Solve the inequality:
2x-1 / 3x-2 < 1
[the < is actually less than or equal to]
I got one of the answers, x is less than or equal to 1, but there's another. How do you find that answer?
Thanks
I mean x is greater than or equal to, not less than. sorry.
To find the other solution for the inequality (2x-1)/(3x-2) ≤ 1, we can start by multiplying both sides of the inequality by 3x-2 to eliminate the denominator:
(2x-1)/(3x-2) * (3x-2) ≤ 1 * (3x-2)
Simplifying this expression:
2x-1 ≤ 3x-2
Next, let's isolate the x term by moving all terms with x to one side:
2x - 3x ≤ -2 + 1
-x ≤ -1
To solve for x, we will divide both sides of the inequality by -1. But since we are dividing by a negative number, the direction of the inequality symbol will change:
x ≥ (-1)/(-1)
x ≥ 1
So, we have found the other solution: x ≥ 1. Therefore, the solutions to the inequality (2x-1)/(3x-2) ≤ 1 are x ≤ 1 and x ≥ 1.
To solve the inequality (2x-1)/(3x-2) ≤ 1, we need to find the values of x that satisfy this inequality.
First, let's look at the numerator and denominator of the expression (2x-1)/(3x-2) separately. We want to avoid dividing by zero, so we need to make sure the denominator is not equal to zero. Therefore, we set the denominator, 3x-2, not equal to zero and solve for x:
3x-2 ≠ 0
Adding 2 to both sides gives us:
3x ≠ 2
Dividing both sides by 3 gives us:
x ≠ 2/3
So, the value x = 2/3 is excluded from the solution set, as it would make the expression undefined.
Now, let's consider the expression (2x-1)/(3x-2) ≤ 1.
One approach to solving this inequality is to subtract 1 from both sides of the equation:
(2x-1)/(3x-2) - 1 ≤ 0
To simplify this, we need to find a common denominator:
((2x-1) - (3x-2))/(3x-2) ≤ 0
Simplifying further:
(2x-1 - 3x + 2)/(3x-2) ≤ 0
Combine like terms:
(-x + 1)/(3x-2) ≤ 0
Now the inequality is in standard form. We can proceed to find the solutions by considering three cases:
1. When (3x-2) > 0:
In this case, x is greater than 2/3.
2. When (3x-2) = 0:
We already found that x ≠ 2/3, so this case is not applicable.
3. When (3x-2) < 0:
In this case, x is less than 2/3.
To determine the sign of the expression (-x + 1)/(3x-2), we can choose a test point within each interval: for example, we can choose x = 0 (which is less than 2/3) and x = 1 (which is greater than 2/3).
For x = 0:
((-0 + 1)/(3(0)-2)) ≤ 0
(1/(-2)) ≤ 0
-1/2 ≤ 0
Since this statement is true, the test point x = 0 satisfies the inequality.
For x = 1:
((-1 + 1)/(3(1)-2)) ≤ 0
(0/(3-2)) ≤ 0
0/1 ≤ 0
0 ≤ 0
Again, this statement is true, so the test point x = 1 satisfies the inequality.
From testing these two points, we can determine that the inequality is satisfied for x ≤ 1 and x < 2/3. Therefore, the solution to the inequality (2x-1)/(3x-2) ≤ 1 is x ≤ 1 and x < 2/3.
I got:
x < 2/3 OR x >= 1
The way I got it was not straightfoward, but there seems to have been method in my madness since my solution seems to work. I checked my answer for some key and other values (x = 2/3, x = 1, x = 3/4, x = 5/6, x = 10, x = -10) and all of them turned out right.
Before we go any farther we need to ruleout the special case where the denominator = 0. That is, we solve denominator = 0 to find the illegal values for x. 3x - 2 = 0 leads to x = 2/3. So when x = 2/3, the denominator is 0, which is undefined. Thus, 2/3 cannot be part of any solution.
Okay, the off-beat part was having to handle the 2 possible cases when multiplying through by 3x - 2 to clear the inequality of fractions. If, for an inequality, you multiply by a negative number you have to reverse the direction of the inequality sign: if you multiply by a positive number, you don't.
Well, is 3x - 2 positive or negative? We don't know. So we have to try both.
I split the problem into 2 separate inequalities: one for 3x - 2 > 0 and one for 3x - 2 < 0. In simplified form they are x > 2/3 and x < 2/3, respectively.
Solving for the first I got x >= 1. Solving for the second I got x <= 1.
Plotting those answers on a number line shows that ALL (real) numbers are a solution (except for 2/3, which we ruled out at the beginning).
That seems fine, but wait. Consider the first branch, where x > 2/3. The answer obtained is x >= 1. So IF x > 2/3, then x >= 1????? What if x = 4/5? Then it is greater than 2/3 but less than 1. A contradiction? Plugging 3/4 in for x I got an invalid solution. So x = 3/4 is not a solution. In fact, any value for x that is greater than 2/3 but less than 1 is not a solution.
Next, I considered the second branch, where x < 2/3. The answer obtained is x <= 1. Looking at it the problem region matched that of the first branch.
Therefore, all numbers are a solution except for 2/3 (which would make the denominator 0) and any numbers greater than 2/3 and less than 1.
Consequently:
x < 2/3 OR x >= 1