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Solve (3x+ 1/ x + 4 ) ≥ 1.
Solution
The first thing that we have to do here is subtract 1 from both of sides and then get everything in a single rational expression.
(3x + 1/ x + 4) -1 ≥ 0
(3x + 1/x+4) - (x + 4/x+4) ≥ 0
(3x +1 - ( x + 4))/ x + 4 ≥ 0
2 x - 3 / x + 4 ≥ 0
In this case there is no factoring to do thus we can go straight to recognizing where the numerator & denominator are zero.
numerator : x = 3/2 denominator : x = -4
Following is the number line for this problem.
Okay, we desire values of x that give positive and/or zero in the rational expression. It looks like the outer two regions as well as x = 3/2 . As with the first instance we will have to avoid x = -4 since that will give a division by zero error.
Then the solution for this problem is,
-∞ < x < -4 and 3/2 ≤ x < ∞
( -∞, -4) and [3 , ∞ )
-16 = n + 1
Qs1=-7+P1 (2) Qd1=15-P1+2P2+P3 (3) Qs1=Qd1
Solve following. 2 x - 4 = 10 Solution There actually isn't much to do other than plug into the formula. As with equations p merely represents whatever is within the a
Example Find out all the zeroes of P ( x ) = x 4 - 7 x 3 + 17 x 2 -17 x + 6 . Solution We found the list of all possible rational zeroes in the earlier example. Follo
have a solution.
I can find what x means I just cant do the interval notation correctly
how do you solve a different fractions
{a|a=9 ,a=N,a
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