Systems of two linear equations and what a solution means, solving by graphing and checking in both equations, systems with infinitely many solutions or none, the consistent and independent vocabulary, and modelling a comparison with a system.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Solve Linear Systems by Graphing
Objectives
Five outcomes. The second is the one that separates a guess read off a graph from an answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 153-159 — the lesson these objectives are drawn from
Warm-up
Lesson 2.3 graphed one line and Lesson 2.8 shaded one region. This lesson asks two conditions at once.
Discussion prompt
The point (3, -4) lies on the line 4x plus y equals 8. Does that make it the answer to anything? What extra information would you need before calling it a solution of a system?
Hint: One line has infinitely many points on it.
Answer:
\[ 4(3) + (-4) = 12 - 4 = 8 \quad \checkmark \]
Not yet — every point on that line satisfies the first equation, so satisfying it is no distinction at all. A solution of a system has to satisfy a second equation too, and that is what narrows infinitely many points down to one.
Concept
A system of two linear equations asks for the ordered pairs that make both statements true. Graphically those are the points lying on both lines, which is why a solution appears as an intersection.
system of two linear equations — Two linear equations in the same two variables, considered together. A solution is an ordered pair that satisfies each of them.
\[ \begin{cases} Ax + By = C \\ Dx + Ey = F \end{cases} \]
Two lines in a plane can cross once, lie on top of each other, or never meet — so a system has exactly one solution, infinitely many, or none. There is no other possibility.
Figure (svg): Two lines crossing at a single point, with that point marked as the solution of the system
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 153-153
Section
Section 1
Concept
Graphing each equation and finding the intersection gives an estimate of the solution. It is an estimate because a point read off a drawing is only as accurate as the drawing.
solution of a system — An ordered pair that satisfies every equation in the system. It corresponds to a point where all the graphs intersect.
\[ \begin{cases} 4x + y = 8 \\ 2x - 3y = 18 \end{cases} \]
Both equations are in standard form, so the intercept method from Lesson 2.3 draws each of them quickly.
Figure (svg): Two lines crossing at a single point, with that point marked as the solution of the system
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 153-153 — Solve a system graphically
Picture it
Example 1: the lines 4x plus y equals 8 and 2x minus 3y equals 18.
Figure (svg): Two lines crossing at a single point, with that point marked as the solution of the system
Every other point of the plane fails at least one equation. The intersection is the only place both conditions hold at once, and that is what makes it the solution.
Worked example
Example 1, the graphing half. The checking half is the next worked example.
\[ \begin{cases} 4x + y = 8 \\ 2x - 3y = 18 \end{cases} \quad \text{Graph and estimate the solution.} \]
Graph the first equation by intercepts
Why: Setting y to zero gives x equal to 2; setting x to zero gives y equal to 8.
\[ (2, 0)\text{ and } (0, 8) \]
Graph the second the same way
Why: Setting y to zero gives x equal to 9; setting x to zero gives y equal to negative 6.
\[ (9, 0)\text{ and } (0, -6) \]
Find where the two lines cross
Why: They appear to meet at the point (3, -4).
\[ \text{estimate } (3, -4) \]
Call it an estimate until it is checked
Why: A point read off a drawing could easily be (3.1, -3.9) instead, so the algebra has to confirm it.
Figure (svg): The solution to Worked example graph and estimate shown as a ladder of expressions, one row per algebraic move
\[ \text{the lines appear to meet at } (3, -4) \]
Verify: confirm both intercept pairs lie on their own lines
Why: For the first line, 4 times 2 plus 0 is 8, and 0 plus 8 is 8 — both correct. For the second, 2 times 9 minus 0 is 18, and 0 minus 3 times negative 6 is 18 — both correct. Confirming the points used to draw each line means any error in the estimate comes from reading the crossing, not from the lines themselves.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 153-153
Sorting
Test each pair against BOTH 4x plus y equals 8 and 2x minus 3y equals 18.
Sort into buckets
Sort each ordered pair.
