5x-7/x <= 4 im getting x<0 and x>=7 sry to ask this stupid q :/
RG11 Member Joined Apr 19, 2012 Messages 102 Location Sydney Gender Male HSC 2014 Apr 30, 2013 #1 5x-7/x <= 4 im getting x<0 and x>=7 sry to ask this stupid q :/
K Kurosaki True Fail Kid Joined Jul 14, 2012 Messages 1,161 Location Tubbytronic Superdome Gender Male HSC 2014 Apr 30, 2013 #2 its 0<=x<=7 I think. Multiply both sides by x^2, so that the sign doesn't change, and draw a quick graph to see which x-values correspond to y-values
its 0<=x<=7 I think. Multiply both sides by x^2, so that the sign doesn't change, and draw a quick graph to see which x-values correspond to y-values
I Indoz Member Joined Jul 4, 2012 Messages 89 Gender Male HSC 2014 Apr 30, 2013 #3 so ... wheres the absolute value?
H HeroicPandas Heroic! Joined Mar 8, 2012 Messages 1,546 Gender Male HSC 2013 Apr 30, 2013 #4 ^Above graph is y = x(x-7) We want x(x-7) <= 0 Which means we pick x-values that give negative y-values when subbed in the curve ( i have indicated the negative sign in the graph) Attachments parabola.png 7.9 KB Views: 135
^Above graph is y = x(x-7) We want x(x-7) <= 0 Which means we pick x-values that give negative y-values when subbed in the curve ( i have indicated the negative sign in the graph)
RG11 Member Joined Apr 19, 2012 Messages 102 Location Sydney Gender Male HSC 2014 Apr 30, 2013 #5 Indoz said: so ... wheres the absolute value? Click to expand... lol sry meant unknown in denominator, but was doing absolute value questions hence i got confused.
Indoz said: so ... wheres the absolute value? Click to expand... lol sry meant unknown in denominator, but was doing absolute value questions hence i got confused.
RG11 Member Joined Apr 19, 2012 Messages 102 Location Sydney Gender Male HSC 2014 Apr 30, 2013 #6 ok so this is how i did it: 5x-7/x <= 4 multiply both sides by x^2 (5x-7)x <= 4x^2 0<=(4x^2)-(5x-7)x 0<=(x)(4x-5x+7) (x)(4x-5x+7) => 0 (x)(-x+7) => 0 hence since x not equal to 0 solution: x<0 and x=>7 edit: crap, didn't realise that it was a negative parabola so you guys are right. solution is 0<x<=7 Last edited: Apr 30, 2013
ok so this is how i did it: 5x-7/x <= 4 multiply both sides by x^2 (5x-7)x <= 4x^2 0<=(4x^2)-(5x-7)x 0<=(x)(4x-5x+7) (x)(4x-5x+7) => 0 (x)(-x+7) => 0 hence since x not equal to 0 solution: x<0 and x=>7 edit: crap, didn't realise that it was a negative parabola so you guys are right. solution is 0<x<=7
RG11 Member Joined Apr 19, 2012 Messages 102 Location Sydney Gender Male HSC 2014 Apr 30, 2013 #7 thanks alot guys