Chapter 1: Linear Functions & Systems of Equations

Algebra 2 Textbook • Estimated reading time: 30 min

Chapter Contents

  1. 1.1 Slope & Linear Equations
  2. 1.2 Graphing Linear Functions
  3. 1.3 Systems of Linear Equations (2 Variables)
  4. 1.4 Systems of Linear Equations (3 Variables)
  5. 1.5 Systems of Linear Inequalities
  6. 1.6 Practice Problems

Linear functions are the starting point of Algebra 2 and form the foundation for every topic that follows. In this chapter, we develop a thorough understanding of slope, the three standard forms of a linear equation, and the relationships between parallel and perpendicular lines. We then move to systems of equations—first in two variables, then in three—and finish with systems of linear inequalities and an introduction to linear programming. Every concept is reinforced with worked examples, interactive Desmos graphs, and practice problems so you build both computational skill and geometric intuition.

1.1 Slope & Linear Equations

The concept of slope captures how steep a line is and in which direction it tilts. Slope is the single most important number associated with a linear function: it tells you the rate at which the output changes per unit change in the input. In applications, slope represents speed, cost per item, temperature change per hour, and countless other rates.

Definition: Slope

The slope of a line passing through two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ is defined as:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x} = \frac{\text{rise}}{\text{run}}$$

The slope measures the vertical change (rise) divided by the horizontal change (run) between any two points on the line. A positive slope means the line rises from left to right; a negative slope means it falls. A slope of zero gives a horizontal line, and an undefined slope (division by zero) gives a vertical line.

Interactive: Drag P₁ or P₂ directly on the graph to any position. The orange segment shows the run (Δx) and the green segment shows the rise (Δy). Watch slope $m = \dfrac{\Delta y}{\Delta x}$ change as you move the points.

Figure 1.1 — Rise, Run, and Slope

The beauty of slope is that it is constant everywhere on a given line. No matter which two points you select, the ratio $\frac{\Delta y}{\Delta x}$ produces the same value. This constant rate of change is precisely what makes a function linear.

Slope-Intercept Form

The most common way to write a linear equation is slope-intercept form:

$$y = mx + b$$

Here $m$ is the slope and $b$ is the $y$-intercept—the point where the line crosses the $y$-axis, i.e., the value of $y$ when $x = 0$. This form is especially useful for graphing: start at the point $(0, b)$ on the $y$-axis, then use the slope to find additional points by moving $m$ units up (or down if $m$ is negative) for every 1 unit to the right.

Example 1.1.1 — Using Slope-Intercept Form

A line passes through $(0, -3)$ with slope $\frac{2}{5}$. Write its equation and find the value of $y$ when $x = 10$.

Solution. Since the line passes through $(0, -3)$, the $y$-intercept is $b = -3$. With $m = \frac{2}{5}$:

$$y = \frac{2}{5}x - 3$$

When $x = 10$: $y = \frac{2}{5}(10) - 3 = 4 - 3 = 1$. So the point $(10, 1)$ lies on the line.

Point-Slope Form

When you know the slope $m$ and one point $(x_1, y_1)$ on the line (but the point is not necessarily the $y$-intercept), the most direct way to write the equation is point-slope form:

$$y - {\color[rgb]{0.573,0.251,0.055}y_1} = {\color[rgb]{0.086,0.639,0.290}m}(x - {\color[rgb]{0.573,0.251,0.055}x_1})$$

This form comes directly from the definition of slope. If $(x, y)$ is any other point on the line, then ${\color[rgb]{0.086,0.639,0.290}m} = \frac{y - {\color[rgb]{0.573,0.251,0.055}y_1}}{x - {\color[rgb]{0.573,0.251,0.055}x_1}}$, and multiplying both sides by $(x - {\color[rgb]{0.573,0.251,0.055}x_1})$ gives the point-slope equation. You can always rearrange point-slope form into slope-intercept form by distributing and solving for $y$.

Interactive: Drag the brown point to set $(x_1,\,y_1)$, then slide $m$ to see the line rotate around that fixed pivot point.

