Unit 1 Equations And Inequalities Homework 3 Solving Equations

Mastering Equation Solving: A Comprehensive Guide to Unit 1 Homework 3

Solving equations is a fundamental skill in algebra, forming the bedrock for more advanced mathematical concepts. Unit 1 Homework 3 typically focuses on honing these essential skills, ensuring a solid understanding of the principles involved. This article provides a comprehensive walkthrough of solving equations, covering various techniques and offering practical examples to solidify your understanding.

Laying the Foundation: Understanding Equations and Their Properties

An equation is a mathematical statement asserting that two expressions are equal. These expressions are connected by an equals sign (=). The goal of solving an equation is to find the value(s) of the variable(s) that make the equation true. These values are called the solutions or roots of the equation.

Before diving into solving techniques, it's crucial to understand the properties of equality that underpin these methods. These properties allow us to manipulate equations without changing their solutions. The key properties are:

Addition Property of Equality: If a = b, then a + c = b + c for any real number c.
Subtraction Property of Equality: If a = b, then a - c = b - c for any real number c.
Multiplication Property of Equality: If a = b, then ac = bc for any real number c.
Division Property of Equality: If a = b, then a/c = b/c for any real number c, where c ≠ 0.
Distributive Property: a(b + c) = ab + ac

These properties form the toolkit we use to isolate variables and solve equations.

Solving Linear Equations: A Step-by-Step Approach

Linear equations are equations where the variable is raised to the power of 1. They can be written in the form ax + b = c, where a, b, and c are constants and x is the variable. Here's a systematic approach to solving linear equations:

Simplify Both Sides: Combine like terms on each side of the equation. Use the distributive property to eliminate any parentheses.
Isolate the Variable Term: Use the addition or subtraction property of equality to move all terms containing the variable to one side of the equation and all constant terms to the other side.
Isolate the Variable: Use the multiplication or division property of equality to isolate the variable. Divide both sides of the equation by the coefficient of the variable.
Check Your Solution: Substitute the solution back into the original equation to verify that it makes the equation true.

Example 1: Solve the equation 3x + 5 = 14.

Subtract 5 from both sides: 3x + 5 - 5 = 14 - 5 which simplifies to 3x = 9.
Divide both sides by 3: 3x / 3 = 9 / 3 which simplifies to x = 3.
Check: Substitute x = 3 back into the original equation: 3(3) + 5 = 9 + 5 = 14. The equation holds true, so the solution is x = 3.

Example 2: Solve the equation 2(x - 4) = -6.

Distribute the 2: 2x - 8 = -6.
Add 8 to both sides: 2x - 8 + 8 = -6 + 8 which simplifies to 2x = 2.
Divide both sides by 2: 2x / 2 = 2 / 2 which simplifies to x = 1.
Check: Substitute x = 1 back into the original equation: 2(1 - 4) = 2(-3) = -6. The equation holds true, so the solution is x = 1.

Tackling Equations with Fractions

Equations containing fractions can seem intimidating, but they can be solved efficiently by eliminating the fractions early in the process. The key is to find the least common denominator (LCD) of all the fractions in the equation. Then, multiply both sides of the equation by the LCD. This will clear the fractions, leaving you with a simpler equation to solve.

Example 3: Solve the equation (x/2) + (1/3) = (5/6).

The LCD of 2, 3, and 6 is 6.
Multiply both sides of the equation by 6: 6 * [(x/2) + (1/3)] = 6 * (5/6).
Distribute the 6: (6x/2) + (6/3) = 5 which simplifies to 3x + 2 = 5.
Subtract 2 from both sides: 3x + 2 - 2 = 5 - 2 which simplifies to 3x = 3.
Divide both sides by 3: 3x / 3 = 3 / 3 which simplifies to x = 1.
Check: Substitute x = 1 back into the original equation: (1/2) + (1/3) = (3/6) + (2/6) = (5/6). The equation holds true, so the solution is x = 1.

Example 4: Solve the equation (2x/5) - (x/3) = 1.

The LCD of 5 and 3 is 15.
Multiply both sides of the equation by 15: 15 * [(2x/5) - (x/3)] = 15 * 1.
Distribute the 15: (30x/5) - (15x/3) = 15 which simplifies to 6x - 5x = 15.
Combine like terms: x = 15.
Check: Substitute x = 15 back into the original equation: (2(15)/5) - (15/3) = (30/5) - (5) = 6 - 5 = 1. The equation holds true, so the solution is x = 15.

