Question

for 13 - 15, b is the mid - point of ac.

1. a 2x - 8 b x + 17 c
2. a x + 6 b 3x - 31 c
3. a 3(3x - 1) b 5(2x + 2) c

Explanation:

Step1: Use mid - point property

Since B is the mid - point of AC, then \(AB = BC\). For problem 13, we set up the equation \(2x - 8=x + 17\).
\[2x - 8=x + 17\]

Step2: Solve for x

Subtract x from both sides: \(2x-x-8=x - x+17\), which gives \(x-8 = 17\). Then add 8 to both sides: \(x=17 + 8=25\).

Step3: Find AB

Substitute \(x = 25\) into the expression for AB: \(AB=2x-8=2\times25 - 8=50 - 8 = 42\).

Step4: Find BC

Since \(BC=x + 17\), substitute \(x = 25\), then \(BC=25+17 = 42\).

Step5: Find AC

\(AC=AB + BC=42+42 = 84\).

For problem 14:

Step1: Set up equation

Since \(AB = BC\), we have \(x + 6=3x-31\).

Step2: Solve for x

Subtract x from both sides: \(x - x+6=3x-x-31\), so \(6 = 2x-31\). Add 31 to both sides: \(6+31=2x\), \(37 = 2x\), then \(x=\frac{37}{2}=18.5\).

Step3: Find AB

\(AB=x + 6=18.5+6=24.5\).

Step4: Find BC

\(BC=3x-31=3\times18.5-31=55.5 - 31 = 24.5\).

Step5: Find AC

\(AC=AB + BC=24.5+24.5 = 49\).

For problem 15:

Step1: Set up equation

Since \(AB = BC\), we have \(3(3x - 1)=5(2x+2)\).

Step2: Expand

Expand both sides: \(9x-3 = 10x+10\).

Step3: Solve for x

Subtract 9x from both sides: \(9x-9x-3=10x-9x + 10\), so \(-3=x + 10\). Subtract 10 from both sides: \(x=-3 - 10=-13\).

Step4: Find AB

\(AB=3(3x - 1)=3[3\times(-13)-1]=3(-39 - 1)=3\times(-40)=-120\).

Step5: Find BC

\(BC=5(2x+2)=5[2\times(-13)+2]=5(-26 + 2)=5\times(-24)=-120\).

Step6: Find AC

\(AC=AB + BC=-120-120=-240\).

Answer:

1. \(x = 25\), \(AB = 42\), \(BC = 42\), \(AC = 84\)
2. \(x = 18.5\), \(AB = 24.5\), \(BC = 24.5\), \(AC = 49\)
3. \(x=-13\), \(AB=-120\), \(BC=-120\), \(AC=-240\)