Figure 1.2 — Point-Slope Form: y − y₁ = m(x − x₁)

Example 1.1.2 — Writing an Equation through Two Points

Find the equation of the line through $(-2, 7)$ and $(4, -5)$.

Solution. First compute the slope:

$$m = \frac{-5 - 7}{4 - (-2)} = \frac{-12}{6} = -2$$

Using point-slope form with the point $(-2, 7)$:

$$y - 7 = -2(x - (-2)) \implies y - 7 = -2(x + 2) \implies y - 7 = -2x - 4$$ $$y = -2x + 3$$

Check with the second point: $y = -2(4) + 3 = -8 + 3 = -5$. Correct.

Standard Form

The standard form of a linear equation is:

$$Ax + By = C$$

where $A$, $B$, and $C$ are integers, and by convention $A \geq 0$. Standard form is useful when working with systems of equations (as we will see in Sections 1.3 and 1.4) because it aligns the $x$ and $y$ terms vertically, making elimination straightforward. To convert from slope-intercept form, move the $x$-term to the left side and clear any fractions.

Example 1.1.3 — Converting Between Forms

Convert $y = \frac{3}{4}x - 2$ to standard form.

Solution. Multiply both sides by 4 to clear the fraction:

$$4y = 3x - 8$$

Rearrange so that $x$ and $y$ are on the left:

$$-3x + 4y = -8$$

Multiply by $-1$ so that $A > 0$:

$$3x - 4y = 8$$

Interactive: Adjust $A$, $B$, $C$ to see how the line $Ax + By = C$ changes. Red dot = x-intercept $(C/A,\,0)$, green dot = y-intercept $(0,\,C/B)$.

Figure 1.3 — Standard Form: Ax + By = C

Parallel and Perpendicular Lines

Parallel and Perpendicular Slopes

Let $\ell_1$ and $\ell_2$ be two non-vertical lines with slopes $m_1$ and $m_2$.

Interactive: Adjust slope $m$ to see how parallel lines (dashed, same slope) and perpendicular lines (red, slope $-\frac{1}{m}$) behave. Notice the slopes multiply to $-1$ for perpendicular lines.

Figure 1.4 — Parallel and Perpendicular Lines

Example 1.1.4 — Parallel and Perpendicular Lines

Find the equation of the line that (a) is parallel to $y = 3x - 1$ and passes through $(2, 5)$; (b) is perpendicular to $y = 3x - 1$ and passes through $(2, 5)$.

Solution.

(a) The given line has slope $m = 3$. A parallel line has the same slope $m = 3$. Using point-slope form:

$$y - 5 = 3(x - 2) \implies y = 3x - 1$$

Interestingly, the parallel line through $(2, 5)$ turns out to be the same line. Let us verify: $y = 3(2) - 1 = 5$. Indeed, $(2, 5)$ lies on $y = 3x - 1$.

(b) The perpendicular slope is $m = -\frac{1}{3}$. Using point-slope form:

$$y - 5 = -\frac{1}{3}(x - 2) \implies y = -\frac{1}{3}x + \frac{2}{3} + 5 = -\frac{1}{3}x + \frac{17}{3}$$
Study Tip: A common error is to use the same slope for perpendicular lines. Remember: perpendicular means negative reciprocal. If the original slope is $\frac{2}{3}$, the perpendicular slope is $-\frac{3}{2}$, not $\frac{2}{3}$ and not $-\frac{2}{3}$.

1.2 Graphing Linear Functions

A linear function $f(x) = mx + b$ produces a straight-line graph. While the equation tells you everything algebraically, the graph provides geometric insight that makes many concepts—especially systems of equations—much easier to understand. In this section, we cover three approaches to graphing linear functions and explore piecewise linear functions.

Graphing Using Intercepts

Every non-vertical, non-horizontal line crosses both axes. The two intercepts give us two points, which is all we need to draw the line.

Example 1.2.1 — Graphing by Intercepts

Graph the line $2x + 3y = 12$.

Solution. Find the intercepts:

Plot $(6, 0)$ and $(0, 4)$ and draw a straight line through them. The slope is $m = \frac{0 - 4}{6 - 0} = -\frac{2}{3}$, confirming the line falls from left to right.

