Exercise 29 Page 18

a If we visualize the given conditions, we will have a straight line with points

R,

S,

and

T,

in that order.

From the exercise, we know that the length of $R S$ can be expressed as $2 x + 10 .$ Let's mark it on our picture.

Now, we will do the same thing with the rest of the information. The length of $S T$ is $x - 4$ and $R T$ is $21 .$

We can use the Segment Addition Postulate to write an equation since the points are collinear. The length of

R T

is the sum of

R S

and

S T .

R T = R S + S T

SubstituteExpressions

Substitute expressions

21 = (2 x + 10) + (x - 4)

RemovePar

Remove parentheses

21 = 2 x + 10 + x - 4

AddTerms

Add terms

21 = 3 x + 6

Now that we have simplified the right-hand side as much as possible, we have an equation that we can solve for

x .

21 = 3 x + 6

SubEqn

$LHS - 6 = RHS - 6$

15 = 3 x

RearrangeEqn

Rearrange equation

3 x = 15

DivEqn

$LHS / 3 = RHS / 3$

x = 5

The value of

x

5 .

Our last step is to use

x = 5

to calculate the different lengths.

Distance	$x = 5$	Length
$R S$	$2 (5) + 10$	$20$
$S T$	$5 - 4$	$1$
$R T$	$21$	$21$

b We will continue the same way as in Part A, by drawing a picture of the points and the lengths between them.

With the Segment Addition Postulate we can form an equation, using the expressions for the lengths.

R T = R S + S T

SubstituteExpressions

Substitute expressions

60 = (3 x - 16) + (4 x - 8)

RemovePar

Remove parentheses

60 = 3 x - 16 + 4 x - 8

AddSubTerms

Add and subtract terms

60 = 7 x - 24

Now that we have simplified the right-hand side as much as possible, we have an equation that we can solve for

x .

60 = 7 x - 24

AddEqn

$LHS + 24 = RHS + 24$

84 = 7 x

RearrangeEqn

Rearrange equation

7 x = 84

DivEqn

$LHS / 7 = RHS / 7$

x = 12

Now that we have solved the equation for

x,

we can use the value to calculate the lengths.

Distance	$x = 12$	Length
$R S$	$3 (12) - 16$	$20$
$S T$	$4 (12) - 8$	$40$
$R T$	$60$	$60$

c This time,

R S = 2 x - 8,

S T = 11,

and

R T = x + 10 .

Let's mark the lengths in the picture.

We can now make an equation using the Segment Addition Postulate with the expressions for the different lengths.

R T = R S + S T

SubstituteExpressions

Substitute expressions

(x + 10) = (2 x - 8) + 11

RemovePar

Remove parentheses

x + 10 = 2 x - 8 + 11

SubTerm

Subtract term

x + 10 = 2 x + 3

Now that we've simplified both sides as much as possible, we have an equation that we can solve for

x .

x + 10 = 2 x + 3

SubEqn

$LHS - x = RHS - x$

10 = x + 3

RearrangeEqn

Rearrange equation

x + 3 = 10

SubEqn

$LHS - 3 = RHS - 3$

x = 7

The value of

x

7 .

We will use this and the given expressions to find the lengths.

Distance	$x = 7$	Length
$R S$	$2 (7) - 8$	$6$
$S T$	$11$	$11$
$R T$	$7 + 10$	$17$

d One last time, let's use the diagram to visualize the relationship.

We can use the Segment Addition Postulate to form an equation that equates the segment lengths.

R S + S T = R T

SubstituteExpressions

Substitute expressions

(4 x - 9) + 19 = (8 x - 14)

RemovePar

Remove parentheses

4 x - 9 + 19 = 8 x - 14

AddTerms

Add terms

4 x + 10 = 8 x - 14

Now that we have simplified the left-hand side as much as possible, we have an equation that we can solve for

x .

4 x + 10 = 8 x - 14

AddEqn

$LHS + 14 = RHS + 14$

4 x + 24 = 8 x

SubEqn

$LHS - 4 x = RHS - 4 x$

24 = 4 x

DivEqn

$LHS / 4 = RHS / 4$

6 = x

RearrangeEqn

Rearrange equation

x = 6

By substituting

x

with

6

in the expressions, we can calculate the lengths.

Distance	$x = 6$	Length
$R S$	$4 (6) - 9$	$15$
$S T$	$19$	$19$
$R T$	$8 (6) - 14$	$34$

Subchapter links

Communicate Your Answer p.11

Monitoring Progress p.12-15

Exercises p.16-18