Systems of Linear Equations: Chapter Review

Vocabulary and notation

Linear equation in two variables, standard form $ax + by = c$

$x$ and $y$ each appear only to the first power: never multiplied together, squared, under a root, or in a denominator, with $a$ and $b$ not both zero.

Ordered pair solution $(x, y)$

A pair making the equation true with the first number put in for $x$ and the second for $y$. Order is part of the claim: $(3, 2)$ and $(2, 3)$ are different pairs.

System of two linear equations

Two equations that must hold at the same time. A solution is a pair satisfying both, a point lying on both lines. A system has exactly one such pair, none, or infinitely many.

Substitution

Replacing a variable in one equation by an equal expression from the other, turning the system into a single one-variable equation.

Back-substitution

Putting a value you just found into an earlier equation or expression, to recover the variable you set aside.

Elimination

Adding or subtracting the two equations, after scaling if needed, so one variable cancels and a single one-variable equation is left.

Independent conditions

Equations that each say something genuinely new. With two, dependence means one is a nonzero multiple of the other; with three, an equation can also repeat a combination of the other two while being a multiple of neither. Pinning down $n$ unknowns takes $n$ independent conditions that do not contradict each other; with fewer, a system has no solution or infinitely many, never exactly one.

Inconsistent system

A system with no solution: no pair (or triple) satisfies every equation at once. With two equations in two variables the lines are parallel and never meet.

Equations describing the same line

One whole equation is a nonzero multiple of the other. Every point of that single line solves the system, so there are infinitely many solutions.

Free variable

In a system with infinitely many solutions, a variable set equal to a parameter $t$ so the rest can be written in terms of it: $(t - 2,\ 8 - 2t,\ t)$.

Linear equation in three variables, standard form $ax + by + cz = d$

Each variable appears only to the first power. Solutions are ordered triples $(x, y, z)$, and three independent equations pin three unknowns down.

Plane

The graph of $ax + by + cz = d$, a flat sheet stretching through space. Three planes in general position meet at the single point that solves the system; planes with no common point give no solution, and planes sharing a whole line or coinciding give infinitely many.

Formulas and theorems

• Generating the solutions of one two-variable equation

  $$y = \frac{c - ax}{b}$$

  Use when From $ax + by = c$ with $b \neq 0$; if $b = 0$ then $a \neq 0$ and you solve for $x$ instead. Every value fed in returns one partner value, so one such equation has infinitely many solutions.

  e.g. $4x + y = 9$ gives $y = 9 - 4x$, so $x = 2$ yields the pair $(2, 1)$.

• The combination elimination is allowed to make

  $$\begin{gathered} m(a_1x + b_1y) + n(a_2x + b_2y) \\ = mc_1 + nc_2 \end{gathered}$$

  Use when Holds for any $m$ and $n$ and any pair satisfying both originals. Keep one original beside the combination, and use $n \neq 0$ on the equation you replace, or the step cannot be undone.

  e.g. $3$ times $2x + 3y = 7$ plus $-2$ times $3x + 2y = 8$ gives $5y = 5$.

• Which method, and add or subtract?

  Substitution when a variable is already isolated or has coefficient $1$ or $-1$; elimination when both equations sit in standard form. Then add when a variable's coefficients are opposites, subtract when they are equal.

  Use when Neither opposite nor equal? Multiply one or both equations by nonzero constants first, scaling that variable's coefficients to their least common multiple. Both methods return the same pair, so the choice is only about which is less work.

• What a vanished variable means

  The variable terms all cancel and leave $0 = k$ with $k \neq 0$: no solution. They leave $0 = 0$: that equation was redundant. Something survives that pins a variable: exactly one solution.

  Use when Read this only after combining the two different originals. With TWO equations, a $0 = 0$ means infinitely many. With THREE or more it means only that one equation was redundant, and the equations still standing decide, all three counts included: $x + y + z = 1$ with $2x + 2y + 2z = 2$ collapses to $0 = 0$ and leaves infinitely many, adding $x + y + z = 5$ gives $0 = 4$ and leaves none, while $x = 1$ with $y = 2$ and $x + y = 3$ collapses to $0 = 0$ yet still pins the single pair $(1, 2)$. A linear system has exactly one solution, none, or infinitely many; no other count is possible.

• Reading the case off the coefficients

  For $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$: if no $k$ makes the second left side $k$ times the first, the lines cross once (exactly one solution). If one does, with $k \neq 0$, then $c_2 = kc_1$ gives one shared line (infinitely many) and $c_2 \neq kc_1$ gives parallel lines (no solution).

  Text description

  Two lines crossing at one point, two parallel lines, and two equations drawing one shared line.