Transformations of $y = x$

The parent linear function is $y = x$, which passes through the origin with slope 1. Every other linear function can be obtained from $y = x$ through transformations:

Explore: Adjust the sliders for $m$ and $b$ to see how they affect the graph of $y = mx + b$

Figure 1.5 — Slope-Intercept Form y = mx + b

In the interactive graph above, the dashed gray line shows the parent function $y = x$ for reference, while the blue line shows $y = mx + b$. Experiment with different values:

Piecewise Linear Functions

A piecewise linear function uses different linear formulas on different intervals of the domain. A common example is the absolute value function $f(x) = |x|$, which can be written as:

$$f(x) = |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$

Piecewise functions arise frequently in real-world modeling. For instance, a cell phone plan might charge \$0.10 per minute for the first 500 minutes and \$0.05 per minute thereafter. To graph a piecewise linear function, graph each piece on its respective interval, using closed dots for included endpoints and open dots for excluded endpoints.

Example 1.2.2 — Piecewise Linear Function

Graph and find the range of:

$$g(x) = \begin{cases} -x + 4 & \text{if } x \leq 2 \\ 2x - 2 & \text{if } x > 2 \end{cases}$$

Solution. For $x \leq 2$, the function is $g(x) = -x + 4$, a line with slope $-1$ and $y$-intercept $4$. At $x = 2$: $g(2) = -2 + 4 = 2$ (closed dot). As $x \to -\infty$, $g(x) \to +\infty$.

For $x > 2$, the function is $g(x) = 2x - 2$, a line with slope $2$. At $x = 2$: $g(2) = 2(2) - 2 = 2$ (open dot, but the value matches). As $x \to +\infty$, $g(x) \to +\infty$.

Both pieces meet at $(2, 2)$, so the function is continuous. The vertex of the V-shape is at $(2, 2)$, so the minimum value is $2$. The range is $[2, \infty)$.

Graph of $g(x)$ from Example 1.2.2: the two linear pieces meet at the vertex $(2,\,2)$, forming a V-shape opening upward.

Figure 1.6 — Piecewise Linear Function g(x)

1.3 Systems of Linear Equations (2 Variables)

A system of linear equations consists of two or more linear equations considered simultaneously. The solution of a system is the set of all ordered pairs $(x, y)$ that satisfy every equation in the system at the same time. Geometrically, solving a system of two linear equations means finding the point(s) where two lines intersect.

Definition: Types of Linear Systems

A system of two linear equations in two variables has exactly one of three outcomes:

Consistent & Independent

One unique solution

Lines intersect at exactly one point. Different slopes.

Figure 1.6a — Lines cross at (2, 1)
Inconsistent

No solution

Lines are parallel. Same slope, different $y$-intercepts.

Figure 1.6b — Parallel lines, never meet
Consistent & Dependent

Infinitely many solutions

Both equations describe the same line.

Figure 1.6c — One line (equations coincide)

CONSISTENT vs. INCONSISTENT — at a glance

CONSISTENT — has at least one solution
INCONSISTENT — has no solution

Within consistent systems: independent = unique solution; dependent = infinite solutions.

Figure 1.6 — Three possible outcomes for a 2-variable linear system

The Substitution Method

The idea behind substitution is to solve one equation for one variable and then substitute that expression into the other equation, reducing the system to a single equation in one variable.

Example 1.3.1 — Solving by Substitution

Solve the system:

$$\begin{cases} y = 2x + 1 \\ 3x + 2y = 16 \end{cases}$$

Solution. The first equation already gives $y$ in terms of $x$. Substitute $y = 2x + 1$ into the second equation:

$$3x + 2(2x + 1) = 16$$ $$3x + 4x + 2 = 16$$ $$7x = 14$$ $$x = 2$$

Back-substitute into $y = 2x + 1$: $y = 2(2) + 1 = 5$.

The solution is $(2, 5)$. Check in the second equation: $3(2) + 2(5) = 6 + 10 = 16$. Correct.

