The algebra to get a solution involves a certain kind of complexity. if the question asked for equality rather than >=, you'd solve it by "clearing fractions", "cross multiplying". You'd multiply both sides by (3x-7)(3-2x), and so would be left with an ordinary linear equation (3-2x) = 4(3x-7). The complicating problem with doing such ordinary algebra when using inequalities is this: if you happen to be multiplying by a negative number, you'll need to flip the inequality, so greater-than-or-equals becomes less-than-or-equals, and visa versa. But in algebra, you can't look and just "see" that the object you're multiplying by is negative. You can still do it "directly" but it introduces complexities to handle this problem. I'll show you how to do it directly, and then tell show you the better way - the much better way, which is how people solve these inequalities in practice. The better way is amazingly quick and simple, and so is a technique worth learning and understanding. It relies on the Intermediate Value Theorem.
One specific thing to note before beginning: x = 3/2 and x = 7/3 won't be in the solution set, since the inequality is undefined gibberish at those two x values. In what follows, I'm always assuming that x isn't either of those two values.
DIRECT APPROACH:
You need to know the sign of (3x-7)(3-2x), but you can't see it. So what you do is just stipulate it, effectively breaking the problem down into separate cases (which are ultimately just doing the problem algebraically in a different way on different intervals of the real line where you know the sign of that product.) It works like this:
Case 1: (3x-7)(3-2x) > 0.
Then 1/(3x-7) >= 4/(3-2x) implies [1/(3x-7)] [(3x-7)(3-2x)] >= [4/(3-2x)] [(3x-7)(3-2x)] implies 3 - 2x >= 4(3x-7) implies 3 - 2x >= 12x - 28 implies 31 >= 14x implies x <= 31/14. In interval notation that's x in ( - infinity, 31/14].
Case 2: (3x-7)(3-2x) < 0.
Then 1/(3x-7) >= 4/(3-2x) implies [1/(3x-7)] [(3x-7)(3-2x)] <= [4/(3-2x)] [(3x-7)(3-2x)] implies ...(same stuff, but flipped inequality)... x >= 31/14. In interval notation that's x in [31/14, infinity).
Not done yet. You now have to find what the intervals are in Case 1 and Case 2. If that product (3x-7)(3-2x) isn't positive, then it's negative, and visa versa, so those two cases will be set complements... where the encompassing set is all reals except 3/2 and 7/3, because the product is never allowed to be zero. The point is, you only need "to do the work" for one of those cases, and you'll have the answer for the other case. It works like this:
If Case1, then (3x-7)(3-2x) > 0, so, since the product of two numbers is positive means that either both numbers are positive, or both number numbers are negative, have
either
Case1A: 3x-7 > 0 and 3-2x > 0
or
Case1B: 3x-7 < 0 and 3-2x < 0.
Now Case 1A says that x > 7/3 and x < 3/2. But that's impossible, there are no such x, since 3/2 < 7/3. (If there were such an x, then 7/3 < x < 3/2, so 7/3 < 3/2, which is absurd). Such "no solutions in this case" cases come up often when using this method, so watch out for them! Case1B says that 3x-7 < 0 and 3-2x < 0, which becomes that x < 7/3 and x > 3/2. Case 1B is thus the interval 3/2<x<7/3, or in interval notation (3/2, 7/3).
So have that Case 1 implies that (3x-7)(3-2x) > 0, so either Case1A (3x-7 > 0 and 3-2x > 0) or Case1B (3x-7 < 0 and 3-2x < 0). But Case1A has no solutions, so Case 1 implies Case 1B implies that x is in (3/2, 7/3).
Thus Case 1 is the interval (3/2, 7/3).
Case 2 will be the complement of Case 1 (excluding 7/3 and 3/2), so Case 2 is (-infinity, 3/2) Union (7/3, infinity).
Although the answer is known, this shows you how you could work it out:
If Case 2, then (3x-7)(3-2x) < 0, so in order for the product to be negative, exactly one is positive, and exactly one is negative. Thus
either
Case 2A: 3x-7 > 0 and 3-2x < 0
or
Case 2B: 3x-7 < 0 and 3-2x > 0.
