Math 107 - Lecture Notes

Solving Linear Equations

Overview

• Focus on solving linear equations with one variable and applications.
• Applications include geometric problems, motion in physics, and simple interest rates.
• Knowledge from these lectures will be used to solve quadratic and higher-order equations later in the course.

Section 1.1: Linear Equations in One Variable

Learning Objectives

• Understanding the vocabulary of equations.
• Solving equations in one variable, including those with fractions and parentheses.
• Solving equations for a specific variable when multiple variables are present.

Vocabulary for Understanding Equations

• Equation: A statement that compares two expressions and asserts that they are equal.

• Key Component: Equal sign (=).

• Terms: Parts of an algebraic expression.

• An equation compares algebraic expressions on the left and right sides and asserts their equality.

• Example: 7 - 5 = 2

  • This statement shows that the expression 7 - 5 is equal to the number 2.

Equations in One Variable

• An equation in one variable involves finding the value(s) of the variable that make the equation true.

• Example: An equation with variable x.

• Solving involves isolating the variable to find its value.

• Solving the example equation:

  • Add 3 to both sides: 2x + 3 = 7 \rightarrow 2x = 4
  • Divide both sides by 2: 2x = 4 \rightarrow x = 2

Domain of an Equation

• The domain is the set of all possible values for the variable that make the equation true.
• Important to consider fractions and radicals when determining the domain.

Finding the Domain: Example 1 (Fraction)

• Equation: 2 = \frac{5}{x - 1}
• Problem: Denominator cannot be zero.
• x - 1 \neq 0 \rightarrow x \neq 1
• Interval Notation: (-\infty, 1) \cup (1, \infty)
• This means x can be any real number except 1.

Finding the Domain: Example 2 (Radical)

• Equation: 5 = \sqrt{x} + 2
• Problem: Radicand (expression inside the square root) must be non-negative.
• x \geq 0
• Interval Notation: [0, \infty)
• This means x must be greater than or equal to 0.

Finding the Domain: Example 3 (No Fraction or Radical)

• Equation: 3x + 1 = 5x - 2
• No fractions or radicals present.
• No restrictions on the variable.
• Domain: (-\infty, \infty)
• x can be any real number.

Summary of Finding Domains

• If an equation has fractions, set the denominator not equal to zero and solve.
• If an equation has radicals, set the radicand greater than or equal to zero and solve.
• If there are no fractions or radicals, the domain is all real numbers.

Examples for Domain Practice

• Example A: Equation with a Fraction
  • Set the denominator not equal to zero: x \neq 2
  • Interval Notation: (-\infty, 2) \cup (2, \infty)
• Example B: Equation with a Radical
  • Set the radicand greater than or equal to zero: x - 1 \geq 0 \rightarrow x \geq 1
  • Interval Notation: [1, \infty)

Linear Equations

Definition

• A linear equation is an equation where the highest exponent of the variable is 1.
• When graphed, linear equations form lines.

Standard Form

• Equations do not always need to be in standard form to be solved.
• As long as you know the steps to isolate the variable, you can solve it. Isolate the the x and perform basic operations to find the solution.

Solving Linear Equations with Fractions

1. Multiply the entire equation by the Least Common Denominator (LCD) to eliminate fractions.
2. Simplify by removing parentheses and combining like terms.
3. Isolate the variable term.
4. Combine like terms.
5. Isolate the variable by dividing both sides by its coefficient.
6. Check the solution in the original equation to avoid mistakes.

