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# MATH HELP: SOLVING INEQUALITIES?

Please help me with this problem:

I have tried so many ways but am still unable to solve it.

### 1 Answer

- llafferLv 71 month ago
This is similar to your other one.

(2x - 1)(x - 3)² / (x - 7) < 0

x can't be 7. if x > 7 the denominator is positive. if x < 7 the denominator is negative and will require us to flip the sign. Multiplying both sides by (x - 7) and splitting up the ranges of each inequlity into two gives us:

(2x - 1)(x - 3)² < 0 when x > 7 and (2x - 1)(x - 3)² > 0 when x < 7

in both cases (x - 3)² will always be positive so we only need to look at the other factor.

if x > 1/2 the product is positive. This is in the solution set of the second inequality

if x < 1/2 the product is negative. This is in the solution set of the first inequality

Since the inequalities are less than or greater than 0, then neither 1/2 nor 3 can be valid values for x. We'll keep this in mind for later.

Making this substitution into each we get:

x < 1/2 when x > 7 and x > 1/2 when x < 7

No values of x will be both greater than 7 and less than 1/2. So we can throw that out:

x > 1/2 when x < 7

There are values between 1/2 and 7 that will satisfy both of these so the answer is:

1/2 < x < 7

Remember when I said that x can't be 3? We have to take 3 out of it now to end up with this:

1/2 < x < 3 and 3 < x < 7

In interval notation: (1/2, 3) U (3, 7)

And in the graph, you would have open circles in three locations: 1/2, 3, and 7 and a lines connecting them between the first two and the last two.