Solve each rational inequality in Exercises 43–60 and graph th...

Question

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (x + 4)/(2x - 1) ≤ 3

Accepted Answer

Solve the following rational inequality. And graph the solution set on the number line X plus six divided by three X minus two is less standard equals to five. And we have four different possible inequalities here. Now to solve this, let's first move everything to one side of the inequality. So X plus six divided by three X minus two. Now we have a five on the right side. So we're gonna move it over by subtracting, we get minus five as lesser and equals to zero. Now we want to combine our fraction together with our negative five. So let's turn our negative five into a fraction by multiplying by a denominator. If we take neither five multiply it by three X minus two, divided by three X minus two, X plus six, divided by three X minus two minus five, multiplied by three, X minus two, divided by three X minus two. We can combine our fractions. Now the X plus six M parenthesis, five multiple battery, X minus two gives us 15 X minus divided by three X minus two which we then get X plus six minus 15 X minus 10, which is now A plus 10 because they are negative divided by three X minus two, which this simplifies even further to negative 14 X plus 16, divided by three X minus two, which is still less than or equals to zero. Now, to solve this, we can set our numerator equal to zero and our denominator equals to zero. So our first one negative 14 X plus 16 equals zero. Solving for X sub attracts 16 on both sides. And then we divide by negative 14, we get X is equals to 8/7. Also set our denominator equals to zero. And solve for that if three, X equals two, divided by three, gives us X equals two dirt. We now have three intervals that we can check our first interval. So we will test intervals. Our first interval will be negative infinity to our smaller number which is two thirds, two thirds to 8/ and finally 8/7 to infinity. So you want to see which one of these intervals has a negative value because you're looking for less equals to zero. So let's find some test values to use for our intervals. Let's say our first one is X equals zero. Our middle interval X is one, our final interval X equals two. If we plug in X equals zero into our equation, we get negative eight. We plug one into our equation. We were kid two. If you plugged in two into our equation, we would get negative three. Now we have two intervals which are negative, negative infinity to two thirds and 8/7 to infinity. That means the answer to our problem will be negative infinity to two thirds and 8/7 to infinity. Finally, because our numerator with 8/7 that will get our square bracket, the two thirds will get a round bracket. So we now have the answer to our problem. If we look at our possible number lens, we determine the star problem as answer. B ok. I hope to help you solve the problem. Thank you for watching. Goodbye.

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (x + 4)/(2x - 1) ≤ 3

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation and Graphing Solutions

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