the original square. Step 2 Assign a variable. Since the length of a side of the original square is to be found, let the variable represent this length. Let x = the length of a side of the original square in centimeters. The length of a side of the new square is 3 cm more than the length of a side of the old square. Then x + 3 = the length of a side of the new square. See Figure 1. Now write a variable expression for the perimeter of the new square. The perimeter of a square is 4 times the length of a side. Thus, 41x + 32 = the perimeter of the new square. Step 3 Write an equation. Translate the English sentence that follows into its equivalent algebraic equation. The new more twice the length of each perimeter is 40 than side of the original square. (11)11* 5 (11111111111111111)11111111111111111* 41x + 32 = 40 + 2x Step 4 Solve the equation. 4x + 12 = 40 + 2x Distributive property 2x = 28 Subtract 2x and 12. x = 14 Divide by 2. Step 5 State the answer. Each side of the original square measures 14 cm. Step 6 Check. Go back to the words of the original problem to see that all necessary conditions are satisfied. The length of a side of the new square would be 14 + 3 = 17 cm. The perimeter of the new square would be 41172 = 68 cm. Twice the length of a side of the original square would be 21142 = 28 cm. Because 40 + 28 = 68, the answer checks. S Now Try Exercise 15. Motion Problems PROBLEM-SOLVING HINT In a motion problem, the three components distance, rate, and time are denoted by the letters d, r, and t, respectively. (The rate is also called the speed or velocity. Here, rate is understood to be constant.) These variables are related by the following equations. d =rt, and its related forms r = d t and t = d r LOOKING AHEAD TO CALCULUS In calculus the concept of the definite integral is used to find the distance traveled by an object traveling at a non-constant velocity.
RkJQdWJsaXNoZXIy NjM5ODQ=