For how many ordered pairs of positive integers (x,y) is 2x+3y<6

1

2

3

5

7

Results 1 to 9 of 9

- Jul 30th 2011, 05:37 PM #1

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- Jul 30th 2011, 05:51 PM #2

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- Jul 30th 2011, 05:56 PM #3

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- Jul 30th 2011, 05:57 PM #4

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- Jul 30th 2011, 05:58 PM #5

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- Jul 30th 2011, 06:07 PM #6

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- Jul 30th 2011, 06:07 PM #7

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## Re: Linear equation question

Passes through origin means line has form y= mx, where m is some real number.

Rearranging the perpendicular line, we get y = -4x + k, so this line has a slope of -4, so the inverse reciprocal slope is 1/4.

Which means, Line 1 has an equation of y = (1/4)x

Line 1: y = (1/4)x

Line 2: y = -4x + k

Since we know they intersect at (t, t+1), this must be a solution to both equations. Since Line 2 would have infinite solutions (since k is unknown), we choose Line 1 to solve for t.

t+1 = (1/4)t

1 = (-3/4)t

(-4/3) = t

So, t = -4/3

- Jul 31st 2011, 02:48 AM #8

- Jul 31st 2011, 04:07 AM #9

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## Re: Inequality question

Hello, RK29;669895!

$\displaystyle \text{For how many ordered pairs of }positive\text{ integers }(x,y)\text{ is }2x+3y \,<\, 6$

. . $\displaystyle (a)\,1 \qquad (b)\;2 \qquad (c)\;3 \qquad(d)\;5 \qquad (e) 7$

Forintegers, there is one pair: $\displaystyle (1,1)$*positive*

Code:| 2* | * 1+ ♠ * | * - - + - + - + - * - - | 1 2 3 |