Can anyone help with this maths question? - XP Math - Forums

 XP Math - Forums Can anyone help with this maths question?

 07-30-2007 #1 environment30 Guest   Posts: n/a Can anyone help with this maths question? Find the smallest integer x that satifies each inequality2/3x > 4/7
 07-30-2007 #2 we_got_the_synergy Guest   Posts: n/a What we do is expand out the fractions to get an integer value of x, and then divide by its coefficient to solve the inequality. So:2/3x > 4/72x > 12/7 We have multiplied both sides by 3x > 6/7 We have divided both sides by 2The question asked for the smallest integer. An integer is a whole number, that is a non fraction. Since x can be any number AFTER 6/7 the first whole number we come to is 1.Therefore the answer is 1
 07-30-2007 #3 Big Bad Wolf Guest   Posts: n/a 2/3x>4/714x>12x>12/14x>6/7 or .86
 07-30-2007 #4 4Yan Guest   Posts: n/a Solution:Consider 12/3) X > 4/7divide both side with (2/3), get(1) X > (4/7) / 2/3X > 6/7So, the equation shown that the smallest integer is 1. #End#Consider 2:2/(3x) > 4/7Both side time (3x), get2 > 4/7 (3x) (if x > 0) ----------------- (1)or 2 < 4/7 (3x) (if x < 0) ----------------- (2)from (1),2 > 4/7 (3x)2 > (12/7) xdivide both side with 12/7, get2 / (12/7) > x7/6 > xAlso x > 0,So, the required smallest integer is 1.from (2),2 < 4/7 (3x)2 < (12/7) xdivide both side with (12/7), get2 / (12/7) < x7/6 < xSince x 7/6, Collision! So no solution from (2).So, we conclude the required smallest integer is 1. #End#

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