Question

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

Explanation:

Step1: Substitute the expressions

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

Step2: Simplify the left - hand side

Simplify \((x + 2)+(x + 3)\) using the commutative and associative properties of addition.
\((x+x)+(2 + 3)=2x+5\). So the inequality becomes \(2x+5>3x - 2\).

Step3: Subtract \(2x\) from both sides

Subtract \(2x\) from both sides of the inequality \(2x+5>3x - 2\).
\(2x+5-2x>3x - 2-2x\), which simplifies to \(5>x - 2\).

Step4: Add 2 to both sides

Add 2 to both sides of the inequality \(5>x - 2\).
\(5+2>x-2 + 2\), so \(7>x\) or \(x<7\).

Also, we need to consider the non - negativity of the side lengths:

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

Taking the intersection of \(x<7\) and \(x>\frac{2}{3}\) (since \(\frac{2}{3}>-2\) and \(\frac{2}{3}>-3\)).

Answer:

\(\frac{2}{3}