Dealing with Equations Involving Decimals

Equations containing decimals can be handled in a similar way to equations with fractions. The goal is to eliminate the decimals by multiplying both sides of the equation by a power of 10. The power of 10 you choose should be large enough to shift the decimal point to the right of all the decimal numbers in the equation.

Example 5: Solve the equation 0.2x + 0.5 = 1.3.

Multiply both sides of the equation by 10: 10 * (0.2x + 0.5) = 10 * 1.3.
Distribute the 10: 2x + 5 = 13.
Subtract 5 from both sides: 2x + 5 - 5 = 13 - 5 which simplifies to 2x = 8.
Divide both sides by 2: 2x / 2 = 8 / 2 which simplifies to x = 4.
Check: Substitute x = 4 back into the original equation: 0.2(4) + 0.5 = 0.8 + 0.5 = 1.3. The equation holds true, so the solution is x = 4.

Example 6: Solve the equation 1.5x - 0.75 = 3.

Multiply both sides of the equation by 100: 100 * (1.5x - 0.75) = 100 * 3.
Distribute the 100: 150x - 75 = 300.
Add 75 to both sides: 150x - 75 + 75 = 300 + 75 which simplifies to 150x = 375.
Divide both sides by 150: 150x / 150 = 375 / 150 which simplifies to x = 2.5.
Check: Substitute x = 2.5 back into the original equation: 1.5(2.5) - 0.75 = 3.75 - 0.75 = 3. The equation holds true, so the solution is x = 2.5.

Solving Equations with Variables on Both Sides

When an equation has variables on both sides, the goal is to consolidate the variable terms to one side and the constant terms to the other. This is achieved by using the addition or subtraction property of equality.

Example 7: Solve the equation 5x - 3 = 2x + 6.

Subtract 2x from both sides: 5x - 3 - 2x = 2x + 6 - 2x which simplifies to 3x - 3 = 6.
Add 3 to both sides: 3x - 3 + 3 = 6 + 3 which simplifies to 3x = 9.
Divide both sides by 3: 3x / 3 = 9 / 3 which simplifies to x = 3.
Check: Substitute x = 3 back into the original equation: 5(3) - 3 = 15 - 3 = 12 and 2(3) + 6 = 6 + 6 = 12. Both sides are equal, so the solution is x = 3.

Example 8: Solve the equation 4(x + 2) = -2(x - 5).

Distribute on both sides: 4x + 8 = -2x + 10.
Add 2x to both sides: 4x + 8 + 2x = -2x + 10 + 2x which simplifies to 6x + 8 = 10.
Subtract 8 from both sides: 6x + 8 - 8 = 10 - 8 which simplifies to 6x = 2.
Divide both sides by 6: 6x / 6 = 2 / 6 which simplifies to x = 1/3.
Check: Substitute x = 1/3 back into the original equation: 4((1/3) + 2) = 4(7/3) = 28/3 and -2((1/3) - 5) = -2(-14/3) = 28/3. Both sides are equal, so the solution is x = 1/3.

Understanding and Solving Inequalities

Inequalities are mathematical statements that compare two expressions using inequality symbols: < (less than), > (greater than), ≤ (less than or equal to), and ≥ (greater than or equal to). Solving inequalities is similar to solving equations, but with one crucial difference: multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality sign.

Example 9: Solve the inequality 2x + 3 < 7.

Subtract 3 from both sides: 2x + 3 - 3 < 7 - 3 which simplifies to 2x < 4.
Divide both sides by 2: 2x / 2 < 4 / 2 which simplifies to x < 2.
The solution is all values of x that are less than 2. This can be represented on a number line or in interval notation as (-∞, 2).

Example 10: Solve the inequality -3x + 5 ≥ 14.

Subtract 5 from both sides: -3x + 5 - 5 ≥ 14 - 5 which simplifies to -3x ≥ 9.
Divide both sides by -3 (and reverse the inequality sign): -3x / -3 ≤ 9 / -3 which simplifies to x ≤ -3.
The solution is all values of x that are less than or equal to -3. This can be represented on a number line or in interval notation as (-∞, -3].

Compound Inequalities: Combining Inequalities

Compound inequalities combine two or more inequalities using the words "and" or "or".

"And" Inequalities: These inequalities require that both conditions be true simultaneously. The solution is the intersection of the solutions to each individual inequality.
"Or" Inequalities: These inequalities require that at least one of the conditions be true. The solution is the union of the solutions to each individual inequality.

