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Find two consecutive odd integers such that three times the smaller one exceeds two times the larger one by $7$.

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  • $\begingroup$ What have you attempted so far? $\endgroup$ – daOnlyBG Nov 7 '14 at 18:45
  • $\begingroup$ Unfortunately nothing :( $\endgroup$ – mohmmed Nov 7 '14 at 18:49
  • $\begingroup$ by my Guess the answer is 11,13 but i don't how i make it in formulas $\endgroup$ – mohmmed Nov 7 '14 at 18:51
  • $\begingroup$ no body can solve it?! $\endgroup$ – mohmmed Nov 7 '14 at 18:59
  • $\begingroup$ I answered a similar quesiton a little while ago, you might try to read that answer to get some more practice for turning words into algebraic expressions. $\endgroup$ – Eric Stucky Nov 7 '14 at 22:25
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Let $x$ be the least of two consecutive odd numbers. So the greater of the two consecutive odd numbers must be $x+2$.

Now, we want [three times the smaller odd number] to exceed [twice the larger odd number] by $7$. So the difference between $3x$ and $2(x+1)$ should equal seven.

$$3x -2(x+2) = 7 \quad\iff \quad x = 11$$

So we have the least of the odd numbers. What is the next consecutive odd number after $x$?

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  • $\begingroup$ thenk you for explien :) $\endgroup$ – mohmmed Nov 7 '14 at 19:59
  • $\begingroup$ by mean = here? $\endgroup$ – mohmmed Nov 7 '14 at 20:00
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"Find two consecutive odd integers..."

OK, that's $2k+1, \space2k+3$, where $k\in \mathbb{Z}$.

"...such that three times the smaller one..."

$$3(2k+1)$$

"...exceeds two times the larger one..."

$$ 2(2k+3)$$

..."by 7."

$$3(2k+1)=2(2k+3)+7$$

Solve for $k=5$, and then plug that into $2k+1$ and $2k+3$ to get $11$ and $13$, respectively.

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  • $\begingroup$ why you put 2k ?? $\endgroup$ – mohmmed Nov 7 '14 at 19:44
  • $\begingroup$ i don't get it , $\endgroup$ – mohmmed Nov 7 '14 at 19:48
  • $\begingroup$ What is an even number? $\endgroup$ – daOnlyBG Nov 7 '14 at 19:50
  • $\begingroup$ I know what the Question mean, but what have to do with that? $\endgroup$ – mohmmed Nov 7 '14 at 19:52
  • $\begingroup$ If you answer me, I can explain to you how I got 2k+1. $\endgroup$ – daOnlyBG Nov 7 '14 at 19:52
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Three times the smaller one exceeds two times the larger one by 7.

$\text{Three times the smaller one} = \text{(two times the larger one)} + 7$.

$3\times\text{the smaller one} = 2\times\text{(the larger one)} + 7$.

$3\times\text{the smaller one} = 2\times(\text{the smaller one}+2) + 7$.

$3x = 2(x+2) + 7$

$\dots$

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  • $\begingroup$ I would also like to add that $x=2t+1$. Sorry about that everyone. $\endgroup$ – John Joy Nov 8 '14 at 22:14

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