Solving Absolute Value Equations

We will calculate the expression by first simplifying what is inside the absolute value. After this, we can remove the absolute value. Since the absolute value of a number is always non-negative, if the simplified expression is negative, it will become positive when the absolute value is removed. If it is already positive, it will remain unchanged.

We found that the expression simplifies to 5.

As in Part A, we start by simplifying the expression inside the absolute value. Then we remove the absolute value and change the sign of the expression to to positive if the simplified expression inside the absolute value is negative.

We found that the expression simplifies to 2.

Let's follow the same procedure to simplify the expression and then remove the absolute value. Remember, the absolute value of an expression is never negative.

We found that the expression simplifies to 1.

When we know two points on a number line and their midpoint, we can write an absolute value equation. The equation is written on the following format. |x- midpoint|= distance to midpoint Let's start by rewriting all equations to match this format. i. & |x-( -2) |= 4 [0.5em] ii. & |x- 4|= 2 [0.5em] iii. & |x-( -4)|= 2 [0.5em] iv. & |x- 2|= 4 Next, we will identify the midpoint and distance to each point in each of the graphs. After that, we can pair the graph with the correct equation. Let's begin with graph A.

With the information of the graph, we can write the absolute value equation of graph A. |x-( -4)| = 2 Therefore, graph A is associated with equation iii. Now let's take a look at graph B.

Now we can use this information to write the equation of graph B. |x-( -2)| = 4 Therefore, graph B is associated with equation i. Now we can go to graph C.

Using the information obtained, we can also write the equation of this graph. |x- 2| = 4 Therefore, graph C is associated with equation iv. Finally, we can look at graph D.

Now we can write the equation. |x- 4| = 2 Graph D is associated with equation ii. Now let's take a final look of all the pairings! A& → -1pt &&iii. B& → -1pt &&i. C& → -1pt &&iv. D& → -1pt &&ii.

Let's start by calculating all of the individual absolute values. Then we can add the results. Remember, since an absolute value is always non-negative, any argument that is negative changes signs and becomes positive.

Now that we found all the absolute values, we can add all the terms! 1+4+11 = 16

Again, we start by calculating all the individual absolute values. Remember that if the argument of an absolute value is negative, we need to change the sign for the argument to become positive.

Now we can add or subtract the results regularly. 99 - 17 + 180 = 262

As in previous parts, we will evaluate the absolute values and simplify the results. But we have to be careful, since if the result of an absolute value has a negative sing, the result is a negative number.

Now we can add and subtract the numbers regularly. - 2 + 11.5 - 7.5 = 2 Now we have finished. Good job!

An absolute value measures an expression's distance from a midpoint on a number line. |m+4|= 7 This equation means that the distance is 7, either in the positive direction or in the negative direction. Algebraically, we can separate the absolute value equation into two cases after the absolute value is removed, one case where 7 is positive and the other where it is negative. |m+4|= 7 ⇒ lm+4= 7 m+4= - 7 To find the solutions to the absolute value equation, we need to solve both of these cases for m.

We can see that m has to possible values. Let's write these answers from least to greatest. m_1 & = -11 m_2 & =3

As in Part A, we get two cases when we remove the absolute value.

Now we can see that x has to possible values. Let's write these answers from least to greatest. x_1 & = -7 x_2 & =23

As in previous parts, we get two cases when we remove the absolute value.

Again we can see that x has to possible values. Let's write these answers from least to greatest. x_1 & = -5 x_2 & =5

As in previous parts, we get two cases when we remove the absolute value.

This time we can also see that b has to possible values. Let's write these answers from least to greatest. b_1 & = -7 b_2 & =12

Before we can solve this equation, we need to isolate the absolute value expression.

An absolute value measures the distance from a midpoint on a number line. Note that a distance cannot be negative. This means the absolute value of a number must also be non-negative. |y-1|≠ -3 This absolute value equation has no solution.

Let's label the percentage of voters that are likely to vote for Candidate A v. The poll tells us that 55 % of voters, including a margin of error of 3 %, are likely to vote for the candidate. A margin of error indicates that the difference between the actual value v and 55 is ± 3. v - 55 = ± 3 The ± symbol indicates that the subtraction can be equal to -3 or 3. Note that we can rewrite this expression as two equations. v-55=-3 v-55= 3 If we look closely at these equations, we can see that this is how we separate the cases of an absolute value equation. Let's write the absolute value equation that results in these cases! |v-55|=3

In order to solve the absolute value equation, we start by to separating the equation back into its cases from Part A. Then we can solve each case equation individually. Let's do it!

Looking at the results, we can see that the least percent of voters who will vote for Candidate A is 52 %, and the greatest percent is 58 %.