How do I solve the inequality (x-2) (x-4) (x-7)/(x+2) (x+4) (x+7)>1?
[math]\dfrac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)} > 1[/math][math]\implies \dfrac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)} - 1 > 0[/math][math]\implies \dfrac{(x-2)(x-4)(x-7) - (x+2)(x+4)(x+7) }{(x+2)(x+4)(x+7)} > 0[/math][math]\implies \dfrac{ (x^3 - 13x^2 + 50x - 56) - (x^3 + 13x^2 + 50x + 56)}{(x+2)(x+4)(x+7)} > 0[/math][math]\implies \dfrac{ -26x^2 - 112 }{(x+2)(x+4)(x+7)} > 0[/math][math]\implies \dfrac{ 26x^2 + 112 }{(x+2)(x+4)(x+7)} < 0 — (1)[/math]For all real values of [math]x[/math], Numerator of [math](1)[/math], will always be [math]> 0[/math]Hence, for [math](1)[/math] to be true, Denominator must be [math]< 0[/math][math]\implies[/math] Range for [math]x = [-∞, -7], [-2, -4], [\ne -2], [\ne -4][/math]
How do you solve the inequality 1/x < 4 ?
(-infinity, 0) and (1/4, infinity) is correct. When solving an inequality with x as a denominator, you know that x cannot be zero. However, when x is negative, then you have to switch the sign of the inequality when you multiply both sides by x. You set up to cases, as shown here: CASE I - x is a positive number 1/x < 4 1 < 4x 1/4 < x CASE II - x is a negative number 1/x < 4, x is not 0. Looking at the equation itself, any negative number you plug in for x will turn the entire quantity into a negative number, which by definition is less than 4. Therefore any negative number works. So your final intervals are (-infinity, 0) and (1/4, infinity)
How to solve inequalities?
I don't really understand inequalities... For example, the question is: 2k^2 - 5k + 2 ≤ 0 Can you talk me through it? Not just give me the answer Thanks =]
How do I solve this inequality 2x − 3/3x − 2 ≤ 1?
This problem can be interpreted in multiple ways due to the lack of parentheses, but I am going to assume you are asking for (2x-3)/(3x-2)<1 (I am using < as less than or equal to because my computer does not have a key for this) (also, if we just used the order of operations on this, we just get 2x-(1/x)-2<1). To solve this, I would probably make a quick chart. The way I would construct this chart is I would mark all 0’s, all asymptotes, and all jumps of a graph on the chart, then draw a negative of a positive between each, depending on whether the numbers in-between them are negative or positive (the idea is that, if there is no a gap in the graph or a jump discontinuity, then the only way to get from - to + is by passing through 0). For this to work, we need to make this relative to 0, so we subtract 1 from both sides and get (-x-1)/(3x-2)<0. I personally, would then multiply both sides by -1 to get out all the negative signs, which yields (x+1)/(3x-2)>0 (remember that we need to flip the sign because we are multiplying/dividing by a negative number). It is then clear that there is an asymptote at x=2/3, the function has no jumps (that aren’t asymptotes), and the function has a 0 at x=-1 (if you can not do this easily, I am just taking the top of the function (x+1 in this case) and solving for 0’s, then taking the bottom of the function (3x-2) and solving for 0’s. When the top is equal to 0, it will be 0/something, which is 0, and when the bottom is equal to 0, it will be something/0, which we call “undefined” (although, technically, it can be defined as every number, but that is for a different time)). Our chart would therefore be (negative infinity) + (-1) - (2/3) + (positive infinity). We can then compare this to our question, which is: (x+1)/(3x-2)>0, or “when is this greater than or equal to 0?”, and we see that (-infinity,-1]U[2/3,infinity) would follow this.
How do you solve this inequality, 4 + 2x > 3, and this inequality, 2 + 3 (x - 4) < 3 (2x - 5)?
Let's solve each inequality one by one.• 4+2x>3Subtracting 4 from both sides2x>-1Dividing by 2 on both sidesx>-1/2• 2+3(x-4)<3Simplifying,3(x-4)<1=> x-4<1/3=> x<13/3So, -1/2
How can I solve this inequality, |3x-11|<-2?
There would be 2 possible answersFirst3x-11 < -23x < 9x < 3Second3x - 11 < 23x < 13x < 13/3
How do you solve this inequality: x>√(1-x)?
TO SOLVE: x > √(1 - x)For real numbers x, , in the above inequation,1 - x should be positive or zero=> x = 1 or less than 1 …………..(1)Now, if x is negative, 1 - x will be positive but in that case LHS will be negative & RHS will be positive. So LHS can not be greater than RHS.=> x can not be negative…………..(2)Now, if we take x=0, LHS won't be greater than RHS=> x can not be zero. ………………(3)Now taking all 3 conditions into considerations..We conclude that x > 0 but less than 1.ie, 0 < x < 1Now, if we substitute different values for x , lying between 0 to 1 in the given inequationWe get the range..0.6180399 <,= x <,= 1 ……..ANS
How do I solve the inequality [(x -1) ÷ (x^2 -4 x +3)] < 1?
I am going to take a more general approach and would not cancel the common factor. Solution consists of 4 steps:Bring everything to one side of inequality and simplify it to express it in p/q form ( rational ).Factorize the numerator and denominator ( solution if numerator and denominator are critical points )Plot the solution with sign for numerator and denominator but use different color until you get a feel of what you are doing.Use the rule related to division of sign to get the results.(x -1) / (x^2 -4x +3 ) < 1=) 1 - (x -1) / (x^2 -4x +3 ) >0=) (x -1) ( x -4 ) / (x -1) ( x -4 ) > 0Solution of Numerator :1, 4Denominator : 1, 3Critical point : x = 1,3,4numerator :-ve when 1 < x < 4,+ve when (- infinity, 1) U (4, infinity )Denominator :-ve when 1 < x < 3+ve when (- infinity, 1) U (3, infinity )Plot these critical point on the number line. Put the sign for numerator and denominator based on their own solution.Now we can clearly see,x belong to (- infinity, 1) U ( (1, 3 ) U (4, infinity )