Example 11: Solve the compound inequality 2 < x + 1 ≤ 5.

This is an "and" inequality. We can rewrite it as two separate inequalities: 2 < x + 1 and x + 1 ≤ 5.
Solve the first inequality: 2 < x + 1 => 1 < x.
Solve the second inequality: x + 1 ≤ 5 => x ≤ 4.
The solution is the intersection of 1 < x and x ≤ 4, which can be written as 1 < x ≤ 4. In interval notation, this is (1, 4].

Example 12: Solve the compound inequality x - 3 < -1 or 2x > 6.

This is an "or" inequality.
Solve the first inequality: x - 3 < -1 => x < 2.
Solve the second inequality: 2x > 6 => x > 3.
The solution is the union of x < 2 and x > 3. In interval notation, this is (-∞, 2) ∪ (3, ∞).

Absolute Value Equations and Inequalities

The absolute value of a number is its distance from zero. Absolute value equations and inequalities require special attention because they involve considering both positive and negative cases.

Absolute Value Equations:

To solve an absolute value equation of the form |ax + b| = c, where c is a positive number, we need to consider two cases:

ax + b = c
ax + b = -c

Example 13: Solve the equation |2x - 1| = 5.

Case 1: 2x - 1 = 5 => 2x = 6 => x = 3.
Case 2: 2x - 1 = -5 => 2x = -4 => x = -2.
The solutions are x = 3 and x = -2.

Absolute Value Inequalities:

Solving absolute value inequalities depends on the inequality symbol:

If |ax + b| < c, then -c < ax + b < c. This is equivalent to an "and" inequality.
If |ax + b| > c, then ax + b < -c or ax + b > c. This is equivalent to an "or" inequality.

Example 14: Solve the inequality |x + 2| ≤ 3.

This is equivalent to -3 ≤ x + 2 ≤ 3.
Subtract 2 from all parts of the inequality: -3 - 2 ≤ x + 2 - 2 ≤ 3 - 2 which simplifies to -5 ≤ x ≤ 1.
The solution in interval notation is [-5, 1].

Example 15: Solve the inequality |2x - 1| > 7.

This is equivalent to 2x - 1 < -7 or 2x - 1 > 7.
Solve the first inequality: 2x - 1 < -7 => 2x < -6 => x < -3.
Solve the second inequality: 2x - 1 > 7 => 2x > 8 => x > 4.
The solution in interval notation is (-∞, -3) ∪ (4, ∞).

Practical Applications: Word Problems

Solving equations and inequalities is not just an abstract exercise; it's a powerful tool for solving real-world problems. Word problems often require translating verbal descriptions into mathematical equations or inequalities. Here are some tips for tackling word problems:

Read Carefully: Understand the problem thoroughly. Identify what is being asked and what information is given.
Define Variables: Assign variables to represent the unknown quantities.
Translate: Translate the verbal statements into mathematical equations or inequalities. Look for keywords like "is," "equals," "less than," "greater than," etc.
Solve: Solve the equation or inequality.
Answer the Question: Make sure your answer addresses the original question. Include units if necessary.
Check Your Answer: Does your answer make sense in the context of the problem?

Example 16: The sum of three consecutive integers is 48. Find the integers.

Let x be the first integer. Then the next two consecutive integers are x + 1 and x + 2.
The equation is x + (x + 1) + (x + 2) = 48.
Simplify and solve: 3x + 3 = 48 => 3x = 45 => x = 15.
The integers are 15, 16, and 17.
Check: 15 + 16 + 17 = 48. The answer is correct.

Example 17: A rectangle has a length that is 3 inches more than its width. If the perimeter of the rectangle is 26 inches, find the length and width.

Let w be the width of the rectangle. Then the length is w + 3.
The perimeter of a rectangle is given by P = 2l + 2w. So, 2(w + 3) + 2w = 26.
Simplify and solve: 2w + 6 + 2w = 26 => 4w + 6 = 26 => 4w = 20 => w = 5.
The width is 5 inches, and the length is 5 + 3 = 8 inches.
Check: 2(8) + 2(5) = 16 + 10 = 26. The answer is correct.

Conclusion: Building Confidence in Equation Solving

Mastering equation solving requires practice and a solid understanding of the underlying principles. By systematically applying the techniques outlined in this article, you can build confidence and proficiency in solving various types of equations and inequalities. Remember to always check your solutions and to practice regularly to reinforce your skills. With consistent effort, you can conquer Unit 1 Homework 3 and lay a strong foundation for future success in mathematics.