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Schach [20]
4 years ago
6

What is the solution of

%204" id="TexFormula1" title="3 + \frac{x - 2}{x - 3} \leqslant 4" alt="3 + \frac{x - 2}{x - 3} \leqslant 4" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
SIZIF [17.4K]4 years ago
7 0

Answer:

x<3

Step-by-step explanation:

you should use Socratic or photo math.

4 x-11/x-3=4

4 x-11=4 x-12 (multiply both sides by x-3)

4 x-11-4 x=4 x-12-4 x (subtract 4 x from both side)

-11=-12

-11+11=-12+11 (Add 11 to both sides)

0=-1

<u>Critical points:</u>

x=3 (Make left denominator equal to 0)

Check intervals in between critical points. (test values in the intervals to see if they work.)

x<3 (Works in original inequality)

x>3 (Doesn´t work in original inequality)

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3 years ago
A two column proof? Help !
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3 years ago
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Convert x - 5 = 0 to a polar equation
dolphi86 [110]

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(

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Adding [1] and [2]:

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2

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4 0
4 years ago
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3 years ago
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agasfer [191]

Answer:

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Step-by-step explanation:

\begin{bmatrix}-8x+y=-32\\ -3x-10=y\end{bmatrix}\\\\\mathrm{Isolate}\:x\:\mathrm{for}\:-8x+y=-32:\quad x=-\frac{-32-y}{8}\\\\\mathrm{Subsititute\:}x=-\frac{-32-y}{8}\\\begin{bmatrix}-3\left(-\frac{-32-y}{8}\right)-10=y\end{bmatrix}\\\\Simplify\\\begin{bmatrix}\frac{3\left(-32-y\right)}{8}-10=y\end{bmatrix}\\\\\mathrm{Isolate}\:y\:\mathrm{for}\:\frac{3\left(-32-y\right)}{8}-10=y:\quad y=-16\\\\\mathrm{For\:}x=-\frac{-32-y}{8}\\\\\mathrm{Subsititute\:}y=-16\\x=-\frac{-32-\left(-16\right)}{8}\\

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3 0
4 years ago
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