The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 7 (Intermediate Algebra) Solving Systems of Equations

Overview

This lecture moves from solving a single linear equation to solving systems of equations — two (or more) equations that must be satisfied by the same values at once. Geometrically, each linear equation is a line, and a solution to the system is a point that lies on every line. The lecture develops three methods for finding that point: graphing, substitution, and elimination. It also classifies the three possible outcomes for a linear system — one solution, no solution, or infinitely many solutions — and connects each outcome to how the lines sit relative to one another. The same ideas extend to systems of three equations in three unknowns, where the algebra grows but no new concept is introduced.

Key Concepts

System of equations — a set of two or more equations whose solution is the set of variable values that satisfy all of them simultaneously. For two equations in $x$ and $y$ , a solution is an ordered pair $(x, y)$ .
Geometric meaning — a linear equation in slope-intercept form $y = mx + b$ graphs as a line, where $m$ is the slope and $b$ is the $y$ -intercept. Every point on the line is a solution of that single equation. The solution of a system is the point (or points) lying on every line at once — i.e. the point of intersection.

Three solution methods

Graphing — put each equation in slope-intercept form, graph both lines, and read off the intersection point. Always verify by substituting the point back into both original equations. Accurate but tedious, so it is rarely the method of choice.
Substitution — solve one equation for one variable, then substitute that expression into the other equation. This collapses two equations in two unknowns into one equation in one unknown, which you solve normally; back-substitute to recover the second variable. This is the most standard, always-works method.
- Watch the circular-reasoning trap: substitute the isolated expression into the equation you did not use to isolate it. Substituting back into the same equation gives a useless identity.
Elimination — multiply one or both equations by constants so that adding the equations cancels (eliminates) one variable. Adding equals to equals keeps a true equation, so this is valid. Best when the equations are arranged so a variable cancels cleanly; it is the "clever" method.

The three possible outcomes

One solution — the lines have different slopes and cross at exactly one point. The algebra yields a definite value such as $x = 2$ .
No solution — the lines are parallel (same slope, different intercepts) and never cross. The algebra collapses to a false statement such as $-15 = 9$ . A statement that is never true means no solution.
Infinitely many solutions — the two equations describe the same line. The algebra collapses to a statement that is always true such as $0 = 0$ or $-8 = -8$ . Every point on the line works.
Three equations, three unknowns — solve one equation for one variable, substitute to reduce to two equations in two unknowns, then solve that smaller system as above. Same concept, messier arithmetic.

Worked Example

Solve the system by substitution:

\begin{aligned} 2x - 3y &= 7, \qquad y = 3x - 7 \\ 2x - 3(3x - 7) &= 7 \\ 2x - 9x + 21 &= 7 \\ -7x &= -14 \\ x &= 2 \\ y &= 3(2) - 7 = -1 \end{aligned}

The solution is $(2, -1)$ . Check in the first equation: $2(2) - 3(-1) = 4 + 3 = 7$ . ✓

Solve the system by elimination:

\begin{aligned} 3x - 4y &= 8, \qquad 5x + 4y = -4 \\ (3x - 4y) + (5x + 4y) &= 8 + (-4) \\ 8x &= 4 \\ x &= \tfrac{1}{2} \\ 3\left(\tfrac{1}{2}\right) - 4y &= 8 \;\Rightarrow\; -4y = \tfrac{13}{2} \;\Rightarrow\; y = -\tfrac{13}{8} \end{aligned}

The $+4y$ and $-4y$ terms cancel on addition, leaving one equation in $x$ alone.

Summary

A solution to a system is a set of values satisfying every equation at once; for two lines it is their point of intersection.
Graphing reads the intersection off a plot, substitution isolates a variable and plugs it into the other equation, and elimination adds (suitably scaled) equations to cancel a variable.
Substitution always works; elimination is fastest when a variable already cancels or can be made to cancel by multiplying through.
A linear system has exactly one of three outcomes: one solution (lines cross), no solution (parallel lines), or infinitely many solutions (same line).
A false final statement ( $-15 = 9$ ) signals no solution; an always-true statement ( $0 = 0$ ) signals infinitely many solutions.
Systems of three equations in three unknowns use the same reduction strategy — eliminate one variable to drop to a two-variable system, then solve that.