Exercise 23 Page 600

Let's draw a diagram to help us visualize the information that we've been given. Remember, the order of the vertices in a congruence statement tells us which angles are congruent in the two shapes.

Now, before we attempt to create a system of equations to solve for x and y, we can use the Triangle Angle-Sum Theorem to find the measure of the remaining angle in △ LMN. 40^(∘)+90^(∘)+m∠ N=180^(∘) Let's solve this equation.

40^(∘)+90^(∘)+m∠ N=180^(∘)

AddTerms

Add terms

130^(∘)+m∠ N=180^(∘)

SubEqn

LHS-130^(∘)=RHS-130^(∘)

m∠ N=50^(∘)

We can add this piece of information to our diagram.

Knowing that ∠ P≅ ∠ L and ∠ R≅ ∠ N, and using the definition of congruent angles, we can equate the measures of these angles. 17x-y=40 2x+4y=50 Let's solve this system using the Elimination Method. By multiplying the first equation by 4, we can eliminate y when we add the equations.

17x-y=40 & (I) 2x+4y=50 & (II)

MultEqn

(I): LHS * 4=RHS* 4

68x-4y=160 2x+4y=50

SysEqnAdd

(II): Add (I)

68x-4y=160 2x+4y+( 68x-4y)=50+ 160

▼

(II): Solve for x

RemovePar

(II): Remove parentheses

68x-4y=200 2x+4y+68x-4y=50+160

AddSubTerms

(II): Add and subtract terms

68x-4y=160 70x=210

DivEqn

(II): .LHS /70.=.RHS /70.

68x-4y=160 x=3

Having solved the system of equations for x, we can substitute this value into the first equation to solve for y.

68x-4y=160 x=3

Substitute

(I): x= 3

68( 3)-4y=160 x=3

▼

(I): Solve for y

Multiply

(I): Multiply

204-4y=160 x=3

SubEqn

(I): LHS-204^(∘)=RHS-204^(∘)

- 4y=- 44 x=3

MultEqn

(I): LHS * (- 1)=RHS* (- 1)

4y=44 x=3

DivEqn

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

y=11 x=3