The Elimination Method

The elimination method (also called the addition method) works by adding or subtracting the equations to eliminate one variable. If necessary, multiply one or both equations by constants first so that the coefficients of one variable are opposites.

Example 1.3.2 — Solving by Elimination

Solve the system:

$$\begin{cases} 4x - 3y = 11 \\ 3x + 2y = 4 \end{cases}$$

Solution. To eliminate $y$, multiply the first equation by 2 and the second by 3:

$$\begin{cases} 8x - 6y = 22 \\ 9x + 6y = 12 \end{cases}$$

Add the equations:

$$17x = 34 \implies x = 2$$

Substitute $x = 2$ into $3x + 2y = 4$: $6 + 2y = 4$, so $2y = -2$ and $y = -1$.

The solution is $(2, -1)$. Verify: $4(2) - 3(-1) = 8 + 3 = 11$ and $3(2) + 2(-1) = 6 - 2 = 4$. Both check out.

Graphical Interpretation

Graphically, solving a system of two linear equations means finding the intersection point of two lines. The interactive graph below lets you adjust the slopes and intercepts of two lines to see how the intersection changes.

Explore: Adjust the slopes and intercepts of two lines to see their intersection point

Figure 1.7 — Systems of Two Lines: Consistent, Inconsistent, Dependent

Try these experiments with the graph above:

Example 1.3.3 — Identifying an Inconsistent System

Solve the system:

$$\begin{cases} 2x - 4y = 6 \\ -x + 2y = 1 \end{cases}$$

Solution. Multiply the second equation by 2: $-2x + 4y = 2$. Add to the first equation:

$$0 = 8$$

This is a false statement. Therefore, the system is inconsistent and has no solution. Rewriting both equations in slope-intercept form: $y = \frac{1}{2}x - \frac{3}{2}$ and $y = \frac{1}{2}x + \frac{1}{2}$. Both lines have slope $\frac{1}{2}$ but different $y$-intercepts, confirming they are parallel.

Example 1.3.4 — Identifying a Dependent System

Solve the system:

$$\begin{cases} 6x - 9y = 12 \\ -2x + 3y = -4 \end{cases}$$

Solution. Multiply the second equation by 3: $-6x + 9y = -12$. Add to the first:

$$0 = 0$$

This is always true, so the system is dependent. The two equations represent the same line. In slope-intercept form, both reduce to $y = \frac{2}{3}x - \frac{4}{3}$. The solution set consists of infinitely many points: $\left\{\left(x,\; \frac{2}{3}x - \frac{4}{3}\right) : x \in \mathbb{R}\right\}$.

When to Use Which Method: Use substitution when one equation is already solved for a variable (or can be easily solved). Use elimination when both equations are in standard form $Ax + By = C$ and the coefficients are easy to manipulate. For large systems, elimination scales better.

1.4 Systems of Linear Equations (3 Variables)

A system of three linear equations in three unknowns has the general form:

$$\begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3 \end{cases}$$

Geometrically, each equation represents a plane in three-dimensional space. The solution is the point (or set of points) where all three planes intersect. A unique solution corresponds to the three planes meeting at a single point. If the system is inconsistent, at least two of the planes are parallel (or the three planes form a triangular prism with no common point). If the system is dependent, the planes share a common line or are all the same plane.

Elimination Method for 3 Variables

The strategy is to reduce the 3-variable system to a 2-variable system, then to a single equation. We do this by choosing one variable to eliminate, then pairing equations to create two new equations that do not contain that variable.

Example 1.4.1 — Solving a 3-Variable System

Solve the system:

$$\begin{cases} x + 2y - z = 3 & (1) \\ 2x - y + 3z = 9 & (2) \\ 3x + y + 2z = 15 & (3) \end{cases}$$

Solution. We will eliminate $y$.

