Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
1. Solving Systems of Linear Equations by Graphing
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Exercise 29 Page 222

The area of a rectangle is l* w and the perimeter is 2w+2l, where w and l are the width and the length, respectively.

Area: A=18x-18
Perimeter: P=6x+6
Solution to the System: (2,18)

Practice makes perfect

We can calculate the area of a rectangle by multiplying the width w by the length l. In the given diagram, we can see that w=3x-3 and l=6.

A=l * w
A= 6( 3x-3)
â–¼
Simplify right-hand side
A=6(3x)-6(3)
A=(6* 3)x-6(3)
A=18x-18
The perimeter is calculated using the formula P=2l+2w. Again, by substituting the known values for width and length, we can find a formula for the perimeter.

P=2l+2w
P=2( 6)+2( 3x-3)
â–¼
Simplify right-hand side
P=2(6)+ 2(3x)-2(3)
P=2(6)+ (2* 3)x-2(3)
P=12+6x-6
P=6x+6

Finally, we can graph these linear equations and find their point of intersection.

The solution, or point of intersection, to this system of linear equations is (2,18). This means that the perimeter and area are both 18 when x=2.