Solve algebraically: 7 / (x-8) < 9 / (5x-2)

Question

Simplify your answer. Write your answer in interval notation. Use integers or fractions for any numbers in the expression.

Answer 1

Problem statement:

Solve algebraically: 7 / (x-8) < 9 / (5x-2)

Algebraic statements with "=" in them are called equations. Algebraic statements like this one with "<" or ">" in them are called inequalities.

To "Solve for x" in an inequality, perform algebraic operations on the inequality that:

1. Does not change the truth of the inequality statement

2. Makes progress towards the solution

3. Leads to solution statements that look like: x = a numeric expression, x < numeric expression, x > numeric expression, or some combination of these statements.

Here are some typical algebraic manipulations that do not change the truth of inequality statements:

1. Add or subtract numeric expressions to both sides of the inequality

2. Multiply or divide both sides of the inequality by a positive numeric expression

3. Multiply or divide both sides of the inequality by a negative numeric expression, and change "<" to ">" or ">" to "<"

There are others, which are not needed to solve this problem.

Solution:

First, notice two important values for x.

If (x - 8) = 0, x = 8. If (5x-2) = 0., x = 2/5.

These are important because when multiplying both sides of the inequality by (x-8) or (5x-2), the sign of these expressions change from negative to positive or from positive to negative when the x value changes from being greater than or less than these numbers to being less than or greater than these numbers.

Also, when multiplying both sides of an inequality by zero, it is no longer an inequality: 0 = 0.

Second, place these numbers x = 2/5 and x = 8 on a number line.

To the left of x = 2/5 write: (x-8)(5x-2) > 0

Between x = 2/6 and x=8 write: (x-8)(5x-2) < 0

To the right of x = 8 write: (x-8)(5x-2) > 0

Multiply both sides of the inequality by (x-8)(5x-2) for values of x other than 2/5 and 8.

There re two cases:

Case 1: (x-8)(5x-2) > 0 [when x <2/5 or x > 8]

7 / (x-8) * (x-8)(5x-2) < 9 / (5x-2) * (x-8)(5x-2)

7 (5x-2) < 9 (x-8)

Use the distributive property to multiply by 7 and 9

35x - 14 < 9x - 72

Add 14 to both sides and subtract 9x from both sides:

26x < -58

Divide both sides by 26

x < -58/26

x < -2 1/13

Since these x values are also < 2/5 they are part of the solution.

Case 2: (x-8)(5x-2) < 0 [when 2/5 < x < 8]

When the inequality is multiplied by (x-8)(5x-2) the < becomes >

7 / (x-8) * (x-8)(5x-2) > 9 / (5x-2) * (x-8)(5x-2)

7 (5x-2) > 9 (x-8)

The math is the same s case 1.

For case 2,

x > -2 1/13

Since x values such that 2/5 < x < 8 are also > -2 1/3 they are part of the solution.

Notice if x = -2 1/13, then 7 / (x-8) = 9 / (5x-2), so it is not part of the solution.

Answer:

x < -2 1/13 or 2/5 < x <8

In interval notation: (-∞,-2 1/13) U (2/5,8), where U represents union.

Interval notation uses parentheses () when the numbers at the ends of the interval are not included. It uses square brackets [ ] when the numbers at the ends of the interval are included.

Answer 2

For this problem, I would get one side = 0, so

1. subtract (9/(5x-2) from both sides).
2. (7/(x-8)) - (9/(5x-2) < 0
3. then you will need a common denominator on the left side to add the numerators, so multiply the 7 by (5x-2) and the -9 by (x-8); the denominator now is the product of (x-8)(5x-2) on the left side.
4. (7(5x-2) - 9(x-8))/(x-8)(5x-2)
5. distribute out and gather like terms in the numerator, and refactor the numerator (as you can)
6. (35x - 14 -9x + 72)/(x-8)(5x-2)
7. (26x + 58)/(x-8)(5x-2)
8. (2(13x + 29))/(x-8)(5x-2)
9. set each expression of x = 0 and solve to get your critical points to test
10. 13x + 29 = 0; x = -29/13
11. x - 8 = 0; x = 8
12. 5x-2 = 0; x = 2/5
13. test x values in in between and on either side of those points to see if you have any invalid solutions; and discard those if you do.
14. Intervals to test points on include: (-∞,-29/13), (-29/13,2/5), (2/5,8) and (8, ∞)
15. I selected these x-values to test on the original equation to determine if it was true or false, for any false answers, that area is not included. x = -3, -1, 3 and 9
16. At x = -1 and x = 9 you will get false answers (.7<-1.29 and 7<.21), therefore these are the remaining intervals that satisfy:
17. (-∞,-29/3) ∪ (2/5, 8)

I hope you have gained some insight on the process for this type of problem based on the above example and it was helpful to you! :-)