Question

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

Answer 1

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)

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

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

Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.

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

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

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

$\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}$

Converting ÷ sign into multiplication we will take reciprocal of the second term

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

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

Answer 2

The given equations are the algebric equations for obtaining solution the equation can be simplified by algebric operations.

1).

$\dfrac{x}{x+3} + \dfrac{x+2}{x+5}$

Taking LCM of denomenator (x+3) and (x+5).

$=\dfrac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}$

Further simplification has been done by multtiplication in the numerator.

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

2).

$\dfrac{x+4}{x^2+5x+6}\times \dfrac{x+3}{x^2-16}$

For further simplification of above equation factorization of $x^2+5x+6$ and $x^2-16$ has been performed.

$= \dfrac{x+4}{(x+3)(x+2)}\times \dfrac{x+3}{(x-4)(x+4)}$

$= \dfrac{1}{(x+2)(x-4)}$

3).

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

For further simplification of above equation factorization of $x^2-5x+6$ and $x^2-9$ has been performed.

$=\dfrac{2}{(x-3)(x+3)}-\dfrac{3x}{(x-3)(x-2)}$

Taking LCM of denomenator (x+3)(x-3) and (x-3)(x-2).

$=\dfrac{(2\times(x-2))-(3x\times(x+3))}{(x-3)(x+3)(x-2)}=\dfrac{2x-4-3x^2-9x}{(x-3)(x+3)(x-2)}$

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

4).

$\dfrac{\dfrac{x+4}{x^2-5x+6}}{\dfrac{x^2-16}{x+3}}$

For further simplification of above equation factorization of $x^2-5x+6$ and $x^2-16$ has been performed.

$=\dfrac{\dfrac{x+4}{(x-3)(x-2)}}{\dfrac{(x-4)(x+4)}{x+3}}=\dfrac{x+3}{(x-3)(x-2)(x-4)}$

For more information, refer the link given below

https://brainly.com/question/21450998