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

−∣−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 #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

−∣−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 −y2≤0-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 #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

−∣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 #6

−∣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 #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 expression−∣3y2∣ -\left|3y^2\right| evaluates to −3y2 -3y^2 .

Answer

−3y2 -3y^2