Question

Point S is between R and T on line segment RT. Use the given information to write an equation in terms of x.

Answer 1

Answer:

Given,

S is the between R and T on line segment RT.

Thus, RS + ST = RT,

RS and ST.

a. If RS = 2x-10. ST = x-4. and RT = 21,

2x - 10 + x - 4 = 21

3x - 14 = 21

3x = 35

$\implies x = \frac{35}{3}$

RS = $2(\frac{35}{3})-10=\frac{70}{3}-10=\frac{40}{3}$

ST = $\frac{35}{3}-4=\frac{35-12}{3}=\frac{23}{3}$

b. RS = 3x-16. ST = 4x-8. and RT= 60,

3x - 16 + 4x - 8 = 60

7x - 24 = 60

7x = 84

$\implies x =\frac{84}{7}$ = 12,

RS = 3(12) - 16 = 36 - 16 = 20,

ST = 4(12) - 16 = 48 - 16 = 32

c. RS= 2x-8. ST = 3x-10. and RT = 17​.

2x - 8 + 3x - 10 = 17

5x - 18 = 17

5x = 35

$\implies x =\frac{35}{5}$ = 7,

RS = 2(7) - 8 = 14 - 8 = 6,

ST = 3(7) - 10 = 21 - 10 = 11

Answer 2

Answer:  $\bold{a)\quad x=\dfrac{35}{3}\qquad RS=13\dfrac{1}{3}\qquad ST=7\dfrac{2}{3}}$

                b)   x = 12         RS = 20            ST = 40

                c)    x = 7          RS = 6               ST = 11

Step-by-step explanation:

Use the Segment Addition Postulate: RS + ST = RT

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

                   3x - 14 = 21

                   3x       = 35

                    x        $=\dfrac{35}{3}$

RS = 2x - 10

    $=2\bigg(\dfrac{35}{3}\bigg)-10$

    $=\dfrac{70}{3}-\dfrac{30}{3}$

    $=\dfrac{40}{3}$

    $=\large\boxed{13\dfrac{1}{3}}$

ST = RT - RS

    $=21-13\dfrac{1}{3}$

    $=\large\boxed{7\dfrac{2}{3}}$

Use the same formula to solve for b and c