9514 1404 393
Answer:
x = 4
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.
![(7)/(x+3)+(3)/(x-3)-(x)/(x-3)=0\\\\ (7(x-3)+(x+3)(3-x))/((x+3)(x-3))=0\\\\((x-3)(4-x))/((x-3)(x+3))=0\\\\ (4-x)/(x+3)=0\qquad x\\e3\\\\x=4 \qquad\text{equate the numerator to zero, add $x$}](https://img.qammunity.org/2022/formulas/mathematics/high-school/5m11ps7i1le62squu2mly09gw0sqijunbm.png)
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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.