Exercise 22 Page 501

Let's find the values of both variables at one time.

Values of x and y

Notice that the angles measuring 2x+2y and 4y+x are both consecutive to the angle measuring x+y. Recall that if an angle of a quadrilateral is supplementary to both of its consecutive angles, then the quadrilateral is a parallelogram. Therefore, the measures of our consecutive angles must add up to 180. (2x+2y)+(x+y)=180 & (I) (4y+x)+(x+y)=180 & (II) We will solve the system by using the Substitution Method. Let's start by isolating the x-variable in Equation (I).

(2x+2y)+(x+y)=180 & (I) (4y+x)+(x+y)=180 & (II)

▼

Solve for x

RemovePar

(I): Remove parentheses

2x+2y+x+y=180 (4y+x)+(x+y)=180

AddTerms

(I): Add terms

3x+3y=180 (4y+x)+(x+y)=180

SubEqn

(I): LHS-3y=RHS-3y

3x=180-3y (4y+x)+(x+y)=180

DivEqn

(I): .LHS /3.=.RHS /3.

x=60-y (4y+x)+(x+y)=180

RemovePar

(II): Remove parentheses

x=60-y 4y+x+x+y=180

AddTerms

(II): Add terms

x=60-y 5y+2x=180

Now that x is isolated in Equation (I), we will substitute x=60-y in Equation (II), and solve for y.

x=60-y 5y+2x=180

Substitute

(II): x= 60-y

x=60-y 5y+2( 60-y)=180

▼

Solve for y

Distr

(II): Distribute 2

x=60-y 5y+120-2y=180

SubTerm

(II): Subtract term

x=60-y 3y+120=180

SubEqn

(II): LHS-120=RHS-120

x=60-y 3y=60

DivEqn

(II): .LHS /3.=.RHS /3.

x=60-y y=20

We found that y=20. Finally, we will find the value of x by substituting this value in Equation (I).

x=60-y y=20

Substitute

(I): y= 20

x=60- 20 y=20

SubTerm

(I): Subtract term

x=40 y=20