Question

points a, b, and c are collinear. point b is between a and c. solve for x if: ac = 3x + 3; ab = -1 + 2x; bc = 11; x =
solve for x. r ---- q ---- p (rq = 22 + 2x, qp = 8, rp = x + 20); x =
points a, b, and c are collinear. point b is between a and c. solve for x if: ac = 22; bc = x + 14; ab = x + 10; x =
solve for x. t ---- u ---- v (tu = 2x - 4, uv = 2x - 7, tv = 17); x =

Explanation:

Problem 8 (Top - Left)

Step1: Use segment addition postulate

Since \( B \) is between \( A \) and \( C \), \( AC = AB + BC \). Substitute the given expressions: \( 3x + 3 = (-1 + 2x) + 11 \)

Step2: Simplify and solve for \( x \)

Simplify the right - hand side: \( 3x + 3=2x + 10 \)
Subtract \( 2x \) from both sides: \( 3x-2x + 3=2x-2x + 10\Rightarrow x + 3 = 10 \)
Subtract 3 from both sides: \( x+3 - 3=10 - 3\Rightarrow x = 7 \)

Step1: Use segment addition postulate

The length of \( RP\) is the sum of \( RQ\) and \( QP\). So, \( (22 + 2x)+8=x + 20 \)

Step2: Simplify and solve for \( x \)

Simplify the left - hand side: \( 30+2x=x + 20 \)
Subtract \( x \) from both sides: \( 30 + 2x-x=x - x+20\Rightarrow30 + x=20 \)
Subtract 30 from both sides: \( x=20 - 30=- 10 \)

Step1: Use segment addition postulate

Since \( B \) is between \( A \) and \( C \), \( AC = AB+BC \). Substitute the given values: \( 22=(x + 10)+(x + 14) \)

Step2: Simplify and solve for \( x \)

Simplify the right - hand side: \( 22 = 2x+24 \)
Subtract 24 from both sides: \( 22-24=2x+24 - 24\Rightarrow - 2 = 2x \)
Divide both sides by 2: \( x=\frac{-2}{2}=-1 \)

Answer:

\( x = 7 \)

Problem 9 (Top - Right)