Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
3. Section 3.3
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Exercise 84 Page 121

Practice makes perfect
a

To solve the equation, we need to apply the definition of absolute value.

|x| = a ⇒ x=a or x= - a By applying it to the equation |x-3|=5 we will get two equations to solve. |x-3|=5 ↙ ↘ x-3= 5 x-3=- 5 Let's begin by solving the left-hand side equation.

x-3=5
x=8

Next, we solve the right-hand side equation.

x-3=-5
x=-2

Checking Our Answer

Checking our Solutions
Let's check our solutions by plugging them into the original equation. If we get a true statement, it will imply that the solution is correct. Otherwise, it will be incorrect. Let's begin with x=-2.

|x-3|=5
| -2-3| ? =5
|-5| ? =5
5=5

Now, we proceed by replacing x=8.

|x-3|=5
| 8-3| ? =5
|5| ? =5
5=5

b

Again, we will use the definition of an absolute value.

|x| = a ⇒ x=a or x= - aThis time, before applying the definition, we can divide both sides of the equation by 5. That way, we isolate |x| on the left side. 5|x|=35 ⇒ |x| = 7 Then, when we apply the definition of absolute value, we instantly get the solutions. |x|=7 ↙ ↘ x= 7 x=- 7

Checking Our Answer

Checking our Solutions
As above, we will check that our solutions are correct by replacing them into the original equation. Let's begin with x=- 7.

5|x|=35
5| -7| ? = 35
5 * 7 ? = 35
35 = 35

Next, we substitute x=7.

5|x|=35
5| 7| ? = 35
5 * 7 ? = 35
35 = 35

c

As before, we need to apply the definition of an absolute value.

|x| = a ⇒ x=a or x= - a As in part A, by applying it to the equation |x+1|=2 we will get two equations to solve. |x+1|=2 ↙ ↘ x+1= 2 x+1 = - 2 Let's begin by solving the left-hand side equation.

x+1=2
x=1

Next, we solve the right-hand side equation.

x+1 = -2
x=- 3

Checking Our Answer

Checking our Solutions
Let's check our solutions by plugging them into the original equation. If we get a true statement, it will imply that the solution is correct. Otherwise, it will be incorrect. Let's begin with x=-3.

|x+1|=2
| -3+1| ? =2
|-2| ? =2
2=2

Now, we proceed by replacing x=1.

|x+1|=2
| 1+1| ? =2
|2| ? =2
2=2

d

This time, notice that the right-hand side of the equation is a negative number. Since the absolute value of a number is always non-negative, it is impossible to solve this equation.

ccc Non-negative & & Negative |x+3| & ≠ & - 2 Therefore, the equation |x+3|=- 2 has no real solution.