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September 30, 2016
Posted by **peter** on Monday, June 14, 2010 at 2:31pm.

4 < 2x+1/x-3 < 7

- math -
**Reiny**, Monday, June 14, 2010 at 6:10pmmultiply by x-3

4(x-3) < 2x + 1 < 7(x-3) , if x-3 > 0 or x > 3

4x-12 < 2x+1 or 2x+1 < 7x - 21

2x < 13 ----- or -5x < -22

x < 6.5 ------ or x > 22/5

So we have 3 critical values, x = 6.5 , x = 4.4 and x = 3

dividing our number line into 4 segments.

I then take any value in each of the segments and test it in the original

left segment , x < 3, e.g let x = 0

4 < 1/-3 < 7 false

segment between 3 and 4.4, x = 4

4 < 9/1 < 7 false

segment between 4.4 and 6.5, x = 5

4 < 11/2<7

4<5.5<7 true

segment > 6.5, x = 10

4 < 21/8<7

4 < 2.625<7 false

so 4.4 < x < 6.5

btw, if you have a graphing calculator

graph

y = (2x+1)(x-3) , y = 4 and y = 7 and you will see that they intersect at (4.4 , 7) and (6.5 , 4)

so the graph is between 4 and 7 for my values