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(x-8) less than or equal to [ (40/(x+4)) - (90/(x+5)) solve for x

Another one is (6x-5 ) < (6/x)

 

Do you solve for x on these the way that you would if it were an equal sign rather than <> signs?

Comments

Yes, you do solve inequality the way you solve equation. There is only one difference: If you multiply or divide both sides of inequality by negative number, you should change the sign of inequality to opposite.

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4 Answers

(x - 8) ≤ 40/(x + 4) - 90/(x + 5) 
important condition: x + 4 ≠ 0 ---> x ≠ -4     and                    x + 5 ≠ 0 ---> x ≠ -5
1. Let's assume (x + 5) > 0 and (x + 4) > 0 ---> x is from (-4, ∞)
    or                (x + 5) < 0 and (x + 4) < 0 ---> x is from (-∞. -5)
(x - 8)(x + 4)(x + 5) ≤ 40(x + 5) - 90(x + 4)
After open parentheses and combine like terms:
x3 + x2 - 2x ≤ 0
x(x - 1)(x + 2) ≤ 0 ---> x is from (-∞, -5)
2. Let's assume that (x + 5)(x + 4) < 0 ---> x is from (-5, -4)
x(x - 1)(x + 2)  0 does not have solution without conflict with (1.)
The final answer is (-∞, -5)

ANOTHER WAY TO SOLVE THIS INEQUALITY:
Divide the numeric line onto intervals (-∞, -5), (-5, -4), (-4, 8) and (8, +∞) then pickup any number from each interval plug it into original inequality, simplify and look at gotten inequality, is it true or false.

Comments

The solution is (-oo, -5) U (-4, -2] U [0, 1].

Attn: Since all the critical points are first degree, you don't need to check every interval. Check x-->oo, it doesn't work, then every the other interval must work.

Well, Robert, may be I can't count :) , but try those:
-3-8=40(-3+4)-90(-3+5)
0-8=40(0+4)-90(0+5)
1-8=40(1+4)-90(1+5) .... -90 is the critical number
P.S. Robert you're perfect from Algebra2 and up, but this is 6-7 grade :)

Sorry, Robert, it's not about count, it's about pay attention :( I forgot that there is division on the left side, I multiplied. Too bad. My apology to you and to student 

Equations & Inequalities are solved in the same way. However, in an inequality if a negative is multiplied or divided, the inequality is flipped.

You could graph the inequality, but to get exact answers I will solve this algebraically.

Original Equation:

         (x-8)  ≤  40/(x+4)  -  90/(x+5)

STEP 1: Multiplying each side of the inequality by the equivalent of 1 (lcd / lcd). STEP 2: Simplify by distribution. Leave the denominators as binomials. The left side is converted to 1 polynomial. STEP 3: Combine like terms. The right side of the inequality is condensed to 1 term. STEP 4: Simplify by adding like terms. The inequality that is left is the last one equivalent to the Original Inequality.

(1) (x-8)(x+4)(x+5)/[(x+4)(x+5)] ≤ 40(x+5)/[(x+4)(x+5)] - 90(x+4)/[(x+4)(x+5)]

(2) [x3+x2-52x-160)/[(x+4)(x+5)] ≤ (40x+200)/[(x+4)(x+5)] - (90x+360)/[(x+4)(x+5)]

(3) [x3+x2-52x-160)/[(x+4)(x+5)] ≤ [40x-90x+200-360]/[(x+4)(x+5)]

(4) [x3+x2-52x-160)/[(x+4)(x+5)] ≤ [-50x-160]/[(x+4)(x+5)]

FLAG: The next step is to multiply both sides by the denominator [(x+4)(x+5)]. This means writing down the values for x that make the expression 0 (-4 & -5). Division by 0 is undefined.

STEP 5: Multiplfy both sides by the denominator. STEP 6: Add the opposite of the right side (50x+60) to both sides. This leaves 0 on the left side. STEP 7: Combine like terms (the x-terms & constants). STEP 8: Simplify the expression. STEP 9: Factor the expression. STEP 10: Give the equality solutions. If the comparison was an "=", this would be the final step. STEP 11: Graph the equation from either step 8 or 9 & determine the range (not the same as Range, which is all the y-values) of values. STEP 12 (finish): Write down all the correct ranges from the previous step, but not including the values mentioned after the FLAG.

(5) [x3+x2-52x-160) ≤ [-50x-160] 

(6) [x3+x2-52x-160) + [50x+160] ≤ 0

(7) [x3+x2+(-52x+50x)+(-160+160)

(8)(x3+x2-2x) ≤ 0

(9) (x+2)(x-0)(x-1) ≤ 0

(10) x = -2, 0, or 1

(11) From (-∞ to -2] && [0 to 1]

(12) The following is the solution: (-∞ to -5)  U  (-5,-4)  U  (-4,-2]  U  [0,1]

I did use a graph & plugging in values from each range to verify the solutions and they work.

To the first answerer:

"=> 6x2- 5x < 6 Mult. both sides by x

=> 6x2 - 5x -6 < 0 Sub. both sides by 6"

is incorrect because if x < 0, you should have 6x2 - 5x -6 > 0

There are several ways to do such kind of problem. One way is to collect all terms in one side,

6x-5 - 6/x = (6x2-5x-6)/x = (3x+2)(2x-3)/x < 0

There are three critical points at x = -2/3, 0, and 3/2.

Check the intervals: (-oo, -2/3), (-2/3, 0), (0, 3/2) and (3/2, +oo)

Answer: (-oo, -2/3) U (0, 3/2)

Comments

Thanks for the correction

The solution to an equation is specific values of 'x', like x=a, x=b for a quadratic equation.

The solution to an inequality is a range of values of 'x' like x<a, a<x<b etc.

The arithmetic operations leading you to the simplified expression is the same however.

For example, (6x-5)<(6/x)

=> 6x2- 5x < 6              Mult. both sides by x

=> 6x2 - 5x -6 < 0          Sub. both sides by 6

Now represent the above in the  form,

(x-a)(x-b)<0

where a and b are the solutions to to f(x)=0; That is, factor the polynomial on the left.

Now look at the number line.

------|-------|--------|-------

       a         0          b

The left and right extremes of the number line are -infinity and +infinity respectively.

Now you have four ranges of values -inf<x<a , a<x<0, 0<x<b, b<x<inf.

Try one value each in these ranges in your expression to see if the inequality is satisfied. The range(s) of values which satisfies your inequality is the solution.

Note: -inf<x<a can be written as (-inf,a) => the range of values from -inf to a, not including -inf and a. (-inf,a] => the range of values from -inf to a, but including a.

 

Comments