Question

using algebra in exercises 24 and 25, solve the inequality ab + ac > bc. 25.

Explanation:

Step1: Substitute the expressions

Substitute \(AB = x + 2\), \(AC=x + 4\), and \(BC = 3x-1\) into the inequality \(AB + AC>BC\).
We get \((x + 2)+(x + 4)>3x-1\).

Step2: Simplify the left - hand side

Combine like terms on the left - hand side: \(x+x+2 + 4>3x-1\), which simplifies to \(2x+6>3x - 1\).

Step3: Move the \(x\) terms to one side

Subtract \(2x\) from both sides: \(2x+6-2x>3x - 1-2x\), resulting in \(6>x - 1\).

Step4: Solve for \(x\)

Add \(1\) to both sides: \(6 + 1>x-1+1\), so \(x<7\).
Also, consider the triangle side - length conditions:

1. \(AB=x + 2>0\), which gives \(x>-2\).
2. \(AC=x + 4>0\), which gives \(x>-4\).
3. \(BC=3x-1>0\), which gives \(x>\frac{1}{3}\).

Taking the intersection of \(x<7\) and \(x>\frac{1}{3}\) (since \(x>\frac{1}{3}\) is more restrictive than \(x>-2\) and \(x>-4\)).

Answer:

\(\frac{1}{3}