Exercise 8 Page 611

Let's analyze the given figure. Since we are given a triangle with a line that is parallel to one of its sides, we can use the Triangle Proportionality Theorem.

The lengths of the segments intercepted by the parallel line are proportional. Let's write a proportion using the expressions for the lengths of the segments. 5x-8/3x+11 = 2y-1/4y-7 Notice that one of the segments is divided in half. Therefore, 5x-8 = 3x+11. Let's solve this equation to find x.

5x-8 = 3x+11

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Solve for x

AddEqn

LHS+8=RHS+8

5x = 3x+19

SubEqn

LHS-3x=RHS-3x

2x = 19

DivEqn

.LHS /19.=.RHS /19.

x=19/2

CalcQuot

Calculate quotient

x=9.5

Since 5x-8 = 3x+11, also 2y-1 = 4y-7. Let's solve it to find y.

2y-1 = 4y-7

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Solve for y

AddEqn

LHS+7=RHS+7

2y+6 = 4y

SubEqn

LHS-2y=RHS-2y

6 = 2y

DivEqn

.LHS /2.=.RHS /2.

3=y

RearrangeEqn

Rearrange equation

y=3

We found that x=9.5 and y=3.