📈College Algebra Unit 11 Review

A system of linear equations is two or more equations that share the same variables. Solving the system means finding the values that make all the equations true at the same time. This comes up constantly in real-world problems where you have multiple conditions to satisfy at once.

Graphing Solutions of Linear Systems

Since each linear equation represents a line on the coordinate plane, a system of two linear equations is really just two lines on the same graph. The solution is the point where those lines intersect, because that's the $(x, y)$ pair that satisfies both equations simultaneously.

To solve by graphing:

Rewrite each equation in slope-intercept form ( $y = mx + b$ ) if it isn't already.
Graph both lines on the same coordinate plane.
Identify the intersection point and read off its coordinates.
Check by plugging the coordinates into both original equations.

The number of solutions depends on how the two lines relate to each other:

One solution: The lines intersect at a single point. The slopes are different ( $m_1 \neq m_2$ ). This is called a consistent and independent system.
No solution: The lines are parallel. They have the same slope but different y-intercepts ( $m_1 = m_2$ and $b_1 \neq b_2$ ). This is an inconsistent system.
Infinitely many solutions: The lines are the same line (coincident). Same slope and same y-intercept ( $m_1 = m_2$ and $b_1 = b_2$ ). This is a consistent and dependent system.

Graphing gives you a nice visual, but it's not always precise. If the intersection lands at something like $(1.3, 2.7)$ , you might not read it accurately off the graph. That's where algebraic methods come in.

Substitution and Elimination Methods

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

Solve one equation for one variable. For example, if $2x + y = 7$ , rewrite as $y = 7 - 2x$ .
Substitute that expression into the other equation. This gives you a single equation with one variable.
Solve for that variable.
Plug the value back into the expression from Step 1 to find the other variable.

Example: Solve $x + y = 5$ and $2x - y = 1$ .

From the first equation: $y = 5 - x$ . Substitute into the second: $2x - (5 - x) = 1$ , so $3x - 5 = 1$ , giving $x = 2$ . Then $y = 5 - 2 = 3$ . Solution: $(2, 3)$ .

Elimination (also called the addition method) works by making the coefficients of one variable opposites so they cancel when you add the equations.

Multiply one or both equations by constants so that one variable has coefficients that are equal in magnitude but opposite in sign.
Add the two equations together. One variable drops out.
Solve for the remaining variable.
Substitute back into either original equation to find the other variable.

Example: Solve $3x + 2y = 12$ and $x - 2y = 4$ .

The $y$ -coefficients are already $+2$ and $-2$ . Add the equations: $4x = 16$ , so $x = 4$ . Substitute into the second equation: $4 - 2y = 4$ , giving $y = 0$ . Solution: $(4, 0)$ .

A practical tip: use substitution when a variable already has a coefficient of 1 or $-1$ . Use elimination when both equations have similar structure and neither variable is easy to isolate.

Graphing solutions of linear systems, Solve a system of linear equations | College Algebra

Consistency of Linear Systems

When you solve a system algebraically, the algebra itself tells you what type of system you have:

If you get a specific value for each variable (like $x = 3, y = -1$ ), the system is consistent and independent with exactly one solution.
If you reach a contradiction (like $0 = 5$ ), the system is inconsistent with no solution. The lines are parallel.
If you reach an identity (like $0 = 0$ ), the system is dependent with infinitely many solutions. The equations describe the same line.

Solutions for Dependent Systems

When a system is dependent, every point on the line is a solution. You express this by writing the solution set in terms of a parameter.

Typically, you let $x = t$ (where $t$ is any real number) and then express $y$ in terms of $t$ using one of the equations. For instance, if both equations simplify to $y = 2x - 1$ , the solution set is $(t,\, 2t - 1)$ for all real numbers $t$ .

Graphing solutions of linear systems, Solve Systems of Linear Equations with Two Variables – Intermediate Algebra

Real-World Applications of Linear Systems

Word problems with two unknowns are where systems really earn their keep. Here's a reliable approach:

Identify the unknowns and assign each a variable.
Write two equations based on the relationships described in the problem.
Solve the system using whichever method fits best.
Interpret and check. Make sure the answer makes sense in context (no negative prices, no fractional people, etc.).

Example: A theater sells adult tickets for $8 and child tickets for $5. One night, 200 tickets were sold for a total of $1,300. How many of each type were sold? Let $a$ = adult tickets and $c$ = child tickets. Equation 1: $a + c = 200$ Equation 2: $8a + 5c = 1300$ Solving (substitution or elimination) gives $a = 100$ and $c = 100$ .

If a real-world system has no solution, it means the given conditions are contradictory. If it has infinitely many solutions, there isn't enough information to pin down a unique answer.

Matrix Methods for Solving Systems

A system of linear equations can be written compactly as an augmented matrix, where each row represents one equation and the columns hold the coefficients and constants.

For the system $3x + 2y = 12$ and $x - 2y = 4$ , the augmented matrix is:

$\begin{bmatrix} 3 & 2 & | & 12 \\ 1 & -2 & | & 4 \end{bmatrix}$

Gaussian elimination uses row operations (swapping rows, multiplying a row by a nonzero constant, adding a multiple of one row to another) to transform this matrix into row echelon form, where you can read off the solution by back-substitution. This method becomes especially useful for larger systems with three or more variables.

📈College Algebra Unit 11 Review