Step 1: Combine equations (1) and (2) to eliminate $y$. Multiply (1) by 1 and (2) by 2, then add:

Actually, let us use a more efficient pairing. Add $(2)$ and $2 \times (1)$ to bring $y$-coefficients to a manageable form. Let us instead add (2) and (3) directly, since they have $-y$ and $+y$:

$$(2) + (3): \quad (2x - y + 3z) + (3x + y + 2z) = 9 + 15$$ $$5x + 5z = 24 \quad \cdots (4)$$

Step 2: Combine equations (1) and (3) to eliminate $y$. Multiply (3) by 2 and subtract (1) from it... or multiply (1) by 1 and add to $2 \times (3)$. Since (1) has $+2y$ and (3) has $+y$, multiply (3) by $-2$ and add to (1):

$$(1) + (-2)(3): \quad (x + 2y - z) + (-6x - 2y - 4z) = 3 + (-30)$$ $$-5x - 5z = -27 \quad \cdots (5)$$

Step 3: Add equations (4) and (5):

$$0 = -3$$

Wait — this is a contradiction. Let us recheck. Equation (4): $5x + 5z = 24$. Equation (5): $-5x - 5z = -27$. Adding: $0 = -3$, which is false.

This means the system is inconsistent and has no solution.

Let us now work through a system that does have a unique solution.

Example 1.4.2 — A Consistent 3-Variable System

Solve the system:

$$\begin{cases} x + y + z = 6 & (1) \\ 2x - y + z = 3 & (2) \\ x + 2y - z = 5 & (3) \end{cases}$$

Solution. Eliminate $z$ by adding equations (1) and (3):

$$(1) + (3): \quad (x + y + z) + (x + 2y - z) = 6 + 5$$ $$2x + 3y = 11 \quad \cdots (4)$$

Now eliminate $z$ by adding equations (2) and (3):

$$(2) + (3): \quad (2x - y + z) + (x + 2y - z) = 3 + 5$$ $$3x + y = 8 \quad \cdots (5)$$

Back-substitution: Solve the 2-variable system {(4), (5)}. From (5): $y = 8 - 3x$. Substitute into (4):

$$2x + 3(8 - 3x) = 11 \implies 2x + 24 - 9x = 11 \implies -7x = -13 \implies x = \frac{13}{7}$$

Then $y = 8 - 3 \cdot \frac{13}{7} = \frac{56 - 39}{7} = \frac{17}{7}$.

From equation (1): $z = 6 - x - y = 6 - \frac{13}{7} - \frac{17}{7} = \frac{42 - 13 - 17}{7} = \frac{12}{7}$.

The solution is $\left(\frac{13}{7},\; \frac{17}{7},\; \frac{12}{7}\right)$.

Check in equation (2): $2 \cdot \frac{13}{7} - \frac{17}{7} + \frac{12}{7} = \frac{26 - 17 + 12}{7} = \frac{21}{7} = 3$. Correct.

Strategy for 3-Variable Systems

To solve a system of three equations in three unknowns:

  1. Choose a variable to eliminate. Pick whichever variable has the simplest coefficients (ideally $1$ or $-1$).
  2. Create two new equations in two variables by pairing the original equations and eliminating the chosen variable.
  3. Solve the resulting 2-variable system using substitution or elimination.
  4. Back-substitute to find the third variable.
  5. Check the solution in all three original equations.

Example 1.4.3 — Application: Finding a Quadratic through Three Points

Find the equation of the parabola $y = ax^2 + bx + c$ that passes through $(1, 6)$, $(2, 11)$, and $(-1, 2)$.

Solution. Substitute each point into $y = ax^2 + bx + c$:

$$\begin{cases} a + b + c = 6 & (1) \\ 4a + 2b + c = 11 & (2) \\ a - b + c = 2 & (3) \end{cases}$$

Subtract (3) from (1): $(a + b + c) - (a - b + c) = 6 - 2$, giving $2b = 4$, so $b = 2$.

Subtract (1) from (2): $(4a + 2b + c) - (a + b + c) = 11 - 6$, giving $3a + b = 5$. Since $b = 2$: $3a + 2 = 5$, so $a = 1$.

From (1): $1 + 2 + c = 6$, so $c = 3$.

The parabola is $y = x^2 + 2x + 3$.

1.5 Systems of Linear Inequalities

A linear inequality in two variables looks like $ax + by < c$ (or with $\leq$, $>$, $\geq$). The solution set of a single linear inequality is a half-plane: one side of the corresponding boundary line $ax + by = c$. A system of linear inequalities asks for the set of points that satisfy all the inequalities simultaneously, which is the intersection of the individual half-planes.