Now Case 2A is that 3x-7 > 0 and 3-2x < 0, which is that x > 7/3 and x > 3/2. In order for both of those conditions to hold (see the word "and"!) you need x to be greater than the larger of the two, so x >7/3. In interval notion, that's (7/3, infinity).
For Case 2B, 3x-7 < 0 and 3-2x > 0, so x <7/3 and x < 3/2. In order for both of those conditions to hold (see the word "and"!) you need x to be less than the smaller of the two, so x < 3/2. In interval notion, that's (- infinity, 3/2).
Thus Case 2 is either Case 2A or Case 2B which is either (7/3, infinity) or (- infinity, 3/2), and so Case 2 is the union: (- infinity, 3/2) Union (7/3, infinity). (This is what we already knew from Case 2 being the complement of Case 1.)
Now, you have to build final solution by combining the cases with the solution given the cases:
Remember that Case 1 implied that x <= 31/14, and Case 2 implied that x >= 31/14. Combining that with Case 1 being the interval (3/2, 7/3), and Case 2 being the set (- infinity, 3/2) Union (7/3, infinity), you're ready to build the full solution to your problem:
(Watch the and's and or's): If 1/(3x-7) >= 4/(3-2x), then
Either [ x in (3/2, 7/3) AND x in ( - infinity, 31/14] ], OR [ x in (- infinity, 3/2) Union (7/3, infinity) AND x in [31/14, infinity) ]. (see how that's Case 1 AND the solution given Case 1 - OR - Case 2 AND the solution given Case 2)
That becomes
x in { (3/2, 7/3) INTERSECT ( - infinity, 31/14] } UNION { ( (- infinity, 3/2) Union (7/3, infinity) ) INTERSECT [31/14, infinity) }.
See how and's and or's become intersections and unions?
Looking at the pieces: { (3/2, 7/3) INTERSECT ( - infinity, 31/14] } = (3/2, 31/14], because 3/2 < 31/14 < 7/3. For the same reason { ( (- infinity, 3/2) Union (7/3, infinity) ) INTERSECT [31/14, infinity) } = (7/3, infinity).
Thus the complete solution is "x is in (3/2, 31/14] UNION (7/3, infinity)".
BEST APPROACH:
1. Transform you inequality, using addition/subtraction (which never flips the inequality sign) into an inequality where one side is being compared to 0 (so one of these >0, >=0, <0, <=0).
2. Find where the function (the expression being compared to 0) is continuous and where it's zero.
Note: any expression/formula in x, all trig functions, absolute values, radicals, exponentials, logarithms, when combined using any arithmetic operations (add/subtract/multiply/divide) and compositions, will be continuous wherever it's defined. So basically anything you know how to write down as a single formula will be continuous everywhere in its domain, meaning everywhere that it's defined. So determining where a simple expression is continuous is the same as just determining its domain, i.e. determining where it's defined.
3. Apply the following fact (which is an immediate corollary of the Intermediate Value Theorem):
If a function is continuous and never zero on an interval, then the function is either always positive or always negative on that interval.
For emphasis, that's requiring 3 things:
i) it's about a specific interval
ii) the function is continuous (on that interval)
iii) the function is never 0 on that interval
4. Use #2 to break down the real numbers into intervals where the function is continuous and never 0. Apply #3 to conclude that the function is always positive or always negative on each of those intervals. Now PICK A SINGLE POINT in each interval, and discover whether the function at that point is positive or negative. Then the function will have the same sign (+/-) as it has at that point over that entire interval! Thus you'll have learned everywhere where the function is positive and where it's negative.
In your case:
STEP1: \(\displaystyle \frac{1}{3x-7} \ge \frac{4}{3-2x}\), so write it as \(\displaystyle \frac{1}{3x-7} - \frac{4}{3-2x} \ge 0\), and define a function f by \(\displaystyle f(x) = \frac{1}{3x-7} - \frac{4}{3-2x}\).
Your problem becomes "Find all x such that f(x) >= 0".
