Honors Algebra II Unit 3 Review: Linear Equations and Inequalities

A system of linear equations is two or more equations involving the same variables. For a system in two variables, the solution set contains every ordered pair that makes all the equations true at once. On a graph, those are the points shared by all the lines.

You have three ways to solve a system. Graphing shows the picture but can miss fractional answers. Substitution works best when one variable is already alone or has a coefficient of 1. Elimination works best when the equations are in standard form and you can scale them so one variable cancels.

Worked example. Solve $3x + 2y = 16$ and $x - y = 2$. The second equation gives $x = y + 2$. Substitute that into the first equation to get $3(y + 2) + 2y = 16$, so $5y + 6 = 16$ and $y = 2$. Then $x = 2 + 2 = 4$. Check in the first equation, where $3(4) + 2(2) = 16$ is true. The solution is $(4, 2)$.

When solving two line equations, a false statement like $0 = 5$ means the lines are distinct and parallel, so there is no solution. A true statement like $0 = 0$ means the equations describe the same line, so there are infinitely many solutions.

For a system of inequalities, the solution is the overlap of all the shaded half-planes. That overlap can be a region, a line, a point, or empty. A point counts only if it satisfies every inequality, and points on a dashed boundary are not included.

Linear programming finds the maximum or minimum value of a linear expression when the variables must follow a set of linear inequalities. Every problem has three parts. The decision variables are the quantities you choose. The constraints are the limits on those quantities. The objective function is the quantity you want to maximize or minimize.

Graph all the constraints, including $x \ge 0$ and $y \ge 0$ when negative amounts make no sense. The overlap of the shaded regions is the feasible region. For a closed feasible region with vertices, the Corner Point Theorem says that an optimal value, if one exists, occurs at a vertex. Testing the vertices finds an optimal value, though other points may also give that value.

Worked example. A bakery makes $x$ trays of muffins and $y$ trays of cookies. Oven space gives $x + y \le 10$, and labor gives $x + 3y \le 18$, with $x, y \ge 0$. Profit is $P = 3x + 5y$. The vertices are $(0, 0)$, $(10, 0)$, $(0, 6)$, and the intersection of $x + y = 10$ with $x + 3y = 18$. Subtracting those equations gives $2y = 8$, so $y = 4$ and $x = 6$. Testing each vertex gives $P = 0$, $30$, $30$, and $38$. The maximum profit is 38 dollars at 6 muffin trays and 4 cookie trays.

If the region is unbounded, a maximum or minimum might not exist, so check before you report one.

A linear equation in two variables graphs as a straight line, and every point on the line is a solution. For two points with $x_1 \ne x_2$, the slope $m = \frac{y_2 - y_1}{x_2 - x_1}$ is the rate of change. Horizontal lines have slope 0, and vertical lines have undefined slope. In context, the slope is a rate like dollars per hour, and the y-intercept is a starting value like a flat fee.

You'll use three forms. Slope-intercept form is $y = mx + b$, point-slope form is $y - y_1 = m(x - x_1)$, and standard form is $Ax + By = C$. A nonvertical line can be written in any of these forms, so pick the form that matches the information you have. A vertical line can be written in standard form, but not in slope-intercept or point-slope form.

Worked example. Find the line through $(2, 5)$ and $(6, -3)$. The slope is $\frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2$. Point-slope form gives $y - 5 = -2(x - 2)$. Distribute to get $y = -2x + 9$, which is $2x + y = 9$ in standard form.

To graph an inequality, draw the boundary line, solid for $\le$ or $\ge$ and dashed for $<$ or $>$. Then test a point not on the line, usually $(0, 0)$. If the test point makes the inequality true, shade its side. If you multiply or divide both sides by a negative number while solving, flip the inequality symbol.