2 9 X 3 8
Best Answer
+33160 Information technology is possible that this equation was meant to exist:
$$\frac{xiv}{x^2-9}+\frac{vi}{x-3}=\frac{eight}{x+iii}$$
in which example, multiply all terms by x2 -9
$$14+\frac{6(10^2-9)}{x-3}=\frac{8(x^2-9))}{x+3}$$
Limited 102 - nine every bit (x+three)(x-3) and abolish as appropriate
$$14+six(x+iii)=8(x-iii)$$
Expand bracketed terms and collect like terms on the aforementioned side
$$xiv+18+24=8x-6x$$
$$56=2x$$
$$ten=28$$
.
+7188 $${\frac{{\mathtt{14}}}{{{\mathtt{10}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{ix}}{\mathtt{\,\modest\textbf+\,}}{\frac{{\mathtt{6}}}{{\mathtt{ten}}}}{\mathtt{\,-\,}}{\mathtt{3}} = {\frac{{\mathtt{8}}}{{\mathtt{10}}}}{\mathtt{\,\small-scale\textbf+\,}}{\mathtt{3}} \Rightarrow \left\{ \begin{array}{fifty}{\mathtt{ten}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{211}}}}{\mathtt{\,\modest\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{15}}}}\\
{\mathtt{10}} = {\frac{\left({\sqrt{{\mathtt{211}}}}{\mathtt{\,-\,}}{\mathtt{ane}}\right)}{{\mathtt{xv}}}}\\
\finish{array} \right\} \Rightarrow \left\{ \begin{assortment}{l}{\mathtt{ten}} = -{\mathtt{one.035\: \!055\: \!936\: \!422\: \!263\: \!3}}\\
{\mathtt{x}} = {\mathtt{0.901\: \!722\: \!603\: \!088\: \!93}}\\
\end{array} \right\}$$
+23200 happy 7 is correct.
If yous need to show some piece of work: 14/x² - 9 + vi/ten - three = 8/x + 3
Multiply each term by x² to get rid of denominators:
(x²)(14/x²) - (ten²)(9) + (x²)(6/ten) - (x²)(iii) = (x²)(8/x) + (x²)(three)
14 - 9x² + 6x - 3x² = 8x + 3x²
14 - 12x² + 6x = 8x + 3x²
14 - 15x² - 2x = 0
-15x² - 2x + fourteen = 0
15x² + 2x - xiv = 0
Using the quadratic equation: x = [ -b ± √(b² -4ac) ] / (2a)
a = 15 b = ii c = -14
x = [ -2 ± √(two² -4·15·-14) ] / (ii·15)
10 = [ -2 ± √(iv + 840) ] / (30)
ten = [ -ii ± √(844) ] / (30)
x = [ -two ± √4√211 ] / thirty
x = [ -2 ± 2√211 ] / 30
x = [ -1 ± √211 ] / 15
+33160 It is possible that this equation was meant to be:
$$\frac{fourteen}{x^2-ix}+\frac{six}{x-iii}=\frac{viii}{x+3}$$
in which case, multiply all terms by x2 -9
$$14+\frac{6(10^2-ix)}{x-3}=\frac{8(x^2-9))}{x+3}$$
Express xii - nine as (10+3)(ten-3) and cancel every bit appropriate
$$14+6(ten+3)=8(ten-3)$$
Expand bracketed terms and collect like terms on the same side
$$xiv+18+24=8x-6x$$
$$56=2x$$
$$x=28$$
.
11 Online Users
2 9 X 3 8,
Source: https://web2.0calc.com/questions/14-x-2-9-6-x-3-8-x-3
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