Welcome to today's mathematics lesson. We are going to explore Linear Inequations in One Variable. By the end of this session, you will understand what inequations are, how to solve them algebraically, and how to represent their solutions on a number line. You will also learn how to combine multiple inequations and work with replacement sets and solution sets.
Let us begin with the fundamental idea. When we compare two quantities, say x and y, they can relate to each other in four possible ways. Either x is greater than y, written as x > y; or x is greater than or equal to y, written as x ≥ y; or x is less than y, written as x < y; or x is less than or equal to y, written as x ≤ y. Each of these statements is called an inequation. Unlike equations where both sides are equal, inequations express an imbalance, a relationship where one side is bigger or smaller than the other.
Now, what exactly is a linear inequation in one variable? Consider real numbers a, b, and c. An expression of the form ax + b > c, ax + b < c, ax + b ≥ c, or ax + b ≤ c is called a linear inequation in one variable. The symbols >, <, ≥, and ≤ are known as signs of inequality. The word linear tells us that the variable x appears only to the first power, just like in linear equations.
To solve a linear inequation means to find all possible values of the variable that make the statement true. There are six essential working rules you must master.
Rule one: when you transfer a positive term from one side of an inequation to the other, its sign becomes negative. For instance, 2x + 3 > 7 becomes 2x > 7 - 3 when we transpose the 3.
Rule two: when you transfer a negative term from one side to the other, its sign becomes positive.
Rule three: if you multiply or divide each term by the same positive number, the sign of inequality remains unchanged. So if p is positive and p is not zero, then x < y ⇒ px < py and x/p < y/p, with similar results for the other inequality signs.
Rule four: this is crucial and often where mistakes happen. If you multiply or divide each term by the same negative number, the sign of inequality reverses. That is, if p is negative, x < y ⇒ px > py and x/p > y/p. For example, x ≤ 6 implies -4x ≥ -24. Always remember: multiplying or dividing by a negative flips the sign.
Rule five: if the sign of each term on both sides is changed, the sign of inequality is reversed. Thus, -x > 5 is equivalent to x < -5.
Rule six: when both sides of an inequation are positive or both are negative, taking their reciprocals reverses the sign of inequality.
That is, if x and y are both positive or both negative, then x > y ⇔ 1/x < 1/y and x ≤ y ⇔ 1/x ≥ 1/y.
Now let us understand replacement sets and solution sets. The replacement set is the collection of numbers from which we are allowed to pick values for our variable. The solution set is the subset of the replacement set containing only those values that actually satisfy the inequation.
Here is an example. Suppose we have x < 3. If our replacement set is the set of natural numbers, then the solution set is {1, 2}. If the replacement set is whole numbers, we get {0, 1, 2}. If the replacement set is integers, the solution set extends infinitely in the negative direction: {..., -2, -1, 0, 1, 2}. And if the replacement set is real numbers, we write the solution in set-builder form as {x : x ∈ R and x < 3}.
Let me walk you through a worked example. Suppose the replacement set is natural numbers, and we need to solve 8 - x ≤ 4x - 2. We begin by gathering like terms. Transposing -x and -4x to the left and -2 and 8 to the right, we get -x - 4x ≤ -2 - 8, which simplifies to -5x ≤ -10. Now we divide by negative five, and because we are dividing by a negative, we must reverse the inequality sign. This gives us x ≥ 2. Since our replacement set is natural numbers, the solution set is {2, 3, 4, 5, 6, ...}.
Alternatively, we could have started by reversing the entire inequation: 4x - 2 ≥ 8 - x. Transposing -x to the left and -2 to the right gives 4x + x ≥ 8 + 2, so 5x ≥ 10, and x ≥ 2. This method avoids dividing by a negative number.
Here is another example with a compound inequation. Solve 2y - 3 < y + 1 ≤ 4y + 7 where y belongs to integers, or where y belongs to real numbers. We split this into two parts. First, 2y - 3 < y + 1 gives us y < 4. Second, y + 1 ≤ 4y + 7 gives us 1 - 7 ≤ 4y - y, which is -6 ≤ 3y, so y ≥ -2. Combining these, we need y to satisfy -2 ≤ y < 4. For integers, the solution set is {-2, -1, 0, 1, 2, 3}. For real numbers, we write it as {y : -2 ≤ y < 4 and y ∈ R}.
