Linnéa M.

asked • 04/27/13# (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?

## 4 Answers By Expert Tutors

Nataliya D. answered • 04/27/13

Patient and effective tutor for your most difficult subject.

(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:

**x**≤ 0

^{3}+ x^{2}- 2x**x(x - 1)(x + 2)**≤ 0 ---> x is from (-∞, -5)

**Let's assume that (x + 5)(x + 4) < 0 ---> x is from (-5, -4)**

*2.*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.

Robert J.

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.

04/28/13

Nataliya D.

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 :)

04/28/13

Nataliya D.

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

04/28/13

Russell C. answered • 06/07/13

Distinguished in Math with a BA and a Master's Student in Education

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) [x^{3}+x^{2}-52x-160)/[(x+4)(x+5)] ≤ (40x+200)/[(x+4)(x+5)] - (90x+360)/[(x+4)(x+5)]

(3) [x^{3}+x^{2}-52x-160)/[(x+4)(x+5)] ≤ [40x-90x+200-360]/[(x+4)(x+5)]

(4) [x^{3}+x^{2}-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) [x^{3}+x^{2}-52x-160) ≤ [-50x-160]

(6) [x^{3}+x^{2}-52x-160) + [50x+160] ≤ 0

(7) [x^{3}+x^{2}+(-52x+50x)+(-160+160)

(8)(x^{3}+x^{2}-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.

Robert J. answered • 04/27/13

Certified High School AP Calculus and Physics Teacher

To the first answerer:

"=> 6x^{2}- 5x < 6 Mult. both sides by x

=> 6x^{2} - 5x -6 < 0 Sub. both sides by 6"

is incorrect because if x < 0, you should have 6x^{2} - 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 = (6x^{2}-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)

Sriram G.

Thanks for the correction

04/27/13

Sriram G. answered • 04/27/13

Your friendly neighborhood TUTOR (Maths & Physics)

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)

=> 6x^{2}- 5x < 6 Mult. both sides by x

=> 6x^{2} - 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.

Sriram G.

04/27/13

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

Nataliya D.

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.

04/27/13