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shepuryov [24]
3 years ago
6

C^2-4/c+3/c+2/3(c^2-9)

Mathematics
1 answer:
Ivahew [28]3 years ago
6 0
<span> (c + 2) • (c - 2)
</span><span> ———————
</span>3 • (c + 3) • (c - 3)
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How would I solve this Rational Equation? <br><br>Solve for all values of x.
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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.

  \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 $x$}

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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.

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