Graphing Linear Inequalities

Steps for Graphing a Linear Inequality

  1. Graph the boundary line: Replace the inequality symbol with $=$ and graph the resulting line. Use a solid line for $\leq$ or $\geq$ (boundary included) and a dashed line for $<$ or $>$ (boundary excluded).
  2. Choose a test point not on the boundary (the origin $(0, 0)$ is usually easiest, provided it is not on the line).
  3. Shade the correct side: If the test point satisfies the inequality, shade the side containing the test point. Otherwise, shade the other side.

Example 1.5.1 — Graphing a System of Inequalities

Graph the solution region for the system:

$$\begin{cases} x + y \leq 8 \\ 2x + y \geq 6 \\ x \geq 0 \\ y \geq 0 \end{cases}$$

Solution. The constraints $x \geq 0$ and $y \geq 0$ restrict us to the first quadrant.

For $x + y \leq 8$: the boundary is the line $x + y = 8$ (intercepts at $(8, 0)$ and $(0, 8)$). Test $(0, 0)$: $0 + 0 = 0 \leq 8$. True, so shade below/left of the line.

For $2x + y \geq 6$: the boundary is the line $2x + y = 6$ (intercepts at $(3, 0)$ and $(0, 6)$). Test $(0, 0)$: $0 + 0 = 0 \geq 6$. False, so shade above/right of the line.

The feasible region is the quadrilateral bounded by these four constraints in the first quadrant.

Finding Vertices

The vertices (corner points) of the solution region are found by solving pairs of boundary equations simultaneously. For the example above, the vertices are:

Checking which vertices satisfy all constraints, the feasible region has vertices at $(3, 0)$, $(8, 0)$, $(0, 8)$, and $(0, 6)$.

Introduction to Linear Programming

Definition: Linear Programming

Linear programming is a method for optimizing (maximizing or minimizing) a linear objective function $P = ax + by$ subject to a system of linear inequality constraints. The feasible region is the set of points satisfying all constraints.

The Corner Point Theorem (Fundamental Theorem of Linear Programming)

If the feasible region is bounded, then the maximum and minimum values of the objective function $P = ax + by$ occur at one or more of the vertices (corner points) of the feasible region. Therefore, to find the optimal value, we need only evaluate the objective function at each vertex and compare.

Example 1.5.2 — Linear Programming

Maximize $P = 5x + 4y$ subject to:

$$\begin{cases} x + y \leq 10 \\ 2x + y \leq 16 \\ x \geq 0 \\ y \geq 0 \end{cases}$$

Solution. First, find the vertices of the feasible region.

Boundary line intersections (checking all pairs):

Check that $(0, 10)$ satisfies $2x + y \leq 16$: $0 + 10 = 10 \leq 16$. Yes.

Evaluate $P$ at each vertex:

The maximum value is $\boxed{P = 46}$ at the point $(6, 4)$.

Interactive: The shaded region is the feasible region for Example 1.5.2. Drag the slider $P$ to sweep the objective line $5x+4y=P$ and find the last vertex it touches — that gives the maximum.

Figure 1.8 — Linear Programming Feasible Region

Linear Programming Strategy: Always sketch the feasible region first. Label all vertices by solving pairs of boundary equations. Then evaluate the objective function at each vertex. The optimal value must occur at a vertex, so you never need to check interior points.

1.6 Practice Problems

Test your understanding of the concepts in this chapter. Each problem has a detailed solution that you can reveal after attempting it yourself. Try to work through each problem completely before checking your answer.

Problem 1. Find the slope of the line through $(-3, 8)$ and $(5, -4)$. Then write the equation in slope-intercept form.

Show Solution

$m = \dfrac{-4 - 8}{5 - (-3)} = \dfrac{-12}{8} = -\dfrac{3}{2}$

Using point-slope form with $(-3, 8)$:

$y - 8 = -\frac{3}{2}(x + 3) = -\frac{3}{2}x - \frac{9}{2}$

$y = -\frac{3}{2}x - \frac{9}{2} + 8 = -\frac{3}{2}x + \frac{7}{2}$

Problem 2. Write the equation of the line perpendicular to $4x - 2y = 10$ that passes through $(6, 1)$.

