The set of all real values of x for which fra | vedclass

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(B) Given the inequality: $\frac{x^2-1}{(x-4)(x-3)} \geq 1$. Subtract $1$ from both sides: $\frac{x^2-1}{(x-4)(x-3)} - 1 \geq 0$. Simplify the express

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$[\frac{13}{7}, 3) \cup (4, \infty)$

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(B) Given the inequality: $\frac{x^2-1}{(x-4)(x-3)} \geq 1$. Subtract $1$ from both sides: $\frac{x^2-1}{(x-4)(x-3)} - 1 \geq 0$. Simplify the expression: $\frac{x^2-1 - (x^2-7x+12)}{(x-4)(x-3)} \geq 0$. $\frac{x^2-1 - x^2+7x-12}{(x-4)(x-3)} \geq 0$. $\frac{7x-13}{(x-4)(x-3)} \geq 0$. Find the critical points: $x = \frac{13}{7}, x = 3, x = 4$. Using the wavy curve method on the number line,we test the intervals: For $x > 4$,the expression is positive. For $3 For $\frac{13}{7} \leq x For $x Thus,the solution set is $[\frac{13}{7}, 3) \cup (4, \infty)$.

The set of all real values of $x$ for which $\frac{x^2-1}{(x-4)(x-3)} \geq 1$ is

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