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klemol [59]
3 years ago
15

Alison has twice as many video games as Kyle. Maurice has 5 more video games than Alison. The total number of video games is les

s than 40. a. Write an inequality to represent this situation.
b. Solve the inequality to determine the greatest number of video games Maurice could have.
Mathematics
1 answer:
belka [17]3 years ago
5 0
So,

a.
Let
1. Alison be "a"
2. Kyle be "k"
3. Maurice be "m"

a = 2k
m = a + 5
m = 2k + 5

40 > 5k + 5

35 > 5k
7 > k
The relationship between k and m:
m = 2k + 5
19 > m 
MAURICE CANNOT HAVE MORE THAN 19 VIDEO GAMES.
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Answer image is attached.

Step-by-step explanation:

Given rational expressions:

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And the rewritten forms:

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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}

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(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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