Notes here, and on pages 369-400 in book

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For any of these, you must (probably) create two equations and then solve by substitution or by addition (elimination).

For instance,

The perimeter of a rectangle is 32 feet. The length is 5 LESS than 2 times the width. What are the dimensions of the rectangle in feet?

First Equation: L+W+L+W=32 Which means 2L + 2W = 32 -OR- L+W=16 (this is easier to use).

Second Equation: L= 2W - 5 (remember that less than means you subtract from what follows!)

Using L+W=16 and L=2W-5, solving by substitution (or addition if you like) is easy:

(2W-5)+W=16

3W -5 = 16

3W=21

W=7

L=2(7)-5 or 9

Perimeter of a rectangle of 7 and 9 is definitely 32 and the length is 5 less than twice the length.