Example: Solving a Linear Equation with Fractions

• Equation: \frac{1}{3}x - \frac{2}{3} = \frac{1}{2}(1 - 3x)

1. Find the LCD: The LCD of 3 and 2 is 6.
2. Multiply the equation by the LCD: 6(\frac{1}{3}x - \frac{2}{3}) = 6(\frac{1}{2}(1 - 3x))
   • Simplify: 2x - 4 = 3(1 - 3x)
3. Distribute: 2x - 4 = 3 - 9x
4. Move all terms with x to one side and constants to the other side:
   • Add 9x to both sides:
   • Add 4 to both sides:
   • 11x = 7
5. Divide everything by 11.
6. Final answer: x = \frac{7}{11}

Check the Solution

• Substitute x = \frac{7}{11} into the original equation to verify the solution.
• Left Side: Calculate \frac{1}{3}(\frac{7}{11}) - \frac{2}{3}
• Simplify to get -\frac{5}{11}
• Right Side: Calculate \frac{1}{2}(1 - 3(\frac{7}{11}))
• Simplify to get \frac{1}{2}(-\frac{10}{11}) = -\frac{5}{11}
• Both sides are equal, so the solution is correct.

Solving Linear Equations with Parentheses

1. Simplify inside the inner most set of parentheses first.
2. Distribute any negative signs to all the terms.
3. Combine like terms on each side of the equation.
4. Add or subtract terms, to isolate the variable on one side.
5. Multiply or divide to isolate the variable completely.
6. Substitute your answer in the original equation to check if you are right.

Example: Solving Linear Equations with Nested Parentheses

• Equation: 6x - [3x - 2(x - 2)] = 11

1. Simplify inside the inner most set of parentheses.

   • Distribute -2 (x-2). This will result in -2x + 4
   • Rewrite the whole structure.
     6x - [3x -2x +4] = 11

2. We can rewrite this as [x + 4]. Distribute the negative in front of the bracket, we have

   • 6x - x - 4 = 11

3. Adding 4 to both sides and combining x, we have

   • 5x = 15

4. Divide everthing by 5 to find your final answer.

   • So the final x is 3.

   Final answer: x = 3
   Check your answer with the original equation for correctness.

Solving Equations for a Specific Variable

• Isolate the desired variable by performing algebraic operations on both sides of the equation.
• This involves treating all other variables as constants and rearranging the equation to solve for the target variable.

Example: Solve for the Value of A and D

1. To find what A equals, move everthing else to the side of F, keeping in mind m is considered constant as well. Divide side by m to find A.

• So A= F/M .
  note Make sure that M is not equal to 0, or else you will have to divide by 0 . It cannot be completed

1. To find what the value of D equals given T.
   • Rewrite the equation with D on the left side.
   • Add A to both sides
   • Factor D's variable's to the right, everything else to the right side.
   • Distribute out D by dividing by (n-1), and isolate D.
   • D = T-A/(n-1).
     note Again, make sure the value of n does not equal 1, so the equation still functions

Applied Problems with Linear Equations

Solving Word Problems

1. Read and understand the problem.
2. Identify what is known and what you need to find.
3. Translate the words into a mathematical expression.
4. Use variables for the the quantity, and build the model for an equation. Then solve it to come to a conclusion. Remember that the conclusion is depended on the problems we're working and solving.

Example: Modeling Presale Price

• A sports complex offers a fitness program to college freshman at a 20% discount. The discounted price is $144. What is the original price?

1. Let x be the original price
2. A freshman basically pays rest, or 80% of the price after 20% off.
   • The discount is 20%, so the discount as a variable is 0.2.
   • After 20% off, or 80% of the price, is the decimal 0.8.
3. So the equation for the fitness plan equals
   • 0. 8x= 144
4. Divide everything to find what x is valued by x = 180
   • Original price of the the fitness plan the students, is $180

Geometry Problem: Rectangle Yard.

1. Draw picture to help understand problem. Note: The length of the soccer side is the the width -30.

2. We need to find: With and and Length and Width.

3. We know that the parameter is 360 yards.

   • So 2 ( Length+ Widith) = 360 yards = Parameter

4. To find relation between two variables re-read the question: The length is the the with - 30 yards. Plug the value find for the the lenth side, to to find an equation with one variable to calculate.
   4W - 60 + 2W =360
   6W=420.
   W= 70

• With variable we can plus the equation for length to calculate that variable.

1. L=2(70) -30 L== 140 -30 L = 110.