9514 1404 393
Answer:
x = 4
Step-by-step explanation:
I like to put these in the form f(x) = 0. We can do that by subtracting the right side. Common factors can be cancelled from numerator and denominator, provided they are not zero.
[tex]\dfrac{7}{x+3}+\dfrac{3}{x-3}-\dfrac{x}{x-3}=0\\\\ \dfrac{7(x-3)+(x+3)(3-x)}{(x+3)(x-3)}=0\\\\\dfrac{(x-3)(4-x)}{(x-3)(x+3)}=0\\\\ \dfrac{4-x}{x+3}=0\qquad x\ne3\\\\x=4 \qquad\text{equate the numerator to zero, add $x$}[/tex]
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If you leave the numerator as (x-3)(4-x), then there are two values of x that make it zero. Because x=3 makes the equation "undefined", it cannot be considered to be a solution.
