Absolute value: Solve absolute value equations with a number outside the absolute value bars

Examples with solutions for Absolute value: Solve absolute value equations with a number outside the absolute value bars

Exercise #1

y2= -\left|-y^2\right|=

Video Solution

Step-by-Step Solution

To solve this problem, let's break it down into the following steps:

  • Step 1: Recognize that inside the absolute value, we have y2-y^2. Since y2y^2 is non-negative, y2-y^2 is less than or equal to zero.
  • Step 2: Apply the absolute value: Since y20-y^2 \leq 0, we use the property a=(a)|-a| = -(-a), resulting in y2=y2\left|-y^2\right| = y^2.
  • Step 3: Apply the outer negative sign, y2-\left|-y^2\right|, which simplifies to y2-y^2.

Therefore, the solution to the problem is correctly identified as y2-y^2.

Answer

y2 -y^2

Exercise #2

42= -\lvert4^2\rvert=

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Compute the expression inside the absolute value.
  • Step 2: Apply the absolute value.
  • Step 3: Apply the negative sign.

Now, let's work through each step:
Step 1: Compute 42 4^2 .
Since 42=16 4^2 = 16 , we have 42=16 \lvert 4^2 \rvert = \lvert 16 \rvert .
Step 2: Apply the absolute value operation. The absolute value of 16 is 16, so 16=16 \lvert 16 \rvert = 16 .
Step 3: Apply the negative sign. We then have 16=16 -\lvert 16 \rvert = -16 .

Therefore, the solution to the problem is 16 -16 .

Answer

16 -16

Exercise #3

18= −\left|-18\right|=

Video Solution

Step-by-Step Solution

The problem requires us to evaluate the expression 18 -\left|-18\right| .

Step 1: Calculate the absolute value of 18-18.

  • The absolute value of a number is the positive version of the number if it's negative. Therefore:
  • 18=18\left|-18\right| = 18

Step 2: Apply the negative sign to the result of the absolute value operation.

  • We attach the negative sign that is outside the absolute value:
  • 18=18-\left|-18\right| = -18

Therefore, the result of the expression 18 -\left|-18\right| is 18 -18 .

Answer

18 -18

Exercise #4

15= -\left|15\right| =

Step-by-Step Solution

To solve the given expression 15 -\left|15\right| , we need to find the absolute value of 15 15 , then apply the negative sign.

The absolute value of a number is the non-negative value of that number without regard to its sign.

Thus, 15=15 \left|15\right| = 15 .

Now apply the negative sign: 15=15 -\left|15\right| = -15 .

Therefore, the answer is 15 -15 .

Answer

15 -15

Exercise #5

7= -\left|7\right| =

Step-by-Step Solution

In the given expression, 7 -\left|7\right| , the absolute value of 7 7 is required.

The absolute value, 7 \left|7\right| , is 7 7 since absolute value denotes a non-negative distance from zero.

Applying the negative sign changes it to 7 -7 .

The final result, therefore, is 7 -7 .

Answer

7 -7

Exercise #6

23= -\left|23\right| =

Step-by-Step Solution

To solve the expression 23 -\left|23\right| , first find the absolute value of 23 23 .

The absolute value of a number is the distance between the number and zero on the number line, so 23=23 \left|23\right| = 23 .

Then apply the negative sign: 23=23 -\left|23\right| = -23 .

Hence, the correct answer is 23 -23 .

Answer

23 -23

Exercise #7

53= -\lvert5^3\rvert=

Step-by-Step Solution

First, calculate the cube of 5: 53=125 5^3 = 125 .

Then, apply the absolute value:
since 125 is positive, 125=125 \lvert 125 \rvert = 125 .

Finally, apply the negative sign outside the absolute value: 125=125 -\lvert 125 \rvert = -125 .

Answer

125 -125

Exercise #8

34= -\lvert3^4\rvert=

Step-by-Step Solution

First, calculate the fourth power of 3: 34=81 3^4 = 81 .

Then, apply the absolute value:
since 81 is positive, 81=81 \lvert 81 \rvert = 81 .

Finally, apply the negative sign outside the absolute value: 81=81 -\lvert 81 \rvert = -81 .

Answer

81 -81

Exercise #9

x3= -\left|-x^3\right|=

Step-by-Step Solution

The expression has an absolute value and a negative sign outside of the absolute value. When you take the absolute value of x3 -x^3 , it results in x3 |x^3| , which is x3 x^3 assumingx x is a real number. The negative sign outside the absolute value inverts it back to x3 -x^3 . Thus, the correct interpretation of the original expression x3 -\left|-x^3\right| is x3 -x^3 .

Answer

x3 -x^3

Exercise #10

2z= -\left|2z\right|=

Step-by-Step Solution

The absolute value function 2z \left|2z\right| simply returns 2z 2z when z z is positive and 2z -2z whenz z is negative, ensuring the result is non-negative. However, the minus sign outside the absolute value 2z -\left|2z\right| negates the result of the absolute value. Therefore, 2z -\left|2z\right| results in 2z -2z for all z z . Hence, the original expression evaluates to 2z -2z .

Answer

2z -2z

Exercise #11

3y2= -\left|3y^2\right|=

Step-by-Step Solution

The absolute value of 3y2 3y^2 is 3y2 3y^2 itself because 3y2 3y^2 is always non-negative regardless of the value of y y since any real number squared is non-negative. The negative sign outside the absolute value indicates that the expression3y2 -\left|3y^2\right| evaluates to 3y2 -3y^2 .

Answer

3y2 -3y^2