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Mathematics

2 answers:

Degger [83]3 years ago

6 0

The answer would be x=-8(negative 8)

N76 [4]3 years ago

4 0

Answer is 4 - x^2 = -16 so the answer is -8 for x.

Read more on transformations;

brainly.com/question/2689696

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5 0

2 years ago

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$ = $\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

$\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}$

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

$\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}$

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$