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?
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?
(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^{3} + x^{2} - 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.
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) [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.
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)
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)
=> 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.
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.