Four of these five points are on one of the lines, which is exactly why passing a single equation proves so little.
Worked example
Example 1's check. The textbook flags this in an Avoid Errors note for a reason.
\[ \text{Verify that } (3, -4) \text{ solves both } 4x + y = 8 \text{ and } 2x - 3y = 18. \]
Substitute into the first equation
Why: Four times three is twelve, plus negative four is eight.
\[ 12 - 4 = 8 \]
Compare with the first right side
Why: Eight equals eight, so the first equation holds.
\[ 8 = 8,\text{ checks} \]
Substitute into the second equation
Why: Two times three is six; minus three times negative four is plus twelve.
\[ 6 + 12 = 18 \]
Compare with the second right side
Why: Eighteen equals eighteen, so the second holds too.
\[ 18 = 18,\text{ checks} \]
Figure (svg): The solution to Worked example check in both equations shown as a ladder of expressions, one row per algebraic move
\[ (3, -4) \quad \text{satisfies both equations} \]
Verify: consider what a single check would have proved
Why: Satisfying only the first equation would place the point somewhere on that line, which infinitely many points do. It is passing BOTH that pins it to the single intersection, which is why checking one equation and stopping proves nothing about a system.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 153-153
Trap
\[ \text{Is } (2, 0) \text{ the solution of } 4x + y = 8, \; 2x - 3y = 18? \]
Substitute into the first equation and stop
Why: The pair satisfies the equation that was tested, and the test ends there.
\[ 4(2) + 0 = 8 \quad \checkmark \]
But the second equation gives 4, not 18. The point is on one line and not on the other, so it solves the first equation and not the system.
\[ \text{Is } (2, 0) \text{ the solution of } 4x + y = 8, \; 2x - 3y = 18? \]
Substitute into EVERY equation before concluding anything
Why: A solution of a system must satisfy all of it, so one success is never enough.
\[ 2(2) - 3(0) = 4 \neq 18 \quad \times \]
The correct solution, (3,-4), passes both. Any candidate that passes one and fails the other is simply a point on one of the two lines.
Prediction
Commit before reasoning.
Predict first
You read an intersection off a hand-drawn graph as (3, -4). How confident should you be?
Correct: Confident only after substituting into both equations.
This is also why Lesson 3.2 exists. Graphing is excellent for seeing WHAT is going on and poor at producing exact answers, especially when the solution has fractional coordinates.
Why: A graph locates the solution but cannot certify it: the true intersection might be at (3.02, -4.03) and look identical at that scale. Substituting is exact arithmetic and settles it. Careful drawing improves the estimate but never turns it into proof, which is why the textbook's procedure ends with an algebraic check rather than with the graph.
Fill the middle
Guided Practice 1: the system 3x plus 2y equals negative 4 and x plus 3y equals 1.
Fill in the blanks
\text-4 (-2, 1): \quad 3(-2) + 2(1) = ___ \quad \text___ \quad -2 + 3(1) = 1
Why: Negative six plus two is negative four, matching the first right side, and negative two plus three is one, matching the second. Both equations hold, so (-2, 1) is the solution. Note that the second check is the easier of the two, which is a reason to do it first when you are testing a candidate quickly.
Explain it to yourself
The link between algebra and geometry here is worth stating.
\[ \begin{cases} 4x + y = 8 \\ 2x - 3y = 18 \end{cases} \]
Discussion prompt
Explain why the solution of a system appears as a point where the graphs cross. What does each line represent, and what is special about a point belonging to both?
Hint: Ask what the graph of a single equation contains.
Answer:
The graph of an equation is the set of ALL points satisfying it — that was Lesson 2.1's definition. So the first line is every solution of the first equation, and the second line is every solution of the second.
A point on both lines therefore satisfies both equations, which is exactly the definition of a solution of the system. The intersection is not a coincidence of the picture; it is the definition drawn.
