Answer: [tex]\bold{a)\quad x=\dfrac{35}{3}\qquad RS=13\dfrac{1}{3}\qquad ST=7\dfrac{2}{3}}[/tex]
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 [tex]=\dfrac{35}{3}[/tex]
RS = 2x - 10
[tex]=2\bigg(\dfrac{35}{3}\bigg)-10[/tex]
[tex]=\dfrac{70}{3}-\dfrac{30}{3}[/tex]
[tex]=\dfrac{40}{3}[/tex]
[tex]=\large\boxed{13\dfrac{1}{3}}[/tex]
ST = RT - RS
[tex]=21-13\dfrac{1}{3}[/tex]
[tex]=\large\boxed{7\dfrac{2}{3}}[/tex]
Use the same formula to solve for b and c
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
[tex]\implies x = \frac{35}{3}[/tex]
RS = [tex]2(\frac{35}{3})-10=\frac{70}{3}-10=\frac{40}{3}[/tex]
ST = [tex]\frac{35}{3}-4=\frac{35-12}{3}=\frac{23}{3}[/tex]
b. RS = 3x-16. ST = 4x-8. and RT= 60,
3x - 16 + 4x - 8 = 60
7x - 24 = 60
7x = 84
[tex]\implies x =\frac{84}{7}[/tex] = 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
[tex]\implies x =\frac{35}{5}[/tex] = 7,
RS = 2(7) - 8 = 14 - 8 = 6,
ST = 3(7) - 10 = 21 - 10 = 11
