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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

5. Systems With Three Variables

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Exercise 44 Page 173

Because the figure is a regular five-pointed star, we have a regular pentagon in the center of the figure.

x=36^(∘)
y=72^(∘)
z=108^(∘)

Practice makes perfect

We have been given a regular five-pointed star.

Because the figure is a regular five-pointed star, we have a regular pentagon in the center of the figure.

Using the angles on the figure, we will write three equations on an organized table.

Verbal Expression	Algebraic Expression
Sum of the interior angles of a triangle is 180^(∘).	x+y+y=180
Sum of the interior angles of a pentagon is 540^(∘).	z+z+z+z+z=540
Sum of two adjacent angles on a straight line is 180^(∘).	y+z=180

Now, there are three equations to write a system. x+y+y=180 & (I) z+z+z+z+z=540 & (II) y+z=180 & (III) To solve the system, we will begin by finding the measure of z. Then, we will use the Substitution Method to find the measures of x and y.

x+y+y=180 z+z+z+z+z=540 y+z=180

(I), (II): Add terms

x+2y=180 5z=540 y+z=180

(II): .LHS /5.=.RHS /5.

x+2y=180 z=108 y+z=180

The measure of z is 108^(∘). Next, we will find measure of y by substituting 108 for z into the third equation.

x+2y=180 z=108 y+z=180

(III): z= 108

x+2y=180 z=108 y+ 108=180

(III): LHS-108=RHS-108

x+2y=180 z=108 y=72

Now that we know the measure of y, we can find the measure of x. Let's find it!

x+2y=180 z=108 y=72

(I): y= 72

x+2( 72)=180 z=108 y=72

(I): Multiply

x+144=180 z=108 y=72

(I): LHS-144=RHS-144

x=36 z=108 y=72

Therefore, x is 36^(∘).

Subchapter links

Lesson Check p.171

Practice and Problem-Solving Exercises p.171-173

Standardized Test Prep p.173

Mixed Review p.173