3.1 Linear Equations and Inequalities in Two Variables

Linear equations and inequalities in two variables are the building blocks for modeling relationships between quantities and visualizing them on a coordinate plane. Mastering these concepts gives you the tools to interpret rates of change, work flexibly across different equation forms, and solve real-world optimization problems.

Graphing linear equations and inequalities

Coordinate plane basics

The coordinate plane is a two-dimensional space defined by a horizontal x-axis and a vertical y-axis that intersect at the origin (0, 0). The x-axis typically represents the independent variable, while the y-axis represents the dependent variable.

Each point is identified by an ordered pair (x, y). The axes divide the plane into four quadrants, numbered counterclockwise starting from the upper right:

• Quadrant I: x > 0, y > 0
• Quadrant II: x < 0, y > 0
• Quadrant III: x < 0, y < 0
• Quadrant IV: x > 0, y < 0

Graphing linear equations

A linear equation in two variables produces a straight line on the coordinate plane. Every point on that line is a solution to the equation.

To graph a linear equation:

1. Find at least two points that satisfy the equation. The intercepts are often the easiest to calculate.
2. Plot those points on the coordinate plane.
3. Draw a straight line through them, extending in both directions.

The x-intercept is where the graph crosses the x-axis (set y = 0 and solve for x). The y-intercept is where it crosses the y-axis (set x = 0 and solve for y).

For example, to find the intercepts of $3x - 4y = 12$: setting y = 0 gives x = 4, and setting x = 0 gives y = -3. Plot (4, 0) and (0, -3), then connect them.

Graphing linear inequalities

Graphing an inequality is like graphing an equation, but with an extra decision about the boundary line and shading.

1. Graph the boundary line (the corresponding equation). Use a solid line for $\leq$ or $\geq$ (points on the line are included). Use a dashed line for $<$ or $>$ (points on the line are not included).
2. Pick a test point not on the line (the origin works well if the line doesn't pass through it).
3. Substitute the test point into the inequality. If it's true, shade the side containing the test point. If it's false, shade the opposite side.

The shaded region represents all points (x, y) that satisfy the inequality.

For $y > 3x - 2$: graph $y = 3x - 2$ as a dashed line, test (0, 0) → $0 > -2$ is true, so shade above the line.

For $2x + 4y \leq 8$: graph $2x + 4y = 8$ as a solid line, test (0, 0) → $0 \leq 8$ is true, so shade the side containing the origin.

Slope and y-intercept interpretation

Slope

The slope (m) measures the rate of change of y with respect to x. It tells you how steep the line is and in which direction it tilts.

$m = \frac{y_2 - y_1}{x_2 - x_1}$

where $(x_1, y_1)$ and $(x_2, y_2)$ are any two distinct points on the line. Think of it as "rise over run."

• Positive slope: line rises from left to right (direct relationship)
• Negative slope: line falls from left to right (inverse relationship)
• Zero slope: horizontal line (y doesn't change)
• Undefined slope: vertical line (x doesn't change; division by zero in the formula)

A slope of 2 means for every 1-unit increase in x, y increases by 2 units. A slope of -0.5 means for every 1-unit increase in x, y decreases by 0.5 units.

Y-intercept

The y-intercept (b) is the value of y when x = 0. Graphically, it's where the line crosses the y-axis.

In slope-intercept form $y = mx + b$, the y-intercept is just the constant term b. You can also find it from any equation form by substituting x = 0 and solving for y.

In real-world problems, the y-intercept often represents an initial value or fixed cost. For $y = 2x + 3$, the graph crosses the y-axis at (0, 3). If this modeled a taxi fare, that 3 would be the base charge before any distance is driven.

Linear equation forms

Slope-intercept form

$y = mx + b$

where m is the slope and b is the y-intercept. This form is the most convenient for graphing quickly: plot (0, b), then use the slope to find a second point.

For $y = -3x + 2$: start at (0, 2), then move right 1 and down 3 to reach (1, -1). Connect the points.

Point-slope form

$y - y_1 = m(x - x_1)$

where $(x_1, y_1)$ is a known point on the line and m is the slope. This form is most useful when you're given a point and a slope (or two points, since you can calculate the slope first).

For a line through (2, 3) with slope 0.5: $y - 3 = 0.5(x - 2)$

To convert to slope-intercept form, distribute and isolate y: $y = 0.5x + 2$

Standard form

$Ax + By = C$

where A, B, and C are integers (by convention), and A and B are not both zero. Standard form is especially handy for finding intercepts and for systems of equations.

To convert to slope-intercept form, solve for y. For $2x + 3y = 6$:

1. Subtract 2x from both sides: $3y = -2x + 6$
2. Divide by 3: $y = -\frac{2}{3}x + 2$

So the slope is $-\frac{2}{3}$ and the y-intercept is 2.

Converting between forms

Converting between forms comes down to algebraic rearrangement. Here are the most common conversions:

• Slope-intercept → point-slope: Pick any point on the line and plug it in. From $y = 2x - 1$, the point (1, 1) satisfies the equation, so: $y - 1 = 2(x - 1)$
• Point-slope → slope-intercept: Distribute the slope and solve for y.
• Point-slope → standard: Distribute, then move all variable terms to one side. From $y - 4 = -3(x - 2)$: distribute to get $y - 4 = -3x + 6$, then rearrange to $3x + y = 10$
• Standard → slope-intercept: Isolate y (shown above).

Real-world applications of linear equations and inequalities

Modeling with linear equations

When a problem describes a constant rate of change, a linear equation is usually the right model.

1. Identify the independent variable (x) and dependent variable (y).
2. Determine the rate of change (slope) and the starting value (y-intercept) from the problem.
3. Write the equation in the form $y = mx + b$.
4. Interpret m and b in context.

Example: A car rental company charges a base fee of $50 plus $0.25 per mile driven. Let x = miles driven and y = total cost. The equation is $y = 0.25x + 50$. The slope (0.25) is the cost per mile, and the y-intercept (50) is the flat fee you pay regardless of distance.

Solving linear equations in context

Once you have a model, you can substitute known values to find unknowns.

Example: Using $y = 0.25x + 50$, if a customer's bill is $100, how many miles did they drive?

1. Substitute y = 100: $100 = 0.25x + 50$
2. Subtract 50: $50 = 0.25x$
3. Divide by 0.25: $x = 200$ miles

Always check that your answer makes sense in context. Negative miles, for instance, would signal an error.

Applying linear inequalities

Inequalities model constraints rather than exact relationships. They define a range of acceptable solutions.

Example: A company produces products A and B with profit margins of $10 and $15, respectively. Product A requires 2 hours to make, and B requires 3 hours. The company has at most 24 hours available.

Let x = units of A and y = units of B. The constraints are:

• $2x + 3y \leq 24$ (time constraint)
• $x \geq 0$, $y \geq 0$ (can't produce negative units)

The objective is to maximize profit: $P = 10x + 15y$

Interpreting solutions in context

When working with inequalities and constraints, the feasible region is the set of all points satisfying every constraint simultaneously. For optimization problems, the optimal solution occurs at a vertex (corner point) of the feasible region.

In the production example above, testing the corner points of the feasible region yields a maximum profit at $x = 6$, $y = 4$:

$P = 10(6) + 15(4) = 60 + 60 = 120$

So the company should produce 6 units of A and 4 units of B for a maximum profit of $120. Always verify that your solution satisfies all constraints: $2(6) + 3(4) = 24 \leq 24$ ✓