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Ket [755]
3 years ago
12

How do I solve this

Mathematics
1 answer:
sdas [7]3 years ago
3 0
\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\
\cfrac{2x}{x^2+2x-24}-\cfrac{x}{x^2-36}\quad 
\begin{cases}
x^2+2x-24\implies (x+6)(x-4)\\
--------------\\
x^2-36\implies x^2-6^2\\
(x-6)(x+6)
\end{cases}

\bf \cfrac{2x}{(x+6)(x-4)}-\cfrac{x}{(x-6)(x+6)}\impliedby 
\begin{array}{llll}
\textit{thus our LCD is}\\
(x-6)(x+6)(x-4)
\end{array}
\\\\\\
\cfrac{[(x-6)2x]~-~[(x-4)x]}{(x-6)(x+6)(x-4)}\implies \cfrac{2x^2-12x-x^2+4x}{(x-6)(x+6)(x-4)}
\\\\\\
\cfrac{x^2-8x}{(x-6)(x+6)(x-4)}
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<h3>Answer: (-infinity, 7]</h3>

=====================================

Explanation:

The first interval (-infinity, 3) describes any number less than 3, so we can write x < 3 in short hand (where x is the unknown number).

The second interval (-1, 7] means we start at -1 and stop at 7. We do not include -1 but include 7. So we say that -1 < x \le 7 (ie x is between -1 and 7; exclude -1, include 7)

If you were to graph each ona number line, you would see that the too intervals have overlapping parts. The right most edge extends out as far as x = 7. There is no left most edge as it goes onforever that direction.

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-----------

It might help to think of it like this: x < 3 and -1 < x \le 7 say "x is some number that is less than 3, or it is between -1 and 7". So x could be anything less than 7, including 7 itself.

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3 years ago
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trasher [3.6K]

Answer:

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Step-by-step explanation:

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