Lesson 11 Absolute Value Equations and Inequalities

Absolute Value Equations and Inequalities

Lesson 11

Lesson 11 is absolute value equations and inequalities, our warm up will begin with a review of two step equations and solving and graphing a two step inequality. We will being lesson 11 with our SOLVE problem; All wrestlers in the light weight division must weigh within in 7 pounds of 140. The equation the absolute value of X minus 140 equals 7, represents the given situation. What are the minimum and maximum weights? We will begin by our underlining the question, for “S” which is studying the problem, our question which is what are the minimum and maximum weights? The second step in “S” is to answer the question what is this problem asking me to find. This problem is asking me to find the minimum and maximum weights.

We will being the lesson by discussing absolute value, we know that a distance always has to be positive, for example no one could ever walk a negative 200 feet or drive a negative 25 miles, you would just be driving in an opposite direction. Our definition for the absolute value is, the distance from zero. Absolute value is the distance from zero. Therefore the absolute value of any number must be positive because the distance is always positive.

If we are going to find the absolute value of 5, we will go to the number 5 on the number line and count how many units it is to zero, 1, 2, 3, 4, 5, so the absolute value of 5 is 5 because the distance is 5, 1,2, 3, 4, 5 units. If we want to find the absolute value of negative 5, we will find negative 5 on the number line and count to see how many units negative 5 is from zero. Remember that the absolute value is the distance from zero, 1, 2, 3, 4, 5; so the absolute value of negative 5 is also equal to 5. 5 is a distance of 5 units from zero, and negative 5 is also a distance of 5 from zero. So the absolute value of 5 is 5 and the absolute value of negative 5 is also 5. When solving absolute value equations, we will actually create two separate equations. Our first equations will be X equals 5. It is the positive value of the expression inside the absolute value our second equations is the opposite of X equals 5. Our second equations is the opposite of the expression inside the absolute value equations. Our first equations in this case is already solved because X is isolated, so our first value for X is 5. In our second equation we have a negative X equals 5; in order to isolate X we must divide each side by negative 1. Leaving us with the answer X equals negative 5. We have two answers, X equals 5 and X equals negative 5. We can check our answers by plugging them back into our original equation. Our original equation is the absolute value of X equals 5. If we plug our first answer of 5 in, the absolute value of 5 equals 5; and this is a true statement because the absolute value of 5 is 5, it is 5 units from zero. Where the distance from zero is 5; so 5 equals 5. We must check our second answer as well, we will plug in negative 5 for X, Is the absolute value of negative 5, 5? Yes, it is, because the distance from zero is 5. 5 equals 5 so both of our answers check.

When you have an equation where the absolute value expression is equal to a negative, your answer is automatically no solution. This is telling us that the absolute value is equal to negative 5, the absolute value of any number or expression can never be equal to a negative because that would be that a distance was negative and we have already said that the absolute value of anything has to be a positive number because the distance can never be negative. You will never walk a negative 50 feet. So anytime an absolute value expression is isolated and equal to a negative your answer is no solution. To solve problem one we will create our two equations. Our first equation will be the positive X plus 3 equal to 5 and our second equation will be the opposite of what’s in side the absolute value bars, so we must us parenthesis, to make sure that we distribute the negative to both terms. If we solve our first equation we want to isolate X. we have a plus 3 on the same side as X, so the opposite of plus 3 is subtract 3 or minus 3, keeping our equation balanced we will subtract 3 from both sides, and X will equal 2. Our second equation, our first step is to distribute the negative, we must distribute the negative to both terms, leaving us with negative X minus 3 equals 5. From here we will need to add 3 to both sides in order to isolate the negative X, the opposite of subtract 3 is to add 3, this will leave us with negative X equal to 5 plus 3 which is 8. Our X is still not isolated so we must divide both sides by negative 1, a negative divided by a negative is a positive X equal to negative 8. Our two answers are X equals 2 and X equals negative 8. From here we will plug both answer back into the original equation to check our answers. Write the original equation, the absolute value of X plus 3 equals 5; if we plug in our first answer the absolute value of 2 plus 3 equals 5; 2 plus 3 is 5 and the absolute value of 5 is 5. so our answer checks. If we want to check our second equation, we write our original equation and plug in our answer of negative 8 for X, negative 8 plus 3, is the absolute value of negative 5, the absolute value of negative 5 is 5, because there is a distance of 5 from zero, so our second answer checks.

