#### Solving Inequalities Algebraically and Graphically

Solving Inequalities Algebraically and Graphically:

Linear Inequalities:

Whenever solving a linear inequality, treat it just like you were resolving an equation with some exceptions.

A) Whenever you multiply or divide both sides of the inequality by a negative constant, then change the sense (and direction) of inequality.

B) When both sides of an inequality are of similar sign then modify the sense of inequality if you take the reciprocal of both the sides.

C) You might chain like inequalities altogether:

•    If a < b and b < c, then a < c (that is, transitive property)
•    If a < b and c < d, then a + c < b + d. In English, that signifies that when you take two things which are smaller and place them together, it will be much smaller than the two bigger things place altogether.

D) You cannot join mixed inequalities.

•    If a < b and b > c, then you cannot state for sure that a < c or c < a.
•    If a < b and c > d, then you cannot state for sure that a + c < b + d or a + c > b + d

E) When you rewrite the whole problem, just switching sides, make sure you modify the sense of the inequality and hence it still points to similar quantity.

F) The given below operations do not modify the sense of inequality.

•    Adding a constant to both the sides of an inequality.
•    Subtracting a constant from both the sides of an inequality.
•    Multiplying or dividing both sides of the inequality by the positive constant.

Double Inequalities:

Sometimes the two inequalities are joined into one. Though, you need to be cautious:

When x > 3 and x < 6 then we can write 3 < x < 6. However, if x < 3 or x > 6, you cannot write 3 > x > 6 as that would imply that 3 > 6 and that will be a false statement and the set would be empty.

To resolve a double inequality, simply apply the operations to all three parts: If 3 < x+2 < 6, then subtract 2 from all the three parts to obtain 1< x < 4.

Absolute Value Inequalities:

This is going to give you trouble. They do not have to, however they will. People just do not obtain absolute values.

I could state you that the main reason people do not get it is as they do not understand limitations. That would be the fact, however just telling you that are not going to help you. The book is going to provide a short cut and people are going to ask ‘why are there too many rules?’ There aren't! There is only one definition of absolute value, and if you know it, and apply the limitations like you are assumed to, then the problems work out each and every time devoid of a need for the additional rules. The extra rules from book are employed to make the problems go fast, just like whenever we omit the absolute value whenever taking the square root of both sides and simply stick in the plus or minus rather. It sounds such as an extra rule, but it is just an application of absolute value which we try to get around as we do not like them.

Absolute Values-in a right way:

Whether the real inequality was a > or < doesn't influence how the problem is done when you are doing it the accurate way. There is an illustration of both, but the method is similar.

Let consider the absolute value inequality |(x - 3)/4| ≤ 5

Start by splitting the absolute value up to its two cases. Keep in mind that the absolute value can be found by dropping the absolute value signs whenever the argument is non-negative (that is, case 1) and by taking the opposite of what was in the absolute values whenever the argument is negative (that is, case 2).

(x-3)/4 ≤ 5 if (x-3)/4 ≥ 0
- (x-3)/4 ≤ 5 if (x-3)/4 < 0

Case 1:

(x - 3)/4 ≤ 5 if (x-3)/4 ≥ 0

At first the function part:

(x-3)/4 ≤ 5
x-3 ≤ 20
x ≤ 23

Now the restriction:

(x-3)/4 ≥ 0
x-3 ≥ 0
x ≥ 3

Be very much sure to pay special attention to the restrictions regarding what is in absolute value being negative and non-negative. Inequalities which take place at similar time similar to the function and its restriction should both be satisfied. That is why the two parts from resolving the first case can be joined and hence x is between 3 and 23 inclusive. It should satisfy both x≥3 and x≤23.

That is, 3 ≤ x ≤ 23

Case 2:

- (x-3)/4 ≤ 5 if (x-3)/4 < 0

At first the function. We will start off by multiplying by -1 to get rid of negative sign, however then we have to modify the sense (that is, direction) of the inequality.

