Dividing both sides by 11 simplifies the equation to (-1)/(x² - 3x - 28) = (1)/(30). Cross-multiplying gives x² - 3x - 28 = -30, which rearranges to x² - 3x + 2 = 0.

$\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30} ; \quad x \neq -4, 7$ Find the value of ' $x$ '.

Answer: $x = 1$ or $x = 2$

Step-by-step solution

Step 1: Combine the fractions on the left-hand side

To solve the equation, we take the common denominator $(x + 4)(x - 7)$ for the terms on the left-hand side. Subtracting the numerators gives $(x - 7) - (x + 4)$ .

Step 2: Simplify the numerator and expand the denominator

Simplifying the numerator gives $x - 7 - x - 4 = -11$ . Expanding the denominator gives $(x + 4)(x - 7) = x^2 - 7x + 4x - 28 = x^2 - 3x - 28$ .

Step 3: Cross-multiply and reduce to standard quadratic form

Dividing both sides by $11$ simplifies the equation to $\frac{-1}{x^2 - 3x - 28} = \frac{1}{30}$ . Cross-multiplying gives $x^2 - 3x - 28 = -30$ , which rearranges to $x^2 - 3x + 2 = 0$ .

Step 4: Factor the quadratic equation

We factor the quadratic polynomial by splitting the middle term $-3x$ into $-1x$ and $-2x$ , since $(-1) \times (-2) = 2$ . This factors into $(x - 1)(x - 2) = 0$ .

Step 5: Solve for x

Setting each factor to zero gives $x = 1$ or $x = 2$ . Both values satisfy the original condition $x \neq -4, 7$ .

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1x+4−1x−7=1130;x≠−4,7\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30} ; \quad x \neq -4, 7x+41​−x−71​=3011​;x=−4,7 Find the value of ' xxx '.