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Murrr4er [49]
3 years ago
14

Find two consecutive odd integers such that their product is 71 more than 3 times their sum.

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
4 0
Let's put this in terms of a system of equations. Let's call the numbers x and y, with x being the larger integer and y the smaller one. Since they are consecutive odd integers, we know that:
x = y + 2
The constraint is:
xy= 3(x+y) + 71         Simplify
xy = 3x + 3y + 71       Plug in (y+2) for x:
(y + 2) y = 3(y + 2) + 3y + 71         Simplify
y² + 2y = 3y + 6 + 3y + 71              Subtract 6y from  both sides, add 71+6
y² -4y = 77
y² - 4y - 77 = 0
Factor:
(y - 11) (y+7) = 0
y = 11 or y=-7

When y = 11, x = 13, since x = y+2
Plug in to make sure:
xy = 3(x+y) + 71
11*13 = 3 (11 + 13) + 71
143 = 3 (24) + 71
143 = 72 + 71
143 = 143               
Our values work! 
Answer: 11, 13
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Step-by-step explanation:

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The "greater than or equal to" and "less than or equal to" symbols, represented by ≥ and ≤, are used to show that a number can also be equal to a set of numbers.

On the number line, open dots are used to represent > and <.

Closed dots are used to represent  ≥ and ≤.

Using this, C, D, and E must match with 1, 2, and 5.

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This lines up with choice B: the line starts at 8 and goes onwards to positive infinity.

This means that 4 matches to A, because it is the only other option with a closed dot.

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This lines up with choice D: The line starts at less than 7 and goes downwards to negative infinity.

Looking at 2 (m + 24 > 20), we can do this the same way. Subtract 24 from both sides to get:

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3 years ago
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Answer:

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Step-by-step explanation:

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Minchanka [31]

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