STEP2: Know that f is going to be continuous everywhere that it's defined, so everywhere except at x = 7/3 and at x = 3/2. So only need to find where f(x) = 0.
To solve f(x) = 0, first clear fractions by multiplying both sides of that equation by (3x-7)(3-2x), giving:
\(\displaystyle f(x) = 0\) implies that \(\displaystyle f(x)(3x-7)(3-2x) = 0\) implies that \(\displaystyle \left( \frac{1}{3x-7} - \frac{4}{3-2x} \right) (3x-7)(3-2x) = 0\) implies that \(\displaystyle (3-2x) - 4(3x-7) = 0\),
implies that \(\displaystyle 3 - 2x - (12x -28) = 0\) implies that \(\displaystyle 3 - 2x -12x + 28 = 0\) implies that \(\displaystyle 31 - 14x = 0\) implies that \(\displaystyle 14x = 31\) implies that \(\displaystyle x = \frac{31}{14}\).
Thus the only possible solution to \(\displaystyle f(x) = 0\) is \(\displaystyle x = \frac{31}{14}\), and checking shows that \(\displaystyle f(31/14) = 0\).
STEP3: f is continuous and never zero on the following intervals: \(\displaystyle \left(-\infty, \frac{3}{2}\right)\), and \(\displaystyle \left(\frac{3}{2}, \frac{31}{14}\right)\), and \(\displaystyle \left(\frac{31}{14}, \frac{7}{3}\right)\), and \(\displaystyle \left(\frac{7}{3}, \infty\right)\).
Thus f is entirely positive or entirely negative on each of those intervals.
STEP4: Pick the (relatively easily computed) values \(\displaystyle x = 0 \in \left(-\infty, \frac{3}{2}\right)\), and \(\displaystyle x = 2 \in \left(\frac{3}{2}, \frac{31}{14}\right)\), and \(\displaystyle x = \frac{32}{14} = \frac{16}{7} \in \left(\frac{31}{14}, \frac{7}{3}\right)\), and \(\displaystyle x = 3 \in \left(\frac{7}{3}, \infty\right)\).
Then \(\displaystyle f(0) = \frac{1}{3(0)-7} - \frac{4}{3-2(0)} = \frac{-1}{7} - \frac{4}{3} <0\),
and \(\displaystyle f(2) = \frac{1}{3(2)-7} - \frac{4}{3-2(2)} = \frac{1}{-1} - \frac{4}{-1} = -1 - (-4) = 4 - 1 = 3 > 0\),
and \(\displaystyle f(16/7) = \frac{1}{3(16/7)-7} - \frac{4}{3-2(16/7)} = \frac{7}{3(16)-7^2} - \frac{4(7)}{3(7)-2(16)}\) \(\displaystyle = \frac{7}{48 - 49} - \frac{28}{21-32} = -7 + \frac{28}{11} = -7 + 2 + \frac{6}{11} < 0\),
and \(\displaystyle f(3) = \frac{1}{3(3)-7} - \frac{4}{3-2(3)} = \frac{1}{2} - \frac{4}{-3} = \frac{1}{2} + \frac{4}{3} > 0\).
Thus f is always negative on \(\displaystyle \left(-\infty, \frac{3}{2}\right)\), and f is always positive on \(\displaystyle \left(\frac{3}{2}, \frac{31}{14}\right)\), and f is always negative on \(\displaystyle \left(\frac{31}{14}, \frac{7}{3}\right)\), and f is always positive on \(\displaystyle \left(\frac{7}{3}, \infty\right)\).
Since the problem is "Find all x such that f(x) >= 0" have that f is always positive on \(\displaystyle \left(\frac{3}{2}, \frac{31}{14}\right) \cup \left(\frac{7}{3}, \infty\right)\), and that f is 0 only when x = 31/14.
Thus the solution to \(\displaystyle \frac{1}{3x-7} \ge \frac{4}{3-2x}\) is the solution to \(\displaystyle f(x) >= 0\), which is: "All x in the set \(\displaystyle \left(\frac{3}{2}, \frac{31}{14}\right] \cup \left(\frac{7}{3}, \infty\right)\)".