Lesson 2.2: Linear Equations in One Variable

New Concept

A linear equation in one variable, like $ax+b=0$, is the foundation for solving many algebraic problems. You'll learn to isolate the variable, a key skill for handling rational equations and analyzing the properties of lines.

What’s next

Get ready to master this concept! You'll begin with interactive examples on solving for $x$, then tackle challenge problems with rational equations and graphing lines.

Property

A linear equation in one variable can be written in the form

where $a$ and $b$ are real numbers, $a \neq 0$. An identity equation is true for all values of the variable. A conditional equation is true for only some values of the variable. An inconsistent equation results in a false statement.

Property

A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.
To solve it:

1. Factor all denominators in the equation.
2. Find and exclude values that set each denominator equal to zero.
3. Find the LCD.
4. Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
5. Solve the remaining equation.
6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.

Examples

• To solve $\frac{5}{2x} - \frac{1}{x} = \frac{3}{4}$, the LCD is $4x$. Multiplying by $4x$ gives $10 - 4 = 3x$, which simplifies to $6 = 3x$, so $x=2$. The excluded value is $x=0$.
• To solve $\frac{5}{x-4} = \frac{7}{x}$, we can cross-multiply to get $5x = 7(x-4)$. This becomes $5x = 7x - 28$, which simplifies to $-2x = -28$, so $x=14$. Excluded values are 0 and 4.
• To solve $\frac{3x}{x^2-9} = \frac{1}{x-3} + \frac{1}{x+3}$, the LCD is $(x-3)(x+3)$. Multiplying gives $3x = 1(x+3) + 1(x-3)$. This simplifies to $3x = 2x$, so $x=0$. Excluded values are 3 and -3.

Explanation

Rational equations have fractions with variables in the bottom. The trick is to wipe out all the denominators! Find the Least Common Denominator (LCD), multiply everything by it, and the fractions vanish, leaving a simpler equation to solve.

Property

The slope of a line, $m$, represents the change in $y$ over the change in $x$. Given two points, $(x_1, y_1)$ and $(x_2, y_2)$, the following formula determines the slope of a line containing these points:

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper.

Property

Given one point and the slope, the point-slope formula will lead to the equation of a line:

Another way that we can represent the equation of a line is in standard form. Standard form is given as

Property

The equation of a vertical line is given as

where $c$ is a constant. The slope of a vertical line is undefined. The equation of a horizontal line is given as

Property

Parallel lines have the same slope and different $y$-intercepts.
Lines that are parallel to each other will never intersect.
Lines that are perpendicular intersect to form a $90^\circ$ angle.
The slope of one line is the negative reciprocal of the other.
We can show that two lines are perpendicular if the product of the two slopes is $-1$: $m_1 \cdot m_2 = -1$.

Examples

• The lines $y = -2x + 1$ and $y = -2x - 5$ are parallel because both have a slope of $m = -2$.
• The lines $y = \frac{3}{4}x + 2$ and $y = -\frac{4}{3}x$ are perpendicular because their slopes are negative reciprocals. Their product is $(\frac{3}{4})(-\frac{4}{3}) = -1$.
• The lines $y = 5x - 2$ and $y = -5x + 3$ are neither parallel nor perpendicular. Their slopes ($5$ and $-5$) are not equal, and their product is not $-1$.

Explanation

Parallel lines are like train tracks—they run forever in the same direction and never cross because they have the exact same slope. Perpendicular lines meet at a perfect right angle, with slopes that are negative reciprocals.

Property

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

1. Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
2. Use the slope and the given point with the point-slope formula.
3. Simplify the line to slope-intercept form and compare the equation to the given line.

Examples

• Find a line parallel to $y = 4x - 1$ passing through $(2, 3)$. The slope is $4$. Use point-slope: $y - 3 = 4(x - 2)$, which simplifies to $y = 4x - 5$.
• Find a line perpendicular to $y = -\frac{1}{2}x + 6$ passing through $(1, 5)$. The original slope is $-\frac{1}{2}$, so the new slope is $2$. Use point-slope: $y - 5 = 2(x - 1)$, which simplifies to $y = 2x + 3$.
• Find a line perpendicular to $3x + y = 7$ passing through $(3, 0)$. Rearrange to $y = -3x + 7$ to find the slope $m=-3$. The perpendicular slope is $\frac{1}{3}$. Use point-slope: $y - 0 = \frac{1}{3}(x-3)$, which simplifies to $y = \frac{1}{3}x - 1$.

Explanation

First, find the slope of the original line. For a parallel line, steal that slope. For a perpendicular line, flip it and reverse the sign. Then, use this new slope and the given point to build your new line's equation.