Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
6. Solving Absolute Value Inequalities
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Exercise 29 Page 92

Use the formulas for the area of a triangle and the area of a rectangle.

Absolute Value Inequality: |1/2*4(x+6)-2*6|<2
Solution: -1

Practice makes perfect

Let A_T be the area of the triangle and A_R be the area of the rectangle. We are told that the difference between the areas of the figures is less than 2. We can write this as an absolute value inequality. |A_T-A_R|< 2 We will now find the area of each figure.

Area of Triangle

The area of a triangle with base b and height h can be found using the following formula. A_T=1/2bh We can see from the given diagram that the base of the triangle is 4 and its height is x+6. Let's substitute these values into the formula for the area of a triangle.
A_T=1/2bh
A_T=1/2* 4( x+6)

Area of Rectangle

The area of a rectangle with length l and width w can be found using the following formula. A_R=l w We can see from the given diagram that the length of the rectangle is 2 and its width is 6. Let's substitute these values into the formula for the area of a rectangle.
A_R=l w
A_R= 2* 6

Absolute Value Inequality

We can now substitute these expressions into our absolute value inequality. |1/2*4(x+6)-2*6|<2 Let's solve this inequality for x.
|1/2*4(x+6)-2*6|<2
|2(x+6)-12|<2
|2x+12-12|<2
|2x|<2

Write as Compound Inequality

-2 < 2x < 2

Divide each expression by 2

-1 < x < 1