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Mathematics 13 Online
OpenStudy (anonymous):

Write an inequality to represent the problem (1 pt.) and then solve the inequality by writing the pairs which solve it (1 pt.). Find all sets of two consecutive positive odd integers whose sum is less than or equal to 18.

OpenStudy (anonymous):

if you call the positive integers \(m\) and \(n\) then there sum is \(m+n\)

OpenStudy (anonymous):

and if the sum is less than or equal to 18, then you can write \[m+n\leq 18\]

OpenStudy (anonymous):

Yeah...

OpenStudy (anonymous):

Sorry, my internet is slow.

OpenStudy (anonymous):

i think the site slowed up for a moment i had to log out and back in

OpenStudy (anonymous):

So it would be like, \[1 + 3 \le 18\] right? Ohh.

OpenStudy (anonymous):

that would be one solution, yes

OpenStudy (anonymous):

Then the next would be \[5 + 7 \le 18\] right?

OpenStudy (anonymous):

sure there are lots

OpenStudy (anonymous):

Yeah, I got it now. Thank you so much!

OpenStudy (anonymous):

oh i did not read carefully!!

OpenStudy (anonymous):

Hm?

OpenStudy (anonymous):

Okay then....

OpenStudy (anonymous):

the inequality should be \(n+n+1\leq 18\) or \(2n+1\leq 18\)

OpenStudy (anonymous):

but the solutions are still the same the pairs can be 1,3 or 3,5 or 5,7 or 7,9

OpenStudy (anonymous):

So, \[1 + 3 + 1 \le 18\] instead?

OpenStudy (anonymous):

no the solution is still pairs of numbers

OpenStudy (anonymous):

like you were writing above i think i wrote a complete list

OpenStudy (anonymous):

Oh, but the inequality would be what you said before, n+n+1≤18 or 2n+1≤18?

OpenStudy (anonymous):

So my answer would be like this, Inequality: \[n+n+1 \le 18\] The solutions would be {(1,3),(5,7),(7,9)} Correct?

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