- (x-3)/ 4 ≤ 5
(x-3)/4 ≥ -5
x-3 ≥ -20
x ≥ -17

Now the restriction:

(x-3)/4 < 0
x-3 < 0
x < 3

Similarly, for the second case, the two parts should both be satisfied at the same time. Also notice that when I multiplied both sides of the inequality by -4, I changed the sense of the inequality so that it was no longer ≤, but now ≥. You could have simplified this in other ways, but I usually find it easier to get rid of the negative right away.

-17 ≤ x < 3

Putting it back together:

The two inequalities within a particular case must both be satisfied at similar time. Though, the inequalities which occur between the two sides might be joined with an ‘or’. Just similar to if (x-2)(x+4) = 0, you would state the answer is x = 2 or x = -4, you would not insist that it be both 2 or -4 and the same time. The same principle holds with the two portions of absolute value. It fits altogether.

Now, put the two answers from Case 1 and Case 2 altogether:

-17 ≤ x < 3 or 3 ≤ x ≤ 23

Such two intervals can be united together to obtain the final answer.

-17 ≤ x ≤ 23

In interval notation, that will be [-17, 23]

Absolute Values in a shortcut way:

Less Than:

Whenever the absolute value is less than the right hand side, then the outcome or answer will be between opposite of right hand side and right hand side.

The absolute value inequality |x-2| < 3 can be written as -3<x-2<3. Applying methods explained in the double inequality section, can be solved by adding 2 to all parts to arrive at -1<x<5.

Greater Than:

Whenever the absolute value is more than the right hand side, then the answer will be in two portions, with an or separating them. This will be more than the right hand side or less than the opposite of right hand side.

The absolute value inequality |x-2|>3 can be written as x-2<-3 or x-2>3. Solving this turns up at x<-1 or x>5.

Note that this can’t be joined to be -1>x>5 as that implies that 1>5 that is just plain false and would be the empty set.

Absolute Values the Geometric Way:

There is no problem in using the geometric approach to solving absolute value inequalities. Not the geometric approach where you place it to the calculator, however the geometric approach where you utilize the geometric definition of absolute value. Geometric definition of absolute value is the distance from 0 on the number line. When you slightly modify it, then |x-a| is the distance from x=a on number line.

Though, this method needs that you know some of properties of absolute values. These are excellent things to know anyway (isn't it all?).

Let's consider similar inequality which we solved before: |(x - 3)/4| ≤ 5

Multiply both the sides by 4 to obtain |x-3| ≤ 20.

This states that the distance from 3 on number line is less than or equal to 20. Therefore, begin at 3 and go to 20 to the left (-17) and 20 to right (23). You want the distance to be less than (or equal to) 20 units away. This would comprise values between -17 and 23. As the equal to is comprised, you comprise the endpoints to obtain the answer (interval notation) of [-17, 23].

A little bit tough problem: |(3-5x)/2| ≥ 3

Multiply both the sides by 2 to obtain |3-5x| ≥ 6.

Here is where such properties of absolute values come into play. The absolute value of opposite of a number is similar as the absolute value of the number. That signifies that we can state |3-5x| = |5x-3| and |5x-3| ≥ 6.

Now divide both sides by 5 to obtain |x- 3/5| ≥ 6/5.

This states that the distance from 3/5 is at least 6/5. Therefore begin at 3/5 and go to 6/5 to the left (-3/5) and 6/5 to the right (9/5). As you require the distance to be more than 6/5 units away from 3/5, you want the values to the left of -3/5 and to the right of 9/5. The answer is thus x≤-3/5 or x ≥ 9/5. In interval notation, you would have to employ the union of two intervals (-∞,-3/5] U [9/5, +∞).

Polynomial Inequalities:

The polynomials are continuous. That signifies that you can draw them devoid of picking up your pencil (there is a more rigorous definition in the calculus, however that definition will work for us, now). If you are going to modify from being less than zero to being more than zero and you cannot pick up your pencil, then at some point, you should cross the x-axis. That signifies that the only place the inequality can modify is at an x-intercept, a zero, a root or a solution.

