1. Find Two consecutive odd integers whose product is 1 less than their sum.

x = 1st odd integer

x + 2 = 2nd odd integer

x(x + 2) = x + x + 2 - 1 {product is one less than sum}

x² + 2x = 2x + 1 {used distributive property and combined like terms}

x² = 1 {subtracted 2x from both sides}

x = 1 or x = -1 {took square root of both sides}

x + 2 = 3 or x + 2 = 1 {substituted 1 and -1 into x + 2}

the two consecutive odd integers are

**1 and 3**.....also

**-1 and 1**3. The length of a rectangle is 3 feet more than twice the width. If the area of the rectangle is 20 square feet, what are the dimensions of the rectangle?

x = width

2x + 3 = length {length is 3 feet more than twice width}

Area of a rectangle is length x width

x(2x + 3) = 20 {area is length x width}

2x² + 3x = 20 {used distributive property}

2x² + 3x - 20 = 0 {subtracted 20 from both sides}

(2x - 5)(x + 4) = 0 {factored into two binomials}

2x - 5 = 0 or x + 4 = 0 {set each factor equal to 0}

2x = 5 or x = -4 {added 5 and subtracted 4, respectively}

x = 5/2 or x = -4 {divided left equation by 2}

*width cannot be negative*x = 5/2

2x + 3 = 8 {substituted 5/2, in for x, into 2x + 3}

**width = 5/2 ft
**

length = 8 ft*- Algebra House*
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