Warm Up Find the discounted price of a tent with a price of $89 and a discount of 15%. Find the final price of a pair of hiking boots with a price of $78, a discount of 10%, and a tax of 6%. On September 1, a stock sold for $46 per share and on October 1 it sold for $48.30 per share. What was the percent of change in the price of the stock? $75.65 $74.41 5% increase

Rectangle/Frame Problems Math 8H Problem Solving Day 3 Rectangle/Frame Problems Algebra 1 Glencoe McGraw-Hill JoAnn Evans

A rectangular swimming pool is 4 meters longer than it is wide A rectangular swimming pool is 4 meters longer than it is wide. The pool is surrounded by a cement sidewalk that is 1 meter wide. The area of the sidewalk is 32 m2. Find the dimensions of the pool. Let Statements: Let x = width of the pool Let x + 4 = length of the pool Let x + 6 = total length Let x + 2 = total width Verbal Sentence: Area of pool + area of sidewalk = total area x (x + 4) + 32 = (x + 6) (x + 2) x + 4 1 + x + 1 x x x + 4 1 + (x + 4) + 1

The pool is 5 meters wide and 9 meters long. x2 + 4x + 32 = x2 + 2x + Equation: x (x + 4) + 32 = (x + 6) (x + 2) 4x + 32 = 8x + 12 -4x -4x 32 = 4x + 12 -12 -12 20 = 4x 5 = x Solution: The pool is 5 meters wide and 9 meters long. x2 + 4x + 32 = x2 + 2x + 6x + 12

Let x = width of the painting Let x + 10 = length of the painting A painting is 10 cm longer than it is wide. It is mounted in a frame that is 1.5 cm wide. The area of the frame is 339 cm2. Find the dimensions of the painting. Let Statements: Let x = width of the painting Let x + 10 = length of the painting Let x + 3 = width of framed painting Let x + 13 = length of framed painting Verbal Sentence: Area of painting + area of frame = total area x ( x + 10) + 339 = (x + 3) (x + 13) x + 10 1.5 + (x+10) + 1.5 x 1.5 + x + 1.5

Equation: x ( x + 10) + 339 = (x + 3) (x + 13) x2 + 10x + 339 = x2 + 13x + 3x + 39 10x + 339 = 16x + 39 -10x -10x 339 = 6x + 39 -39 -39 300 = 6x 50 = x Solution: The painting is 50 cm wide and 60 cm long.

A rectangle is 4 meters longer than it is wide A rectangle is 4 meters longer than it is wide. If the length and width are both increased by 5 meters, the area is increased by 115 m2. Find the original dimensions. Let Statements: Let x = original width of rectangle Let x + 4 = original length of rectangle Let x + 5 = new rectangle width Let x + 9 = new rectangle length Verbal Sentence: original area + 115 = new larger rectangle area x x + 4 x + 5 (x + 4) + 5 x (x + 4) + 115 = (x + 5) (x + 9)

x2 + 4x + 115 = x2 + 9x + 5x + 45 4x + 115 = 14x + 45 -4x -4x Equation: x (x + 4) + 115 = (x + 5) (x + 9) x2 + 4x + 115 = x2 + 9x + 5x + 45 4x + 115 = 14x + 45 -4x -4x 115 = 10x + 45 -45 -45 70 = 10x 7 = x Solution: The original dimensions were 7 m wide and 11 m long.

A Mini Cooper and a truck heading for San Pedro on the same freeway left the Redwood Middle School parking lot at the same time. The Mini Cooper, which drove 20 mi/h faster than the truck, arrived in San Pedro after two hours. The truck arrived in San Pedro one hour later than the car. Find the rate of the car. RMS distance = distance San Pedro Let Statements: Let x = rate of Mini Cooper Let x - 20 = rate of truck Verbal Sentence: distance Mini Cooper drives = distance truck drives

Mini’s distance = Truck’s distance Equation: Solution: The car’s rate was 60 mi/h. Mini’s distance = Truck’s distance rt = rt x(2) = (x – 20)3 2x = 3x – 60 -3x -3x -x = -60 x = 60