Exercise 24 Page 578

Let's analyze the given figure. We will start by finding x.

Notice that since the segments formed on the left part of the diagram are congruent, we can solve the equation 20-5x = 2x+6 to find x.

20-5x = 2x+6

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Solve for x

AddEqn

LHS+5x=RHS+5x

20=7x+6

SubEqn

LHS-6=RHS-6

14=7x

DivEqn

.LHS /7.=.RHS /7.

2=x

RearrangeEqn

Rearrange equation

x=2

Since we are given three parallel lines that intersect two transversals, we can use the Corollary to the Triangle Proportionality Theorem.

The lengths of the segments intercepted on the transversals are proportional. Let's write a proportion using the expressions for the lengths of the segments. 20-5x/2x+6=y/35y+2 Since we know that two segments on the left part of the diagram are congruent, we know that two segments on the right side are also congruent. Let's solve the equation y = 35y+2 to find y.

y = 3/5y+2

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Solve for y

MultEqn

LHS * 5=RHS* 5

5y = 3y+10

SubEqn

LHS-3y=RHS-3y

2y = 10

DivEqn

.LHS /2.=.RHS /2.

y=5

We found that x=2 and y=5.