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find 3 consecutive integers such that the product of the first 2 is 16 more than the product of the last two​

Find 3 Consecutive Integers Such That The Product Of The First 2 Is 16 More Than The Product Of The Last Two class=

Sagot :

LeviTheMathMajoridn

Answer:

4, 5, and 6

Step-by-step explanation:

  • let x, (x+1), and (x+2) be the consecutive integers.

  • It says that the product of the first two [x and (x+1)] is 10 less than the product of the last two [(x+1) and (x+2)]

  • Translate to Mathematical statement.

x (x + 1) = (x + 1)(x + 2) - 10

x² + x = x² + 3x + 2 - 10

  • Combine Like Terms.

x² + x = x² + 3x - 8

x² - x² + x = x² - x² + 3x - 8 (SPE)

x = 3x - 8

x - 3x = 3x - 3x - 8 (SPE)

-2x = -8

-2x/-2 = -8/-2 (DPE)

x = 4

  • Use the value of x to get the other two integers.

x = (4) = 4

(x + 1) = (4 + 1) = 5

(x + 2) = (4 + 2) = 6

  • Checking:

x (x + 1) = (x + 1)(x + 2) - 10

4 (4 + 1) = (4 + 1)(4 + 2) - 10

4 (5) = (5)(6) - 10

20 = 30 - 10

20 = 20 ✓

Therefore, the three consecutive integers are:

4, 5, and 6

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