Exercise 27 Page 228

x+y+90=180 & (I) x+2=3y & (II) Let's solve it using the Substitution Method. To do so, we will start by isolating the x-variable in Equation (I).

x+y+90=180 & (I) x+2=3y & (II)

SubEqn

(I): LHS-90=RHS-90

x+y=90 x+2=3y

SubEqn

(I): LHS-y=RHS-y

x=90-y x+2=3y

Let's now substitute 90-y for x in Equation (II).

x=90-y x+2=3y

Substitute

(II): x= 90-y

x=90-y 90-y+2=3y

▼

(II): Solve for y

AddTerms

(II): Add terms

x=90-y 92-y=3y

AddEqn

(II): LHS+y=RHS+y

x=90-y 92=4y

DivEqn

(II): .LHS /4.=.RHS /4.

x=90-y 23=y

RearrangeEqn

(II): Rearrange equation

x=90-y y=23

Finally, to find the value of x, we will substitute 23 for y in Equation (I).

x=90-y y=23

Substitute

(I): y= 23

x=90- 23 y=23

SubTerm

(I): Subtract term

x=67 y=23

The solution to the system of equations, which is the point of intersection of the lines, is (67,23). In the context of the problem, this means that the measures of the missing angles are 67^(∘) and 23^(∘).