The key then, to finding out the solution set for a polynomial inequality, is to find out the zeros of the inequality (pretend it was an equation), place them on the number line, and picking a test point in each and every region.

a) Write the polynomial inequality in standard form and hence the right side is zero.

b) Determine the real solutions (avoid complex solutions comprising i) to the inequality any way which you want to. Factoring is favored; however you can use the quadratic formula when you can get it down to the quadratic factor. Such values are termed as ‘critical numbers’ (not to be confused with similar however slightly different ‘critical values’ from the calculus).

c) Place the zeros of polynomial (that is, critical numbers) on the number line. Be sure to place them in order from minimum to maximum. This is not significant to label any other value on number line. Some of the people were taught that you always place zero on the number line. That is not essential here.

d) Pick a test point in each and every interval. You will encompass one more interval than the number of test points. Plug that test point back to either factored inequality or the original inequality. When the test point provides you a true statement, then any point in that interval will work, and you wish to comprise that interval in answer. When the test point provides you a false statement, then all the points in that interval will not work and you don’t want to comprise that interval.

e) Comprise the endpoints when the inequality comprises equal to and do not comprise the endpoints when the inequality doesn’t comprise the equal to.

Rational Inequalities:

Rational inequalities are identical to polynomials; however there is an additional temptation and an extra place where critical numbers could take place.

Temptation: Yield not to temptation, for yielding is sin.

Here is a problem. When you multiply by a constant, it is pretty clear whether it is negative or positive therefore you know whether or not to modify the inequality. Though, with rational expressions, there is going to be a variable in the denominator. The variables can take on distinct values – that is why they are termed as variable. At times an expression is positive, however for other values of x, the expression is negative. Therefore, if you multiply by the least common denominator, you do not know whether you’re multiplying by a positive or a negative number unless you keep track of the restrictions! Therefore, you have your choice - fractions which you hate or restrictions which you hate? In this condition, fractions are the lesser of two evils.

Continuity:

The Polynomial functions are continuous everywhere. Though, rational functions are not. They are continuous everywhere apart from where undefined and that takes place when the denominator is zero.

If you are moving all along and you cannot pick up your pencil and you modify from negative to positive, then it has to occur at zero of the function, similar as with polynomial. However, if the function is undefined, then you encompass to pick up your pencil. When the pencil is up, there is nothing that states you cannot move to a completely different place, perhaps on the other side of x-axis whenever you put it back down.

We now encompass two places which critical numbers can take place. One is at zero of the function; and the other is where the function is undefined.

a) Write rational inequality in the standard form and hence the right side is zero.

b) Obtain a common denominator by multiplying top and bottom of terms. Keep in mind you cannot multiply via and get rid of the fractions as you do not know if what you are multiplying by is positive or negative (unless you wish to mess with restrictions - and you do not in this case).

c) Find out the critical numbers. In simple terms a critical number is anything which makes the numerator or denominator zero.

• Find out the places where the function is undefined as of division by zero. These will be critical points which can’t be comprised in the final answer even if the equal to be comprised as it would cause division by zero.
• Find the real solutions (ignore complex solutions involving i) to the function any way that you want to. To find the zeros, all you need to worry about is the numerator.

d) Place the critical numbers on number line.

e) Pick a test point in each and every interval and find out if the interval works or does not work.

f) Comprise the endpoints if the inequality comprises the equal to and don’t involve the endpoints if the inequality doesn’t involve the equal to. Be sure you do not comprise any endpoint which would cause the division by zero whenever included.

Actually, transforming the entire thing to an equation and resolving to find out the critical numbers is not that bad of a route. The book and most of the mathematics teachers suppose that you are going to keep the inequality in the problem in all way to the end. If you are willing to plug the values back to the original problem, you can change over to an equivalent sign, get rid of the fractions, and find out the critical numbers. You still have to watch the limitations, however now the restrictions are of the form x≠2 rather than x<2, and much simpler to deal with.

However, learn it this way, because when we get to chapter 3, there is going to be a fundamental concept which will greatly speed finding the solutions to these problems, and it requires (sort of) that you look at positives and negatives.

Graphing Utilities:

The other way to solve inequalities is to graph the left hand side of inequality as y1 and the right hand side as y2 and then find out intersection point. Intersection point will be your critical number. By looking at graph, you can state whenever the inequality is satisfied and record that interval. Keep in mind that whenever you are giving intervals, only the x-coordinate is essential. There is no y in the original problem; it was something that was added to make the graphing well-situated.

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