Respuesta :

gmany

[tex]Domain:\\x+4\neq0\ \wedge\ x\neq0\\\\\boxed{D:x\neq-4\ \wedge\ x\neq0}\\------------------------\\\\\dfrac{x-11}{x+4}=2-\dfrac{4}{x}\\\\\dfrac{x-11}{x+4}=\dfrac{2x}{x}-\dfrac{4}{x}\\\\\dfrac{x-11}{x+4}=\dfrac{2x-4}{x}\qquad\text{cross multiply}\\\\(x+4)(2x-4)=x(x-11)\qquad\text{use distributive property}\\\\(x)(2x)+(x)(-4)+(4)(2x)+(4)(-4)=(x)(x)+(x)(-11)\\\\2x^2-4x+8x-16=x^2-11x\\\\2x^2+(-4x+8x)-16=x^2-11x\\\\2x^2+4x-16=x^2-11x\qquad\text{subtract}\ x^2\ \text{from both sides}[/tex]

[tex]x^2+4x-16=-11x\qquad\text{add}\ 11x\ \text{to both sides}\\\\x^2+15x-16=0\\\\x^2+16x-x-16=0\\\\x(x+16)-1(x+16)=0\\\\(x+16)(x-1)=0\iff x+16=0\ \vee\ x-1=0\\\\\boxed{x=-16\in D\ \vee\ x=1\in D}\\\\Answer:\ B.\ \{-16,\ 1\}[/tex]

SonOfZeus

Answer:

B.) {-16,1}

Step-by-step explanation:

First let us combine like terms so let us move the -4/x to the right side:

[tex]\frac{x-11}{x+4}=2-\frac{4}{x}[/tex]

[tex]\frac{x-11}{x+4} +\frac{4}{x}=2[/tex]

Let us find a common denominator for the left side by using : [tex]\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}[/tex] and so:

[tex]\frac{x(x-11)+4(x+4)}{x(x+4)} = 2\\ \\\frac{x^2-11x+4x+16}{x^2+4x}=2\\[/tex]

Now lets get rid of that fraction by multiplying both sides by [tex]x^2+4x[/tex] and we obtain:

[tex]x^2-7x+16=2(x^2+4x)\\\\x^2-7x+16=2x^2+8x[/tex]

I'm going to move everything on the left side to the right by: [tex]-(x^2-7x+16)[/tex] and so:

[tex]0=2x^2+8x-x^2+7x-16\\\\0=x^2+15x-16[/tex]

Now let's factor.  We can factor by multiplying outer coefficients, so 1 x -16 = -16.  Now let's list all the factors of 16:  1×16,8×2,4×4.  From these factors try to find two that if you add or subtract them they will return the middle term of 15.  So, 16 - 1  = 15.  Therefore,

[tex]0=x^2+15x-16\\\\0=x^2+16x-x-16\\\\0=x(x+16)-(x+16)\\0=(x-1)(x+16)[/tex]

Now lets solve for both cases:

Case 1:

x-1=0

x=1

Case 2:

x+16=0

x=-16

and so your solutions are:

x={1,-16} or x={-16,1}



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