Exercise 29 Page &nbsp; 394

Substitute expressions

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

Remove parentheses

21=2x+10+x-4

AddTerms

Add terms

21=3x+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=3x+6

LHS-6=RHS-6

15=3x

Rearrange equation

3x=15

DivEqn

.LHS /3.=.RHS /3.

x=5

The value of x is 5. Our last step is to use x=5 to calculate the different lengths.

Distance	x=5	Length
RS	2( 5)+10	20
ST	5-4	1
RT	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.

RT=RS+ST

Substitute expressions

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

Remove parentheses

60=3x-16+4x-8

AddSubTerms

Add and subtract terms

60=7x-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=7x-24

AddEqn

LHS+24=RHS+24

84=7x

Rearrange equation

7x=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
RS	3( 12)-16	20
ST	4( 12)-8	40
RT	60	60

c This time, RS=2x-8, ST=11, and RT=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.

RT=RS+ST

Substitute expressions

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

Remove parentheses

x+10=2x-8+11

SubTerm

Subtract term

x+10=2x+3

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

x+10=2x+3

LHS-x=RHS-x

10=x+3

Rearrange equation

x+3=10

LHS-3=RHS-3

x=7

The value of x is 7. We will use this and the given expressions to find the lengths.

Distance	x=7	Length
RS	2( 7)-8	6
ST	11	11
RT	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.

RS+ST=RT

Substitute expressions

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

Remove parentheses

4x-9+19=8x-14

AddTerms

Add terms

4x+10=8x-14

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

4x+10=8x-14

AddEqn

LHS+14=RHS+14

4x+24=8x

LHS-4x=RHS-4x

24=4x

DivEqn

.LHS /4.=.RHS /4.

6=x