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point s is between points r and t on $overline{rt}$. use the informatio…

Question

point s is between points r and t on $overline{rt}$. use the information to draw a picture and write an equation in terms of x. then solve the equation and find rs, st, and rt.

  1. rs = 2x + 10

st = x - 4
rt = 21

  1. rs = 2x - 8

st = 11
rt = x + 10

  1. rs = 3x - 16

st = 4x - 8
rt = 60

  1. rs = 4x - 9

st = 19
rt = 8x - 14

Explanation:

Step1: Use segment - addition postulate

Since point S is between R and T on $\overline{RT}$, we have $RS + ST=RT$.

Step2: Substitute the given expressions

For problem 10: Substitute $RS = 2x + 10$, $ST=x - 4$, and $RT = 21$ into $RS+ST = RT$. We get the equation $(2x + 10)+(x - 4)=21$.

Step3: Simplify the left - hand side of the equation

Combine like terms: $2x+x+10 - 4=21$, which simplifies to $3x + 6=21$.

Step4: Solve for x

Subtract 6 from both sides: $3x=21 - 6=15$. Then divide both sides by 3: $x = 5$.

Step5: Find RS, ST, and RT

$RS=2x + 10=2\times5+10=20$.
$ST=x - 4=5 - 4 = 1$.
$RT = 21$ (given).

For problem 11: Substitute $RS = 2x-8$, $ST = 11$, and $RT=x + 10$ into $RS+ST = RT$. We get $(2x-8)+11=x + 10$.
Simplify the left - hand side: $2x+3=x + 10$.
Subtract x from both sides: $2x-x+3=x - x+10$, so $x+3 = 10$.
Subtract 3 from both sides: $x=7$.
$RS=2x-8=2\times7-8 = 6$.
$ST = 11$ (given).
$RT=x + 10=7 + 10=17$.

For problem 12: Substitute $RS = 3x-16$, $ST = 4x-8$, and $RT = 60$ into $RS+ST = RT$. We get $(3x-16)+(4x-8)=60$.
Combine like terms: $3x+4x-16 - 8=60$, so $7x-24 = 60$.
Add 24 to both sides: $7x=60 + 24=84$.
Divide both sides by 7: $x = 12$.
$RS=3x-16=3\times12-16=20$.
$ST=4x-8=4\times12-8 = 40$.
$RT = 60$ (given).

For problem 13: Substitute $RS = 4x-9$, $ST = 19$, and $RT=8x-14$ into $RS+ST = RT$. We get $(4x-9)+19=8x-14$.
Simplify the left - hand side: $4x + 10=8x-14$.
Subtract 4x from both sides: $4x-4x+10=8x-4x-14$, so $10=4x-14$.
Add 14 to both sides: $10 + 14=4x$, $24=4x$.
Divide both sides by 4: $x = 6$.
$RS=4x-9=4\times6-9 = 15$.
$ST = 19$ (given).
$RT=8x-14=8\times6-14=34$.

Answer:

For problem 10: $x = 5$, $RS = 20$, $ST = 1$, $RT = 21$.
For problem 11: $x = 7$, $RS = 6$, $ST = 11$, $RT = 17$.
For problem 12: $x = 12$, $RS = 20$, $ST = 40$, $RT = 60$.
For problem 13: $x = 6$, $RS = 15$, $ST = 19$, $RT = 34$.