Show Solution

First, find the slope of $4x - 2y = 10$. Solve for $y$: $y = 2x - 5$, so $m = 2$.

The perpendicular slope is $m_\perp = -\frac{1}{2}$.

$y - 1 = -\frac{1}{2}(x - 6) \implies y = -\frac{1}{2}x + 3 + 1 = -\frac{1}{2}x + 4$

Problem 3. Convert $y = -\frac{5}{3}x + 7$ to standard form with integer coefficients.

Show Solution

Multiply both sides by 3: $3y = -5x + 21$.

Rearrange: $5x + 3y = 21$.

Since $A = 5 > 0$, this is already in standard form.

Problem 4. Solve by substitution:

$$\begin{cases} 3x - y = 7 \\ x + 2y = 0 \end{cases}$$

Show Solution

From the second equation: $x = -2y$.

Substitute into the first: $3(-2y) - y = 7 \implies -6y - y = 7 \implies -7y = 7 \implies y = -1$.

Then $x = -2(-1) = 2$.

Solution: $(2, -1)$.

Check: $3(2) - (-1) = 7$ and $2 + 2(-1) = 0$. Both correct.

Problem 5. Solve by elimination:

$$\begin{cases} 5x + 3y = 1 \\ 2x - 5y = 19 \end{cases}$$

Show Solution

Multiply the first equation by 5 and the second by 3:

$25x + 15y = 5$ and $6x - 15y = 57$.

Add: $31x = 62 \implies x = 2$.

Substitute into $5(2) + 3y = 1$: $10 + 3y = 1 \implies 3y = -9 \implies y = -3$.

Solution: $(2, -3)$.

Check: $2(2) - 5(-3) = 4 + 15 = 19$. Correct.

Problem 6. Determine whether the following system is consistent, inconsistent, or dependent. If consistent, find the solution.

$$\begin{cases} 6x - 4y = 10 \\ -9x + 6y = -15 \end{cases}$$

Show Solution

Multiply the first equation by 3 and the second by 2:

$18x - 12y = 30$ and $-18x + 12y = -30$.

Add: $0 = 0$. This is always true.

The system is dependent. Both equations simplify to $3x - 2y = 5$, or $y = \frac{3}{2}x - \frac{5}{2}$.

The solution set is $\left\{\left(x,\; \frac{3x - 5}{2}\right) : x \in \mathbb{R}\right\}$.

Problem 7. Solve the 3-variable system:

$$\begin{cases} x + y + z = 6 \\ x - y + 2z = 5 \\ 2x + y - z = 1 \end{cases}$$

Show Solution

Add equations (1) and (2): $(x + y + z) + (x - y + 2z) = 6 + 5$, giving $2x + 3z = 11$ ... (4)

Add equations (2) and (3): $(x - y + 2z) + (2x + y - z) = 5 + 1$, giving $3x + z = 6$ ... (5)

From (5): $z = 6 - 3x$. Substitute into (4): $2x + 3(6 - 3x) = 11 \implies 2x + 18 - 9x = 11 \implies -7x = -7 \implies x = 1$.

$z = 6 - 3(1) = 3$. From (1): $1 + y + 3 = 6$, so $y = 2$.

Solution: $(1, 2, 3)$.

Check in (2): $1 - 2 + 6 = 5$. In (3): $2 + 2 - 3 = 1$. Both correct.

Problem 8. A movie theater sold 345 tickets for a total of \$2,550. Adult tickets cost \$10 and child tickets cost \$5. How many of each type were sold?

Show Solution

Let $a$ = number of adult tickets and $c$ = number of child tickets.

$\begin{cases} a + c = 345 \\ 10a + 5c = 2550 \end{cases}$

From the first equation: $c = 345 - a$. Substitute:

$10a + 5(345 - a) = 2550 \implies 10a + 1725 - 5a = 2550 \implies 5a = 825 \implies a = 165$.

