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valentina_108 [34]
3 years ago
5

Match the following rational expressions to their rewritten forms.

Mathematics
1 answer:
Nookie1986 [14]3 years ago
7 0

Answer:

Answer image is attached.

Step-by-step explanation:

Given rational expressions:

1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}

And the rewritten forms:

(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}

We have to match the rewritten terms with the given expressions.

Let us consider the rewritten terms and let us solve them one by one by taking LCM.

(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}

So, correct option is 3.

(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}

So, correct option is 1.

(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}

So, correct option is 2.

(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}

So, correct option is 4.

The answer is also attached in the answer area.

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Asked on December 27, 2019 byShweta Rudravaram

Solve: 3x+4y=10,2x−2y=2 by the method of elimination.

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ANSWER

3x+4y=10 …………..(1)

2x−2y=2 (or) x−y=1 …………(2)

(1) ⇒(3x+4y=10)1

(2) ⇒(x−y=1)3

                     _____________________

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