Question

point s is between r and t on rt. find the length of rs if: rs = 2x + 10, st = x - 4, rt = 21. rs= \boxed{ } 15

Explanation:

Step1: Use segment addition postulate

Since S is between R and T, \( RS + ST = RT \). Substitute the given expressions: \( (2x + 10) + (x - 4) = 21 \).

Step2: Simplify and solve for x

Combine like terms: \( 3x + 6 = 21 \). Subtract 6 from both sides: \( 3x = 15 \). Divide by 3: \( x = 5 \).

Step3: Find RS length

Substitute \( x = 5 \) into \( RS = 2x + 10 \): \( RS = 2(5) + 10 = 10 + 10 = 20 \)? Wait, but the image has 15. Wait, maybe miscalculation. Wait, recheck Step1: \( (2x +10)+(x -4)=21 \) → \( 3x +6 =21 \) → \( 3x=15 \) → \( x=5 \). Then \( RS=2*5 +10=20 \). But the image shows 15 on the right. Wait, maybe the problem has a typo? Or maybe I misread. Wait, no, let's check again. Wait, maybe the given RT is 21? Wait, if RS is 15, then 15 + ST =21 → ST=6. Then ST=x-4=6 → x=10. Then RS=2x+10=30, which is not 15. Wait, maybe the original problem has different numbers? Wait, the user's image: RS=2x+10, ST=x-4, RT=21. Let's solve again. \( 2x +10 +x -4 =21 \) → \( 3x +6=21 \) → \( 3x=15 \) → \( x=5 \). Then RS=25+10=20. But the box is for RS, and 15 is on the right. Maybe the image's 15 is a distractor. So following the math, RS=20? Wait, no, maybe I made a mistake. Wait, segment addition: RS + ST = RT. So 2x+10 +x-4 = 3x +6 =21. 3x=15, x=5. Then RS=25+10=20. So the answer should be 20.

Answer:

20