  Use when Both equations in standard form, with $a$ and $b$ not both zero in each.

  e.g. $2x - 6y = 4$ is $2$ times $x - 3y = 2$, one line; but $4x + 2y = 3$ against $2x + y = 5$ doubles the left side and not the $5$, so no solution.

• Count and value model

  $$x + y = N, \qquad px + qy = V$$

  Use when $x$ and $y$ count the two kinds of item, $p$ and $q$ their unit values, $N$ the total count, $V$ the total value. Keep the value equation in one unit, all cents or all dollars.

  e.g. $32$ coins worth $250$ cents: $n + d = 32$ and $5n + 10d = 250$, giving $14$ nickels and $18$ dimes.

• Mixture model

  $$x + y = T, \qquad rx + sy = mT$$

  Use when $x$ and $y$ are the amounts of the two ingredients, $r$ and $s$ their strengths and $m$ the blend's, all written the same way (all decimals or all percents). One equation adds amounts, the other content.

  e.g. $12$ liters of $30\%$ acid from $20\%$ and $50\%$: $x + y = 12$ and $0.20x + 0.50y = 3.6$, so $x = 8$ and $y = 4$.

• Distance with a current or wind

  $$\begin{gathered} d = rt \\ \text{downstream:}\ r = b + c \\ \text{upstream:}\ r = b - c \end{gathered}$$

  Use when $b$ is the speed in still water or still air and $c$ the speed of the current or wind, with $b > c$ for the upstream leg to make headway. One equation per leg, with that leg's own distance and time.

  e.g. $36$ miles downstream in $2$ hours and back in $3$ hours: $b + c = 18$ and $b - c = 12$, so $b = 15$ and $c = 3$.

• Break-even

  $$\begin{gathered} \text{cost:}\ y = vx + f \\ \text{revenue:}\ y = px \\ \text{break-even:}\ px = vx + f \end{gathered}$$

  Text description

  A revenue line steeper than a cost line that starts at the fixed cost f, crossing it exactly once.

  Use when $x$ items, $f$ the fixed cost, $v$ the cost per item, $p$ the price per item, with $f > 0$. A break-even then needs $p > v$: if $p = v$ the lines are parallel and never meet, and if $p < v$ the crossing sits at a negative count.

  e.g. Fixed cost $900$, $6$ per unit, sold at $15$: $15x = 6x + 900$, so $x = 100$ units and both sides are $1500$ dollars.

Problem types, step by step

Exam traps

• Trap Assuming two equations in two unknowns must cross at one point, so every system has a single answer.

  Fix Only when neither left side is a multiple of the other. A restatement of the same line gives infinitely many solutions, a parallel line none.

• Trap Reading a vanished variable as a value: taking $0 = 7$ to mean $x = 0$, or treating it as proof of an arithmetic slip.

  Fix $0 = k$ with $k \neq 0$ is the finished answer, no solution. $0 = 0$ says that equation was redundant: with two equations that leaves infinitely many, but with three or more the equations still standing decide. Neither says a variable equals zero.

• Trap Substituting an expression back into the equation it came from and reading the resulting $6 = 6$ as infinitely many solutions.

  Fix That identity is an artifact of the move, not a fact about the system. Substitute into the other equation: only a $0 = 0$ from combining the two different originals means infinitely many.

• Trap Solving for one variable and stopping there, reporting $x = 4$ as the answer.

  Fix A system's answer is the whole pair $(x, y)$, or triple $(x, y, z)$. Back-substitute for the ones you set aside, then check them in every original equation, not only the ones you combined.

• Trap Subtracting one equation from another without distributing the minus to every term.

  Fix Subtracting flips every sign: $(3x - y) - (x - y) = 2x$, not $2x - 2y$.

• Trap Adding two equations when the target variable's coefficients are neither opposite nor equal, hoping it cancels.

  Fix Adding cancels only opposites, subtracting only equals. Add $2x + 3y = 7$ to $3x + 2y = 8$ and you get $5x + 5y = 15$, with both variables still present. Scale to the least common multiple first, then combine.

• Trap Checking a word-problem answer only by putting it back into your own two equations.

  Fix That tests the arithmetic, never the translation. A difference written backwards solves its own system perfectly, so read the answer back into the problem's sentences.

• Trap In a mixture problem, writing the amount equation twice: totalling amounts again where the content belongs.

  Fix Two separate columns: $x + y = T$ adds volumes, $rx + sy = mT$ adds acid, salt, or value. The blend's content is its strength times the total amount, never the sum of the strengths.

• Trap In a three-variable system, cancelling a different variable in the second pair, or reusing a pair you have already combined.

  Fix Cancel the same variable both times, involving all three equations. Reusing a pair discards a fact and can manufacture a false "infinitely many".