2.2 Linear Equations in One Variable

Solving Linear Equations and Understanding Lines

A linear equation in one variable is an equation that can be written in the form $ax + b = 0$, where $a$ and $b$ are real numbers and $a \neq 0$. Solving these equations is one of the most fundamental skills in algebra, and nearly every topic you'll encounter later builds on it.

Key Components of Linear Equations

Before solving anything, make sure you can identify the parts of an equation:

• Variable: The unknown quantity, typically represented by a letter like $x$ or $y$
• Coefficient: The number multiplied by a variable. In $3x + 2$, the coefficient of $x$ is 3
• Constant: A fixed numerical value with no variable attached. In $3x + 2 = 7$, both 2 and 7 are constants

Solving Linear Equations

The core strategy is to isolate the variable on one side of the equation using inverse operations. Here's the process:

1. Simplify each side by distributing and combining like terms. For example, $3(x + 2) = 9$ becomes $3x + 6 = 9$
2. Move constant terms to one side by adding or subtracting from both sides. So $3x + 6 = 9$ becomes $3x = 3$
3. Move coefficient terms by dividing or multiplying both sides by the coefficient. So $3x = 3$ becomes $x = 1$
4. Check your solution by substituting it back into the original equation. Here, $3(1 + 2) = 3(3) = 9$. It checks out.

The value that makes the equation true is called the solution. A linear equation in one variable will always have exactly one solution (assuming it's not an identity or a contradiction).

Rational Equations and Extraneous Solutions

A rational equation contains at least one fraction with a variable in the denominator. The strategy is to eliminate the fractions first, then solve.

1. Find the least common denominator (LCD) of all fractions. For $\frac{2}{x+1} + \frac{3}{x-1} = 1$, the LCD is $(x+1)(x-1)$

2. Multiply every term on both sides by the LCD to clear the denominators: $2(x-1) + 3(x+1) = (x+1)(x-1)$

3. Expand and simplify: $2x - 2 + 3x + 3 = x^2 - 1$, which gives $5x + 1 = x^2 - 1$, then $x^2 - 5x - 2 = 0$

4. Solve the resulting equation using appropriate methods (factoring, quadratic formula, etc.)

5. Check every solution in the original equation. Any value that makes a denominator equal to zero is extraneous, meaning it emerged from the algebra but doesn't actually satisfy the original equation. For instance, $x = -1$ would make $\frac{3}{x-1}$ undefined, so it must be rejected.

Note that multiplying by the LCD can produce a quadratic (or higher-degree) equation, not just a linear one. The resulting equation type depends on the original problem.

Slope and Y-Intercept

The slope-intercept form of a linear equation is $y = mx + b$, where:

• Slope ($m$) measures the rate of change, or how steep the line is. It's calculated as $m = \frac{y_2 - y_1}{x_2 - x_1}$ for any two points $(x_1, y_1)$ and $(x_2, y_2)$ on the line.
  • Positive slope: the line rises from left to right
  • Negative slope: the line falls from left to right
  • Zero slope: the line is horizontal
  • Undefined slope: the line is vertical (and can't be written in slope-intercept form)
• Y-intercept ($b$) is the point where the line crosses the y-axis. You find it by setting $x = 0$ and solving for $y$. In $y = 2x + 3$, the y-intercept is 3, meaning the line passes through $(0, 3)$.

Parallel vs. Perpendicular Lines

Parallel lines have the same slope but different y-intercepts. They never intersect.

• $y = 2x + 1$ and $y = 2x - 3$ are parallel because both have slope 2

Perpendicular lines have slopes that are negative reciprocals of each other. Their slopes multiply to $-1$.

• $y = 2x + 1$ and $y = -\frac{1}{2}x + 3$ are perpendicular because $2 \times (-\frac{1}{2}) = -1$

Constructing Parallel and Perpendicular Equations

To write an equation parallel to $y = mx + b$ through a given point $(x_1, y_1)$:

1. Keep the same slope $m$

2. Use point-slope form: $y - y_1 = m(x - x_1)$

3. Simplify to slope-intercept form

For example, a line parallel to $y = 2x + 1$ passing through $(3, 4)$: $y - 4 = 2(x - 3)$, which simplifies to $y = 2x - 2$.

To write an equation perpendicular to $y = mx + b$ through a given point $(x_1, y_1)$:

1. Find the negative reciprocal of the slope: $m' = -\frac{1}{m}$. For $y = 2x + 1$, that's $m' = -\frac{1}{2}$

2. Use point-slope form with the new slope: $y - y_1 = m'(x - x_1)$

3. Simplify. Through $(1, 3)$: $y - 3 = -\frac{1}{2}(x - 1)$, giving $y = -\frac{1}{2}x + \frac{7}{2}$