Question

(x-3)/(x-4)>(x-4)/(x-3)

Answer 1

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Answer 2

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Answer 3

To solve the inequality $\frac{x-3}{x-4} > \frac{x-4}{x-3}$, we need to find the range of values of $x$ that satisfy the inequality. Let's go step by step:

Step 1: Eliminate the denominators by multiplying both sides by $(x-3)(x-4)$ (since the denominators are non-zero, we don't have to worry about changing the direction of the inequality):

$(x-3)^2 > (x-4)^2$

Step 2: Expand both sides:

$x^2 - 6x + 9 > x^2 - 8x + 16$

Step 3: Simplify the inequality by canceling out $x^2$ on both sides:

$-6x + 9 > -8x + 16$

Step 4: Bring all $x$ terms to one side of the inequality:

$-6x + 8x > 16 - 9$

Step 5: Combine like terms:

$2x > 7$

Step 6: Finally, isolate $x$ by dividing both sides by 2:

$x > \frac{7}{2}$

So, the solution to the inequality is $x > \frac{7}{2}$, which means any value of $x$ greater than $\frac{7}{2}$ will satisfy the inequality.