Using algebra, find the set of values of x for which...?

Question

Using algebra, find the set of values of x for which:

2x-5 > 3/x

I know it's not just x>3, but how do I get the other solutions?

Thanks for any help.

Answer 1

2x - 5 > 3/x
2x^2 - 5x > 3
2x^2 - 5x - 3 > 0
2x^2 - 6x + x - 3 > 0
2x( x-3 ) +1 ( x-3 ) > 0
( 2x + 1 )( x-3 ) > 0

Therefore,

2x + 1 > 0 and x-3 > 0

OR

2x+1 < 0 and x-3 < 0

Consider the first case.
2x + 1 > 0 and x - 3 > 0
x > -1/2 and x > 3

==> x > 3

Consider the second case.
2x+1 < 0 and x-3 < 0
x < -1/2 and x < 3

==> x < -1/2

Therefore,

-infinity < x < -12
3 < x < infinity

Answer 2

1) 2x - 5 > 3/x (*)

2x^2 -5x > 3

2x^2 -5x -3 > 0

(2x + 1)(x - 3) > 0

2x + 1 > 0 => x > -1/2

x - 3 > 0 => x > 3

We have x > -1/2 and x > 3 => x > 3 because only ''x > 3''
verify perfectly (*). So,

Answer : x > 3 in R numbers.

Verification:

x = -1 => 2(-1) - 5 > 3/(-1) => -7 > -3 (false)

x = 0 => 2(0) -5 > 3/0 (impossible,because ''3/0'' does'nt exit)

x = 1 => 2(1) - 5 > 3/1 => -3 > 3 (false)

x = 2 => 2(2) - 5 > 3/2 => -1 > 1.5 (false)

x = 3 => 2(3) - 5 > 3/3 => 1 > 1 (false)

x = 3.1 => 2(3.1) - 5 > 3.1/3 => 1.2 > 0.9 (true)

x = 4 => 2(4) - 5 > 3/4 => 3 > 0.75 (true)

x = 10 => 2(10) - 5 > 3/10 => 5 > 0.3 (true)

etc,...

Answer 3

Soumyo almost got it right:

2x-5 > 3/x
multiply by x

2x^2 - 5x > 3

2x^2 - 5x -3 > 0

(2x+1)(x-3) > 0

solutions are x > -1/2 and x > 3

There's no "y" in this equation so there are no "critical points"

Since everything greater than -1/2 is also greater than 3, the real solution is x > 3.

Answer 4

I don't like the look of that one ... 3/x divided by 3 doesn't equal x...

Ok, try this:

2x-5>3/x
Multiply by x:


If x is POSITIVE, 2x²-5x>3
2x²-5x-3>0
(2x+1)(x-3)>0
x<-½ isn't a valid solution because this is only for positive values of x, so the only solution of positive values of x is x>3.

If x is NEGATIVE, 2x²-5x<3
2x²-5x-3<0
(2x+1)(x-3)<0
x<3 is a solution, but only for negative values of x, so x can only be less than 0.

Those are the two solutions, x<0 and x>3.

Answer 5

Multiply by x

2x^2-5x > 3

subtract 3

2x^2-5x-3 > 0

factor

(2x+1)(x-3) > 0

Set both > then 0

2x+1 > 0

x -3 > 0

Solve

x > -1/2

x > 3

when x is less that 3 the equation doesn't work so your only answer would be x > 3.

Answer 6

use the critical points. bring all in one side to get (2x^2-5x-3)/x>0 so (2x+1)(x-3)/x>0the critical points are (-1/2),0,3 so the required set of values are {x>3}U{0>x>-1/2}.exclude 0
for this type of problems make sure you NEVER cross multiply with a variable.make coefficient of x positive.

p.s. a little mistake that kenneth made was to multiply with x which might be negative.he got 1 as a possible solution .check to see that it is not so.

Answer 7

that's an extremely accepted linear algebra difficulty stated as a "least squares" difficulty. in case you have a collection of archives, you need to use the least squares approach to locate a line of perfect in good condition (a linear function of the form y=mx+b) that could be the perfect approximation on your archives- on your case, no linear function exists that could fulfill your archives, so so you might use least squares a) so what you wanna do is aid you linear equation be y=c1+c2x, so applying your given coordinates: a million=(a million)c1+(0)c2 2=(a million)c1+(a million)c2 5=(a million)c1+(2)c2 8=(a million)c1+(3)c2 8=(a million)c1+(4)c2 So the 1st column of A would be [a million,a million,a million,a million,a million], the 2d column would be [0,a million,2,3,4], A would be a 5x2 matrix b. For A^T A, you're taking A and make each row a column and each and each column a row, you're purely transposing the unique matrix A, as an occasion- the 1st column of A^T would be [a million,0], 2d column is [a million,a million], 0.33 is [a million,2], fourth is [a million,3], 5th and final column is [a million,4], then you definately purely multiply the two matrices, so as A^T is 2x5, and A is 5x2, the consequent matrix would be a sq. 2x2 matrix (i assume you realize the thank you to multiply matrices- if no longer you will discover a youtube video showing how, its no longer too difficult) c. comparable approach as above different than your multiplying A^T and the vector y, it is the vector [a million,2,5,8,8], so multiplication ought to grant you with a 2 dimensional vector d. create the augmented matrix with A^TA interior the matrix section, and the vector A^T y interior the augmented section, row decrease (might/possibly will ought to apply a calculator), then you definately've your least squares answer vector, as an occasion- if your least squares answer vector comes out to be [c,d] then your perfect in good condition linear approximation to the archives is y=c+dx assuming you realize the thank you to do something once you're in a math sort of this point

Answer 8

2x-5 > 3/x
2x^2-5x-3>0
(2x+1)(x-3) >0
So x >3 and x < -.5
(-infinity,-.5) U (3, +infinity)

Answer 9

2x-5 > 3/x
Mult. by 3

6x-30>x


x cannot be 5 b/c 6x5=30 and so the statement wouldn't be true. so x has to be < 5

x<5