In problem one our first step is to isolate the absolute value expression, we cannot create our two equations until the absolute value expression is isolated on one side. So in order to do this we have the absolute value of X plus 5, we want to isolate the absolute value of X, so the opposite of plus 5 is subtract 5 or minus 5, we must subtract 5 from both sides in order to isolate the absolute value of X, and 9 minus 5 is 4. we now have the absolute value of X isolated, so now we can create our two equations. Our first equation is a positive of what’s inside the absolute value expression equal to 4, and our second equations is the opposite, of what’s inside the absolute value expression, equal to 4. We don’t have to solve anything on our first equation because X is already isolated. On our second equation we have negative X or negative 1 times X, which gives us negative X equal to 4, and in order to isolate the X we are going to divide both sides by negative 1, X equals negative 4. so our 2 answers are X equals 4 and X equals negative 4. Now we are going to check our answers. We will first write our original equation, the absolute value of X plus 5 equals 9. If we plug 4 in for X, we have the absolute value of 4 plus 5 equals 9, the absolute value of 4 is 4, so 4 plus 5 equals 9 and 4 plus 5 is 9 so our answer checks. We will check our second answer by plugging into the original equation, the absolute value of negative 4 plus 5 equals 9, the absolute value of negative 4 is 4, and 4 plus 5 is equal to 9, so our second answer checks. We have two answers, X equals 4 and X equals negative 4. In problem two our absolute value expression is not isolated, so we must first isolate our absolute value expression, in order to do this we have a multiply by 5 next to the absolute value expression, so we must divide each side by 5, 5 divided by 5 is 1, they cancel which leaves us with the absolute value of X; and 40 divided by 5 is 8. We will create our two equations, where X equals 8 or the positive of what’s inside the absolute value expression, and our second is the opposite of X equals 8. If X equals 8 the X is isolated so we do not need to do anything further. We must multiply the X by negative 1 leaving us with negative X equals 8. In order for us to isolate the 8 we must divide both sides by negative 1, leaving us with X equals negative 8. We now have two answers that we need to check, to check we will begin by writing down the original equation, we have 5 times the absolute value of 8 equals 40, 5 times, and the absolute value of 8 is 8, and 5 times 8 is 40, so our first answer checks. To check our second answer we will write down our original equation, and plug in negative 8 for X it is 5 times the absolute value of negative 8, and the absolute value of negative 8 is 8, and 5 times 8 is 40 and 40 equals 40, our second answer checks. So we have one answer of X equals 8 and another answer of X equals negative 8.

We will solve our absolute value inequality using the same method we used with equations. The only different is instead of having an equal to sign we now have a less than or equal to sign. This will give us an inequality for, two inequalities as our answer. And we will be able to graph our solution on the number line. The first problem the absolute value of X plus 4 is less than or equal to 7, is a less than problem. If the problem is less than, you hear the word and at the end, so we know our two equations are X plus 4 is less than or equals to 7 and the opposite of X plus 4 is less than or equal to 7. To solve our first inequality we must isolate the X, we have a plus 4 and the opposite of plus 4 is subtract 4, we see that our answer will be X is less than or equal to 3. In our second equations we must first distribute the negative to both terms inside the expression, we will have a negative X minus 4 is less than or equal to 7, from here we want to isolate the X so we must add 4 to both sides, the opposite of add 4 is subtract 4, leaving us with negative X less than or equal to 11, and we still have a negative X so we will divide both sides by negative 1. When you divide by a negative in inequalities you must flip your inequality sign. So now we have X is greater than or equal negative 11, and X is less than or equal to 3. When we graph our solution we have less than or equal to so we will have a solid dot on 3 and we have greater than or equal to negative 11 so we will also have a solid dot on negative 11. Since there is an and it is greater than negative 11 and less than 3, an and will have a bar bell looking graph. We can choose any number in our solution to check and make sure we have the correct solution. Suppose we want to try zero, zero is in the solution. If we want to check we rewrite the original inequality, which is the absolute value of X plus 4 is less than or equal to 7, if we plug zero in we have zero plus 4 is less than or equal to 7, zero plus 4 is 4, and the absolute value of 4 is 4. So we ask ourselves, is 4 less than 7? Yes, it is so zero is a correct solution and this is our correct graph. We know that X less than or equal to 3 AND X is greater than or equal to negative 11.