$c = 345 - 165 = 180$.

165 adult tickets and 180 child tickets.

Check: $10(165) + 5(180) = 1650 + 900 = 2550$. Correct.

Problem 9. Graph the solution region for the system of inequalities and identify all vertices:

$$\begin{cases} x + 2y \leq 12 \\ 3x + y \leq 15 \\ x \geq 0 \\ y \geq 0 \end{cases}$$

Show Solution

The feasible region is in the first quadrant, bounded by:

  • $x + 2y = 12$: intercepts at $(12, 0)$ and $(0, 6)$.
  • $3x + y = 15$: intercepts at $(5, 0)$ and $(0, 15)$.

Find the intersection of $x + 2y = 12$ and $3x + y = 15$:

From the first: $x = 12 - 2y$. Substitute: $3(12 - 2y) + y = 15 \implies 36 - 6y + y = 15 \implies -5y = -21 \implies y = \frac{21}{5}$.

$x = 12 - 2 \cdot \frac{21}{5} = 12 - \frac{42}{5} = \frac{18}{5}$.

Vertices: $(0, 0)$, $(5, 0)$, $\left(\frac{18}{5}, \frac{21}{5}\right)$, $(0, 6)$.

Note: $(12, 0)$ does not satisfy $3x + y \leq 15$, and $(0, 15)$ does not satisfy $x + 2y \leq 12$, so they are not vertices of the feasible region.

Problem 10. Maximize $P = 8x + 5y$ subject to the constraints in Problem 9.

Show Solution

Evaluate $P$ at each vertex from Problem 9:

  • $P(0, 0) = 0$
  • $P(5, 0) = 40$
  • $P\!\left(\frac{18}{5}, \frac{21}{5}\right) = 8 \cdot \frac{18}{5} + 5 \cdot \frac{21}{5} = \frac{144}{5} + \frac{105}{5} = \frac{249}{5} = 49.8$
  • $P(0, 6) = 30$

The maximum value is $\boxed{P = 49.8}$ at the point $\left(\frac{18}{5}, \frac{21}{5}\right) = (3.6, 4.2)$.

Problem 11. Find the equation of the line parallel to $y = -4x + 9$ that passes through the midpoint of the segment joining $(-2, 3)$ and $(6, -1)$.

Show Solution

Midpoint: $\left(\frac{-2 + 6}{2}, \frac{3 + (-1)}{2}\right) = (2, 1)$.

Parallel slope: $m = -4$.

$y - 1 = -4(x - 2) \implies y = -4x + 8 + 1 = -4x + 9$.

The midpoint $(2, 1)$ actually lies on the original line: $-4(2) + 9 = 1$. So the line is $y = -4x + 9$ itself.

Problem 12. A farmer has 240 acres and \$18,000 to invest. Crop A costs \$100 per acre and yields \$300 profit per acre. Crop B costs \$50 per acre and yields \$200 profit per acre. How many acres of each crop should the farmer plant to maximize profit?

Show Solution

Let $x$ = acres of Crop A and $y$ = acres of Crop B.

Constraints:

$\begin{cases} x + y \leq 240 & \text{(land)} \\ 100x + 50y \leq 18000 & \text{(budget)} \\ x \geq 0,\; y \geq 0 \end{cases}$

Simplify the budget constraint: $2x + y \leq 360$.

Objective: Maximize $P = 300x + 200y$.

Vertices:

  • $(0, 0)$: $P = 0$
  • $(180, 0)$: from $2x + y = 360$ with $y = 0$. Check: $180 + 0 = 180 \leq 240$. Valid. $P = 54{,}000$.
  • $(120, 120)$: intersection of $x + y = 240$ and $2x + y = 360$. Subtract: $x = 120$, $y = 120$. $P = 36{,}000 + 24{,}000 = 60{,}000$.
  • $(0, 240)$: Check: $2(0) + 240 = 240 \leq 360$. Valid. $P = 48{,}000$.

Maximum profit is $\boxed{\$60{,}000}$ by planting 120 acres of each crop.

Next: Ch 2 — Quadratic Functions →