Question 1091315
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Let x = the lesser of two positive consecutive even integers.
Then x+2 = the greater of the two positive consecutive even 
integers.
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5 times lesser of two positive consecutive even integers 
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That's 5 times x, which is 5x
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is at most...
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That means "is less than or equal to",

written {{{5x}}}{{{""<=""}}}
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...four times the greater?

That's 4 times (x+2), which is written 4(x+2).

So altogether we now have:

{{{5x}}}{{{""<=""}}}{{{4(x+2)}}} and 
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What are the possible 4 pairs of positive consecutive even integers?
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We solve the inequality:

{{{5x}}}{{{""<=""}}}{{{4(x+2)}}}

Distribute to remove the parentheses on the right:

{{{5x}}}{{{""<=""}}}{{{4x+8}}}

Subtract 4x from both sides:

{{{x}}}{{{""<=""}}}{{{8}}}

Since the lesser, x, must be even, x=2, x=4, x=6, or x=8  

So there are 4 possibilities for x, namely, 2, 4, 6, and 8.

The four pairs of positive consecutive even integers 
fulfilling the given requirements are:

{2,4}, {4,6}, {6,8}, {8,10}

Edwin</pre>