In problem 2, we have the inequality the absolute value of 2X is greater than 10, because it is greater we see that we will have two equations and our answer will be connected with an OR. Because greater ends in the word OR. Our first inequality will be 2X is greater than 10. and our second inequality will be negative 2X, or the opposite of 2X, is greater than 10. In order to isolate the X in the first inequality we will divide both sides by 2, leaving us with X is greater than 5. In the second inequality we will multiply or find the opposite of 2X, so it’s negative 2X is greater than 10, in order to isolate X we will divide both sides by negative 2, and when you divide by a negative in inequalities you must flip your inequality sign, from greater than to less than, and 10 divided by negative 2 is negative 5. When we graph our solution we must use open circles on 5 and negative 5 because we have greater than and less than. To graph the first inequality of X is greater than 5, we will draw an arrow to the right because all of the numbers to the right are greater than 5. To graph the solution X is less than negative 5, we will draw an arrow to the left, because all of the numbers to the left are less than negative 5. If we want to check our answer we write the original inequality, the absolute value of 2X is greater than 10. We can choose any number in our solution which is less than negative 5 or greater than 5. After negative 5 the next integer is negative 6, let’s try negative 6 to make sure it is a solution to our inequality. If I plug in negative 6 for X we see that 2 times negative 6 is negative 12, so we have the absolute value of negative 12 is greater than 10. The absolute value of negative 12 is 12 and 12 is greater than 10. So we have a correct solution. Our final answer is X is greater than 5 or X is less than negative 5.

We will now go back to our SOLVE question from the beginning of the lesson, we already have studied the problem and we know that this problem is asking us to find the minimum and maximum weights. In “O” we are going to organize the facts, first we have to identify the facts. All wrestlers in the light weight division must weigh within 7 ponds of 140, first fact; the equation of the absolute value of X minus 140 equals 7 represents the given situation is our second fact, our first fact we will record it in “O”. Our second fact is also important so we will write the equation. In “L” we will line up a plan, we will choose our operation or operations and line up our plan by writing what our plan of action will be. We will use addition on one equation and multiplication and subtraction on one equation. Our plan will be to create 2 equations and solve them using opposite operations. In “V” we will verify our plan with action, our first step in “V” is to estimate. We know that one of our weights will be greater than 140 and one will be less than 140, so this will be our estimate. To carry out our plan we will create our two equations from the equation that was given in the problem. The absolute value of X minus 140 equals 7. Our first equation will be the positive X minus140 equals 7 and our second equation will be the opposite of X minus 140 equals 7. We will solve our first equation using addition we will add 140 to both sides, X will equal 147. In our second equation we must first distribute the negative to both terms, so we have negative X plus 140 equals 7; we have a plus 140 so we will subtract 140 from both sides, leaving us with negative X equals negative 133. We will divide both sides by negative 1 to isolate the X; and X equals 133. In “E” we will examine our results. The first questions we ask is does our answer make sense, if we go back to our question what are the minimum and maximum weights, 147 and 133 could be weights. The second to ask is, is your answer reasonable, we go to our estimate and our estimate says, one will be greater than 140, 147, and one will be less than 140, 133 is less than 140. We also have to ask ourselves, is our answer accurate? We can check our answer by plugging into the original absolute value equation. Our absolute value of X minus 140 equals 7. We write the original equation and we plug in 147. The absolute value of 147-140 is equal to the absolute value 7 equals 7; and the absolute value of 7 is 7. so our first answer checks We will write our original equation to check our second answer we have 133 minus 140 equals 7, the absolute value of 133 minus 140 is the absolute value of negative 7 equals 7; and the absolute value of negative 7 is 7. Both of our answers check so now we will write our final answer as a complete sentence. The minimum weight is 133 and the maximum weight is 147.

We will close our lesson by reviewing the essential questions. Our first questions is, does every number have an absolute value? Yes! The absolute value of all positive numbers is itself. The absolute value of all negative numbers is its opposite, making it a positive number, and the absolute value of zero is zero. A distance must always be positive so the absolute value of any number is positive; except for zero because the distance from zero is zero. Question 2, how many solutions do absolute value equations have? Absolute value equations can have 1 solution, 2 solutions, or no real solution. Question 3, how many solutions do absolute value inequalities have? Absolute value inequalities have infinite many solutions there points or many solutions that we could plug in that will make the inequality true.