Now let us turn to representing solutions on the number line. This gives us a powerful visual way to understand inequalities. The convention is simple but important. A hollow circle marks the end of a range with a strict inequality, meaning less than or greater than, where the endpoint is not included. A filled or darkened circle marks the end where equality is included, so less than or equal to, or greater than or equal to.
For example, to show x < 2, x ∈ R, we place a hollow circle at 2 and shade all numbers to the left. To show x ≤ 2, we fill in that circle at 2. For a range like -2 < x ≤ 4, x ∈ R, we use a hollow circle at -2, a filled circle at 4, and shade everything in between.
Let me demonstrate with a complete solution. Solve and graph -1 ≤ 3 + 4x < 23 where x is real. We split this compound inequation. From -1 ≤ 3 + 4x, we get -4 ≤ 4x, so -1 ≤ x. From 3 + 4x < 23, we get 4x < 20, so x < 5. Combined: -1 ≤ x < 5. On the number line, this is a filled circle at -1, a hollow circle at 5, with the line shaded between them.
When we combine inequations, we use the words and and or with precise mathematical meanings. For and, we seek the intersection: values that satisfy both conditions simultaneously. For or, we seek the union: values that satisfy at least one of the conditions.
Consider solving 3x + 6 ≥ 9 and -5x > -15 where x is real. The first gives x ≥ 1. The second, after dividing by negative five and flipping the sign, gives x < 3. The solution satisfying both is 1 ≤ x < 3: a filled circle at 1, hollow at 3, shaded between.
Now consider -2 < 2x - 6 or -2x + 5 ≥ 13. The first part yields x > 2. The second yields x ≤ -4. Since these do not overlap, the solution is two separate rays: all points to the left of and including -4, together with all points strictly to the right of 2.
Let us apply this to sets. Given P = {x : 5 < 2x - 1 ≤ 11, x ∈ R} and Q = {x : -1 ≤ 3 + 4x < 23, x ∈ I}, where I denotes integers. For P, solving gives 3 < x ≤ 6. For Q, solving gives -1 ≤ x < 5, but since x must be an integer, Q = {-1, 0, 1, 2, 3, 4}. The intersection P ∩ Q contains only elements common to both. From P's interval, the integers are 4, 5, and 6, but recall that 3 is excluded. Only 4 appears in Q. Therefore, P ∩ Q = {4}.
Finally, let us touch upon word problems. The key is translating English phrases into mathematical inequalities. Phrases like at least or minimum become greater than or equal to. Phrases like at most or maximum become less than or equal to.
Suppose we seek three smallest consecutive whole numbers where the difference between one-fourth of the largest and one-fifth of the smallest is at least 3. Let the numbers be x, x + 1, and x + 2. The condition becomes (x+2)/4 - x/5 ≥ 3. Multiplying through by 20 gives 5(x+2) - 4x ≥ 60, which simplifies to x + 10 ≥ 60, so x ≥ 50. Since the smallest value of x satisfying x ≥ 50 is 50, the required three consecutive whole numbers are 50, 51, and 52.
Let me recap the essential points from today's lesson.
First, a linear inequation in one variable has the form ax + b > c or similar, where a is not zero and the variable appears to the first power only.
Second, when solving, remember the golden rule: multiplying or dividing by a negative number reverses the inequality sign.
Third, always identify your replacement set, as this determines whether your solution is a finite list of numbers or an interval of values.
Fourth, on the number line, hollow circles exclude endpoints for strict inequalities, while filled circles include endpoints when equality is permitted.
Fifth, and combines inequations through intersection, while or combines them through union.
Sixth, word problems require careful translation of English phrases into precise mathematical inequalities.
Linear inequations are powerful tools for modeling real-world situations where exact equality is not required, but rather boundaries and ranges. Master these techniques, and you will handle a wide variety of problems with confidence. Keep practicing, stay curious, and I look forward to seeing you in the next lesson.