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rewona [7]
3 years ago
8

darnel’s retirement party cost $33, plus an additional $1 for each guest he invites. what is the maximum number of guests there

can be if darnel can afford to spend a total of $42 on his retirement party?
Mathematics
1 answer:
creativ13 [48]3 years ago
5 0

Answer:

20

Step-by-step explanation:

He's going to need a very sharp pencil to cross out the people he cannot afford to invite but would like to.

Equation

y = 33 + 1x

y = 42

42 = 33  + x

Solution

42 = 22 + x               Subtract 22 from both sides.

42 - 22 = 22 -22 + x    

20 = x

This means that if each extra person costs $1.00 then he can invite 20 more people (20 * 1 = 20)

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<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>

<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>

<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>

<u>2</u><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u>

<u>x =  \frac{ - 6 + 3y}{2}</u>

<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>

<u>2y - 3 =  \frac{ - 6 + 3y}{2}</u>

4y - 6 = -6 + 3y

4y - 3y = -6 + 6

y = 0

Putting value of y in ( iii )

x = 2y - 3

x = 2 ( 0 ) - 3

x = -3

4 0
3 years ago
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