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.
(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
(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.