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Troyanec [42]
3 years ago
9

find two positive even consecutive integers such that a square of the smaller integer is 10 more than the larger integer

Mathematics
1 answer:
nalin [4]3 years ago
6 0

Answer:

the two positive consecutive integers are 4 and 6.

Step-by-step explanation:

Let the smaller integer be s; then s^2 = (s + 2) + 10.

Simplifying, s^2 - s - 2 - 10 = 0, or

s^2 - s - 12 = 0.

Solve this by factoring:  (s - 4)(s + 3) = 0.

Then s = 4 and s = -3.

If the first even integer is 4, the next is 6.  We omit s = -3 because it's not even.

The smaller integer is 4.  Does this satisfy the equation s^2 = (s + 2) + 10?

4^2 = (4 + 2) + 10  True or False?

16 = 6 + 10 = 16.

True.

So the two positive consecutive integers are 4 and 6.

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Answer:

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Step-by-step explanation:

3x-2>5x+10

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Simplify

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Divide both sides by 2

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Divide the numbers: \frac{2}{2} =1

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Apply fraction rule: \frac{-a}{b} =-\frac{a}{b}

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Step-by-step explanation:

<em><u>Given:</u></em>

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<em><u>Solution:</u></em>

(1) Let's move all of components that do not contain x (our target) to the right side (notice the change of sign of c from negative to positive):

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(2) Let's divide both sides of equation by 4:

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(3) Now, we simplify the left side:

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