help please...anyone? :)

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find the set of values of x for which

a) 4x - 3 > 7 - x

b) 2x^2 - 5x - 12 < 0

c) both 4x - 3 > 7 - x and 2x^2 - 5x - 12 < 0

thanx people ^_^

Posted Thu 1st November, 2012 @ 15:55 by Sabah

6 Answers

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itz okay thanx anywayyz edward pinches ^__^ im sure itz fyn

Answered Thu 1st November, 2012 @ 16:14 by Sabah
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Sabah wrote:

find the set of values of x for which

a) 4x - 3 > 7 - x

b) 2x^2 - 5x - 12 < 0

c) both 4x - 3 > 7 - x and 2x^2 - 5x - 12 < 0

thanx people ^_^

a) 4x-3>7-x (bring -x to the other side)

3x-3>7 (bring -3 to the other side)

3x>10 (divide by 3

x>10/3

b) 2x^2 - 5x - 12<0

(factorise it)

(2x+3)(x-4)=0 (make it equal to 0 to solve for x)

2x+3=0, x-4=0

therefore, x=-3/2 and x=4

Then, draw the graph and plot the co-ordinates (-3/2,0) and (4,0)

(it's a u shaped curve as a>0), you should notice the range of values after drawing the graph are x>-3/2 and x<4

c) draw an inequality line (use a dot and and arrow to show the range of values both satisfy)

then, you should notice that both equations satisfy x> 10/3

I hope this helps!  

Answered Sat 3rd November, 2012 @ 21:25 by gaffer dean
Edited by gaffer dean on Sat 3rd November, 2012 @ 21:26
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thank you sooo much gaffer dean XDDD

Answered Tue 6th November, 2012 @ 20:42 by Sabah
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no problem. Happy to help.

Answered Wed 7th November, 2012 @ 15:59 by gaffer dean
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^_^ XDD

Answered Wed 7th November, 2012 @ 17:59 by Sabah
  • -1 votes

a) 4x-3 >7-x

get x terms all to one side, and number all to one side

4x+x>7+3

5x>10

x>10/5

x>2 = answe

b) 2x^2-5x-12<0

quad equation

(2x     )(x       )<0

(2x+3(x-4)<0            

x<-3/2 or x<4

c) x>2 x<-3/2 x<4 worked out all ready

put in one expresison

    X<-3/2 2<x<4

think all this is write nut i id this a long time ago.. x

Answered Thu 1st November, 2012 @ 16:03 by Edward Pinches