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4 years ago

Simplify the difference (b^2-2b-8)/(b^2+b-2)-(6)/(b-1)

Mathematics

1 answer:

ruslelena [56]4 years ago

3 0

Answer:

$\frac{b-10}{b-1}$

Step-by-step explanation:

We start by factoring out all polynomials in the expression:

$\frac{b^2-2b-8}{b^2+b-2} -\frac{6}{b-1} =\\=\frac{(b-4)(b+2)}{(b+2)(b-1)}- \frac{6}{b-1}=\\$

where we can simplify the factor (b+2) that appears in numerator and denominator of the first rational expression, to reduce it to:

$\frac{b-4}{b-1} -\frac{6}{b-1}$

Now we can combine both rational expressions since they share the same denominator:

$\frac{b-4}{b-1} -\frac{6}{b-1} =\frac{b-4-6}{b-1} =\frac{b-10}{b-1}$

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GIVING OUT BRAINLIEST TO WHOEVER GETS ALL OF THEM RIGHT

Thepotemich [5.8K]

Answer:

4) $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$ is equivalent to $\frac{1}{x-3}$ for all $x \ne 3$ . (Answer: None)

7) $\frac{x^{2}+3\cdot x -4}{x+4}$ is equivalent to $x - 1$ . (Answer: None)

8) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$ is equivalent to $\frac{4}{3\cdot a^{3}}$ for all $a\ne 0$ . (Answer: A)

Step-by-step explanation:

We proceed to simplify each expression below:

4) $\frac{x}{7\cdot x +x^{2}}$

(i) $\frac{x}{7\cdot x +x^{2}}$ Given

(ii) $\frac{x}{x\cdot (7+x)}$ Distributive property

(iii) $\frac{1}{7+x} \cdot \frac{x}{x}$ Distributive property

(iv) $\frac{1}{7+x}$ Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$7+x = 0$

$x = -7$

Hence, we conclude that $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$

(i) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ Given

(ii) $\frac{x^{3}\cdot (-14)}{x^{3}\cdot (1-5\cdot x)}$ Distributive property

(iii) $\frac{x^{3}}{x^{3}} \cdot \left(-\frac{14}{1-5\cdot x} \right)$ Distributive property

(iv) $-\frac{14}{1-5\cdot x}$ Commutative property/Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$1-5\cdot x = 0$

$5\cdot x = 1$

$x = \frac{1}{5}$

Hence, we conclude that $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$

(i) $\frac{x+7}{x^{2}+4\cdot x - 21}$ Given

(ii) $\frac{x+7}{(x+7)\cdot (x-3)}$ $x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})$

(iii) $\frac{1}{x-3}\cdot \frac{x+7}{x+7}$ Commutative and distributive properties.

(iv) $\frac{1}{x-3}$ Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$x-3 = 0$

$x = 3$

Hence, we conclude that $\frac{x+7}{x^{2}+4\cdot x - 21}$ is equivalent to $\frac{1}{x-3}$ for all $x \ne 3$ . (Answer: None)

7) $\frac{x^{2}+3\cdot x -4}{x+4}$

(i) $\frac{x^{2}+3\cdot x -4}{x+4}$ Given

(ii) $\frac{(x+4)\cdot (x-1)}{x+4}$ $x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})$

(iii) $(x-1)\cdot \left(\frac{x+4}{x+4} \right)$ Commutative and distributive properties.

(iv) $x - 1$ Existence of additive inverse/Modulative property/Result

Polynomic function are defined for all value of $x$ .

$\frac{x^{2}+3\cdot x -4}{x+4}$ is equivalent to $x - 1$ . (Answer: None)

8) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$

(i) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$

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Rational functions are undefined when denominator equals 0. That is:

$3\cdot a^{3} = 0$

$a = 0$

Hence, $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$ is equivalent to $\frac{4}{3\cdot a^{3}}$ for all $a\ne 0$ . (Answer: A)

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Step-by-step explanation:

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