3.2 Systems of Linear Equations and Inequalities

Systems of linear equations and inequalities let you find values that satisfy multiple conditions at once. They're central to Honors Algebra II because they connect algebraic techniques to real-world modeling, and they show up constantly in later math courses.

This section covers three methods for solving systems of equations, how to classify systems by their number of solutions, graphing systems of inequalities, and applying all of it to word problems.

Solving Systems of Equations

Graphical Method

Graph both equations on the same coordinate plane. The solution is the point (or points) where the lines intersect. You read the $x$- and $y$-coordinates directly from the intersection point.

This method is great for visualizing what's happening, but it's not always precise. If the intersection falls between grid lines, you'll need an algebraic method to get an exact answer.

Substitution Method

Substitution works best when one equation already has a variable isolated (or can be rearranged easily).

1. Solve one equation for one variable.
2. Substitute that expression into the other equation.
3. Solve the resulting single-variable equation.
4. Plug the value back into either original equation to find the other variable.

Example:

Given:

• $2x + y = 7$
• $x - y = 1$

Solve the second equation for $x$: $x = 1 + y$

Substitute into the first equation: $2(1 + y) + y = 7$

Distribute and combine: $2 + 2y + y = 7$, so $3y = 5$, giving $y = \frac{5}{3}$

Back-substitute: $x = 1 + \frac{5}{3} = \frac{8}{3}$

The solution is $\left(\frac{8}{3},\ \frac{5}{3}\right)$.

Elimination Method

Elimination (also called the addition method) works by adding the two equations together so that one variable cancels out.

1. If the coefficients of one variable are already opposites, add the equations directly.
2. If not, multiply one or both equations by constants so that one variable's coefficients become opposites.
3. Add the equations to eliminate that variable.
4. Solve for the remaining variable.
5. Substitute back into either original equation to find the other variable.

Example:

Given:

• $3x + 2y = 11$
• $2x - y = 1$

Multiply the second equation by 2: $4x - 2y = 2$

Add to the first equation: $(3x + 2y) + (4x - 2y) = 11 + 2$, so $7x = 13$

Solve: $x = \frac{13}{7}$

Substitute into the second equation: $2\left(\frac{13}{7}\right) - y = 1$, so $\frac{26}{7} - y = 1$, giving $y = \frac{19}{7}$

The solution is $\left(\frac{13}{7},\ \frac{19}{7}\right)$.

Number of Solutions for Systems

Every system of two linear equations falls into one of three categories. Recognizing which category you're in saves time and prevents confusion when you get an unexpected result during solving.

One Solution (Consistent and Independent)

The lines have different slopes, so they cross at exactly one point. This is the most common case. You'll get a unique $(x, y)$ pair.

No Solution (Inconsistent)

The lines are parallel: same slope, different $y$-intercepts. They never intersect. When you solve algebraically, the variables cancel and you're left with a false statement like $0 = 5$. That false statement is your signal that no solution exists.

Infinitely Many Solutions (Consistent and Dependent)

The lines are coincident, meaning they're the same line. Same slope, same $y$-intercept. Algebraically, the variables cancel and you get a true statement like $0 = 0$.

Example: Consider:

• $2x + 3y = 6$
• $4x + 6y = 12$

The second equation is just the first multiplied by 2. Every point on $2x + 3y = 6$ satisfies both equations, so there are infinitely many solutions.

Graphing Systems of Inequalities

Graphing a Linear Inequality

1. Rewrite the inequality in slope-intercept form if needed.

2. Graph the boundary line:

   • Dashed line for strict inequalities ($<$ or $>$) because points on the line are not included.
   • Solid line for non-strict inequalities ($\leq$ or $\geq$) because points on the line are included.

3. Shade the correct side:

   • Shade above the line for $y >$ or $y \geq$.
   • Shade below the line for $y <$ or $y \leq$.
   • If you're unsure, test a point (the origin is convenient when it's not on the line). If the point satisfies the inequality, shade that side.

Identifying the Solution Set

Graph each inequality on the same coordinate plane. The solution set is the region where all shaded areas overlap.

• Bounded solution set: the overlapping region is enclosed (forms a polygon).
• Unbounded solution set: the overlapping region extends infinitely in some direction.
• Empty solution set: no overlap exists, meaning no point satisfies all inequalities simultaneously.

To find the vertices of the solution region, solve the systems of equations formed by pairs of boundary lines. These vertices matter because, in optimization problems, the best answer always occurs at a vertex.

Example:

Given:

• $y \leq 2x + 1$
• $y > -x + 3$

Graph $y = 2x + 1$ as a solid line and shade below. Graph $y = -x + 3$ as a dashed line and shade above. The solution set is the unbounded region where the shading overlaps.

Real-World Applications of Systems

Modeling with Systems of Linear Equations

The setup process for word problems follows a consistent pattern:

1. Define your variables clearly (what does $x$ represent? what does $y$ represent?).
2. Translate each condition in the problem into an equation.
3. Solve using substitution or elimination.
4. Check that your answer makes sense in context.

Common scenarios include:

• Break-even analysis: finding where revenue equals cost
• Mixture problems: combining solutions of different concentrations
• Rate problems: two objects moving at different speeds

Modeling with Systems of Linear Inequalities

The setup is similar, but constraints become inequalities instead of equations. These problems often ask you to find a feasible region rather than a single answer.

• Linear programming problems ask you to maximize or minimize an objective function (like profit or cost) subject to inequality constraints.
• The optimal solution always occurs at a vertex of the feasible region, so you only need to evaluate the objective function at each vertex.

Interpreting Solutions

Always check that your mathematical answer makes sense in context. Negative quantities of a product, fractional numbers of people, or values that violate a constraint all signal that you need to re-examine your work.

Example:

A bakery sells cupcakes and cookies. Each cupcake uses 2 units of flour and 1 unit of sugar. Each cookie uses 1 unit of flour and 2 units of sugar. The bakery has 50 units of flour and 60 units of sugar. Profit is $\$2$ per cupcake and $\$1$ per cookie.

Let $x$ = number of cupcakes, $y$ = number of cookies.

Constraints:

• $2x + y \leq 50$ (flour)
• $x + 2y \leq 60$ (sugar)
• $x \geq 0,\ y \geq 0$ (can't produce negative quantities)

Objective function: Maximize $P = 2x + y$

Find the vertices of the feasible region by solving pairs of boundary equations:

• $(0, 0)$: $P = 0$
• $(25, 0)$: $P = 50$
• $(0, 30)$: $P = 30$
• Intersection of $2x + y = 50$ and $x + 2y = 60$: solving gives $(x, y) = \left(\frac{40}{3},\ \frac{70}{3}\right) \approx (13.3,\ 23.3)$, so $P \approx 50$

Evaluating at each vertex, the maximum profit of $\$50$ occurs at $(25, 0)$: produce 25 cupcakes and 0 cookies.