Question

In Exercises 67–70, find all values of x such that y = 0. ﻿y=x+63x−12−5x−4−23y = \frac{x + 6}{3x - 12} - \frac{5}{x - 4} - \frac{2}{3}
y=3x−12x+6​−x−45​−32​

Accepted Answer

Start with the given equation: $$y = \frac{x + 6}{3x - 12} - \frac{5}{x - 4} - \frac{2}{3}. We $$want to find all values of $$x$$ such that $$y = 0$$, so set the equation equal to zero: $$\frac{x + 6}{3x - 12} - \frac{5}{x - 4} - \frac{2}{3} = 0. $$Notice that $$3x - 12$$ can be factored as $$3(x - 4). $$Rewrite the equation using this factorization: $$\frac{x + 6}{3(x - 4)} - \frac{5}{x - 4} - \frac{2}{3} = 0. To $$combine the fractions, find the least common denominator (LCD). The denominators are $$3(x - 4)$$, $$x - 4$$, and $$3. $$The LCD is $$3(x - 4). $$Rewrite each term with this common denominator: $$\frac{x + 6}{3(x - 4)} - \frac{5 \cdot 3}{3(x - 4)} - \frac{2(x - 4)}{3(x - 4)} = 0. $$Combine the numerators over the common denominator: $$\frac{(x + 6) - 15 - 2(x - 4)}{3(x - 4)} = 0. $$Since a fraction equals zero only when its numerator is zero (and the denominator is not zero), set the numerator equal to zero and solve for $$x$$: $$(x + 6) - 15 - 2(x - 4) = 0. $$Then simplify and solve the resulting linear equation.

In Exercises 67–70, find all values of x such that y = 0. $y = \(\frac{x + 6}{3x - 12}\) - \(\frac{5}{x - 4}\) - \(\frac{2}{3}\)$

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Key Concepts

Solving Rational Equations

Domain Restrictions and Excluded Values

Setting the Function Equal to Zero

In Exercises 67–70, find all values of x such that y = 0. y=x+63x−12−5x−4−23y = \(\frac{x + 6}{3x - 12}\) - \(\frac{5}{x - 4}\) - \(\frac{2}{3}\)y=3x−12x+6−x−45−32