Unit 1 Equations And Inequalities Homework 3 Answers

Unit 1 Equations and Inequalities Homework 3 Answers

When students reach the first unit of an Algebra 1 course, they encounter the foundational skills of solving equations and inequalities. Think about it: homework 3 typically consolidates these ideas by presenting a mix of one‑step, two‑step, and multi‑step problems, along with simple absolute‑value and compound‑inequality tasks. Having a clear, step‑by‑step guide to the unit 1 equations and inequalities homework 3 answers not only helps learners verify their work but also reinforces the logical processes that underlie more advanced algebraic reasoning. Below is an in‑depth walkthrough of the concepts covered, detailed solutions to representative problems, common pitfalls to avoid, and practical strategies for mastering the material.

Introduction

Unit 1 lays the groundwork for all future algebraic manipulation. Students learn to isolate variables, apply the properties of equality, and translate word problems into mathematical statements. Homework 3 is deliberately designed to test fluency across these skills before moving on to functions and systems. By reviewing the unit 1 equations and inequalities homework 3 answers, students can identify where their understanding is solid and where additional practice is needed.

Understanding the Homework Assignment

What Homework 3 Usually Includes

The answer key for homework 3 will list the exact numeric or interval solutions for each of these categories. Below we solve a few representative items in full detail, showing the reasoning that leads to the final answer.

Key Concepts Covered

Before diving into the solutions, it helps to recall the essential principles that govern each problem type.

Properties of Equality

• Addition Property: If (a = b), then (a + c = b + c).
• Subtraction Property: If (a = b), then (a - c = b - c).
• Multiplication Property: If (a = b), then (ac = bc).
• Division Property: If (a = b) and (c \neq 0), then (\frac{a}{c} = \frac{b}{c}).

These properties make it possible to perform the same operation on both sides of an equation without changing its truth value.

Properties of Inequality * Adding or subtracting the same number on both sides preserves the direction of the inequality.

• Multiplying or dividing both sides by a positive number preserves the direction.
• Multiplying or dividing both sides by a negative number reverses the inequality sign.

Absolute Value Definition

For any real number (a), (|a| = a) if (a \ge 0) and (|a| = -a) if (a < 0). This means (|x| = k) (with (k \ge 0)) yields (x = k) or (x = -k). An inequality such as (|x| < k) becomes (-k < x < k), while (|x| > k) becomes (x < -k) or (x > k).

Step‑by‑Step Solutions to Typical Problems

Below are six problems that mirror the variety found in homework 3, each solved with explicit justification. After each solution, the final answer is highlighted in bold for quick reference.

Problem 1 – One‑Step Equation

Solve: (x + 9 = 20)

1. Subtract 9 from both sides (Subtraction Property).
   [ x + 9 - 9 = 20 - 9 ]
2. Simplify.
   [ x = 11 ]

Answer: (\mathbf{x = 11})

Problem 2 – Two‑Step Equation

Solve: (4x - 7 = 21)

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1. Add 7 to both sides.
   [ 4x - 7 + 7 = 21 + 7 ;\Longrightarrow; 4x = 28 ]
2. Divide both sides by 4.
   [ \frac{4x}{4} = \frac{28}{4} ;\Longrightarrow; x = 7 ]

Answer: (\mathbf{x = 7})

Problem 3 – Multi‑Step Equation with Distribution

Solve: (3(2x - 4) + 5 = 29)

1. Distribute the 3.
   [ 6x - 12 + 5 = 29 ]
2. Combine like terms on the left.
   [ 6x - 7 = 29 ]
3. Add 7 to both sides.
   [ 6x = 36 ]
4. Divide by 6. [ x = 6 ]

Answer: (\mathbf{x = 6})

Problem 4 – Equation with Variables on Both Sides

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Problem 4 – Equation withVariables on Both Sides

Solve: (5x + 3 = 2x - 4)

1. Subtract (2x) from both sides (to collect the variable terms on one side).
   [ 5x - 2x + 3 = -4 ;\Longrightarrow; 3x + 3 = -4 ]
2. Subtract 3 from both sides (to isolate the term with (x)).
   [ 3x = -4 - 3 ;\Longrightarrow; 3x = -7 ]
3. Divide both sides by 3 (Division Property of Equality).
   [ x = -\frac{7}{3} ]

Answer: (\mathbf{x = -\dfrac{7}{3}})

Problem 5 – Solving a Linear Inequality

Solve: (-2(3x - 5) \ge 4 + x)

1. Distribute the (-2) on the left side.
   [ -6x + 10 \ge 4 + x ]
2. Add (6x) to both sides to bring all (x)-terms to the right.
   [ 10 \ge 4 + 7x ]
3. Subtract 4 from both sides to isolate the term with (x).
   [ 6 \ge 7x ]
4. Divide both sides by 7 (a positive number, so the inequality direction stays the same).
   [ \frac{6}{7} \ge x \quad\text{or equivalently}\quad x \le \frac{6}{7} ]

Answer: (\mathbf{x \le \dfrac{6}{7}})

Problem 6 – Absolute‑Value Equation and Inequality

Solve: (|2x - 1| = 7) and then solve (|2x - 1| < 7).

Equation

1. By definition of absolute value, (|A| = k) ((k\ge0)) implies (A = k) or (A = -k).
   [ 2x - 1 = 7 \quad\text{or}\quad 2x - 1 = -7 ]
2. Solve each branch:
   • (2x - 1 = 7 ;\Rightarrow; 2x = 8 ;\Rightarrow; x = 4)
   • (2x - 1 = -7 ;\Rightarrow; 2x = -6 ;\Rightarrow; x = -3)

Answer (equation): (\mathbf{x = 4 \text{ or } x = -3})

Inequality

1. (|A| < k) ((k>0)) is equivalent to (-k < A < k).
   [ -7 < 2x - 1 < 7 ]
2. Add 1 to all three parts:
   [ -6 < 2x < 8 ]
3. Divide by 2 (positive, direction unchanged):
   [

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