Math Problem Statement

Solve the exercises provided in the image: simplify the expressions (1–8) and solve the equations (9–18).

Solution

Let's break down the image to solve each of the exercises step by step. The image contains two main sections: one for simplifying expressions (1–8) and one for solving equations (9–18).

Exercises 1–8: Simplify the Expressions

  1. 9|-9| =9= 9

  2. 15-|15| =15= -15

  3. 1414|14| - |-14| =1414=0= 14 - 14 = 0

  4. 3+3-| -3| + |3| =3+3=0= -3 + 3 = 0

  5. 5(7)-|-5| - (-7) =5+7=2= -5 + 7 = 2

  6. 0.8+0-0.8 + |0| =0.8+0=0.8= -0.8 + 0 = -0.8

  7. 273\frac{27}{-3} =9= -9

  8. 124\frac{-12}{4} =3= -3


Exercises 9–22: Solve the Equation

  1. r=2|r| = 2 r=2orr=2r = 2 \quad \text{or} \quad r = -2

  2. x=13.4|x| = 13.4 x=13.4orx=13.4x = 13.4 \quad \text{or} \quad x = -13.4

  3. m+3=10|m + 3| = 10 m+3=10orm+3=10m + 3 = 10 \quad \text{or} \quad m + 3 = -10 Solving both: m=7orm=13m = 7 \quad \text{or} \quad m = -13

  4. q8=14|q - 8| = 14 q8=14orq8=14q - 8 = 14 \quad \text{or} \quad q - 8 = -14 Solving both: q=22orq=6q = 22 \quad \text{or} \quad q = -6

  5. x2=6\frac{|x|}{2} = 6 x=12sox=12orx=12|x| = 12 \quad \text{so} \quad x = 12 \quad \text{or} \quad x = -12

  6. 3.5q=15.4|3.5q| = 15.4 3.5q=15.4or3.5q=15.43.5q = 15.4 \quad \text{or} \quad 3.5q = -15.4 Solving both: q=4.4orq=4.4q = 4.4 \quad \text{or} \quad q = -4.4

  7. 4b5=11|4b - 5| = 11 4b5=11or4b5=114b - 5 = 11 \quad \text{or} \quad 4b - 5 = -11 Solving both: 4b=16sob=4orb=324b = 16 \quad \text{so} \quad b = 4 \quad \text{or} \quad b = -\frac{3}{2}

  8. 8m+1=5|8m + 1| = 5 8m+1=5or8m+1=58m + 1 = 5 \quad \text{or} \quad 8m + 1 = -5 Solving both: m=12orm=34m = \frac{1}{2} \quad \text{or} \quad m = -\frac{3}{4}

  9. 8w+6=76-8w + 6 = 76 Solving for ww: 8w=70sow=354=8.75-8w = 70 \quad \text{so} \quad w = -\frac{35}{4} = -8.75

  10. 3y27=3\left| \frac{3y - 2}{7} \right| = 3 Solving both cases: 3y27=3or3y27=3\frac{3y - 2}{7} = 3 \quad \text{or} \quad \frac{3y - 2}{7} = -3 Solving both: 3y2=21soy=2333y - 2 = 21 \quad \text{so} \quad y = \frac{23}{3} 3y2=21soy=1933y - 2 = -21 \quad \text{so} \quad y = -\frac{19}{3}


Would you like to review any of the solutions in more detail, or have further questions on any steps?

5 Related Questions:

  1. How do you solve more complex absolute value equations?
  2. What happens if there are two variables inside the absolute value expression?
  3. How can you check your solution in an absolute value problem?
  4. How does division affect solving absolute value equations like in question 13?
  5. Can you have no solution in an absolute value equation?

Tip:

Always consider both the positive and negative cases when solving absolute value equations, as each absolute value can result in two possible solutions.

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Math Problem Analysis

Mathematical Concepts

Absolute Value
Basic Algebra
Simplification of Expressions
Linear Equations

Formulas

Absolute value: |x|
Basic operations with fractions and integers
Linear equation: ax + b = c

Theorems

Absolute Value Theorem: |x| = x if x ≥ 0, and |x| = -x if x < 0
Solving linear equations

Suitable Grade Level

Grades 6-8