Section
Section 2
Concept
If two equations reduce to the same line, every point on it satisfies both, so the system has infinitely many solutions. The second equation contributed nothing the first did not already say.
\[ \begin{cases} 4x - 3y = 8 \\ 8x - 6y = 16 \end{cases} \]
The quickest test is to put both into slope-intercept form: identical slope and identical intercept means one line. Here the second equation is just the first multiplied by two.
Figure (svg): Two equations written in slope-intercept form turning out identical, so their graphs coincide
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 154-154 — Solve a system with many solutions
Picture it
Example 2: 4x minus 3y equals 8, and 8x minus 6y equals 16.
Figure (svg): Two equations written in slope-intercept form turning out identical, so their graphs coincide
Doubling every term of an equation produces a different-looking equation with exactly the same solutions. Spotting that multiple is often faster than solving for y.
Worked example
Example 2. The check the textbook suggests is to compare slope-intercept forms.
\[ \begin{cases} 4x - 3y = 8 \\ 8x - 6y = 16 \end{cases} \quad \text{Solve and classify.} \]
Solve the first equation for y
Why: Subtracting 4x and dividing by negative 3 gives four thirds x minus eight thirds.
\[ y = (\frac{4}{3}) x - \frac{8}{3} \]
Solve the second equation for y
Why: Subtracting 8x and dividing by negative 6 gives the same thing.
\[ y = (\frac{4}{3}) x - \frac{8}{3} \]
Compare the two forms
Why: Identical slope and identical intercept, so the graphs are the same line.
State the conclusion
Why: Every point on the line satisfies both equations, so there are infinitely many solutions.
Figure (svg): The solution to Worked example infinitely many solutions shown as a ladder of expressions, one row per algebraic move
\[ \text{infinitely many solutions: consistent and dependent} \]
Verify: test two different points on the line
Why: Take (2, 0): 4 times 2 minus 0 is 8, and 8 times 2 minus 0 is 16 — both check. Take (5, 4): 20 minus 12 is 8, and 40 minus 24 is 16 — both check as well. Two different solutions is already proof that there is not exactly one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 154-154
Discrimination
Compare the three numbers of each equation.
Sort into buckets
Sort each system by whether one equation is a constant multiple of the other.
Worked example
Guided Practice 4. Faster than solving for y, once you look for it.
\[ \begin{cases} 2x + 5y = 6 \\ 4x + 10y = 12 \end{cases} \quad \text{Solve and classify.} \]
Compare the coefficients term by term
Why: Four is twice two, ten is twice five, and twelve is twice six.
Conclude the second is a multiple of the first
Why: Multiplying an equation by a nonzero constant does not change its solutions — that is the multiplication property of equality from Lesson 1.3.
Say what that means for the graphs
Why: Two names for one line, so the graphs coincide.
Classify
Why: At least one solution makes it consistent; infinitely many makes it dependent.
Figure (svg): The solution to Worked example spot the multiple shown as a ladder of expressions, one row per algebraic move
\[ \text{consistent and dependent} \]
Verify: check the constant scales too
Why: The coefficients scale by two and so does the constant, which is what makes it the same equation. Had the constant been 13 instead of 12, the coefficients would still double but the constant would not — giving parallel lines and no solution, which is the next section's case.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 155-155
Trap
\[ \begin{cases} 4x - 3y = 8 \\ 8x - 6y = 16 \end{cases} \]
Eliminate x and see everything vanish, then report no solution
Why: A result with no variables left is read as a failure.
\[ -8x + 6y = -16 \;\text{ plus }\; 8x - 6y = 16 \;\Longrightarrow\; 0 = 0 \]
But 0 equals 0 is TRUE, and a true statement means every candidate works, not that none does.
\[ \begin{cases} 4x - 3y = 8 \\ 8x - 6y = 16 \end{cases} \]
Read the vanished result as a statement and judge whether it is true
Why: Zero equals zero is true for every x and y, so every point on the line is a solution.
\[ 0 = 0 \;\Longrightarrow\; \text{infinitely many solutions} \]
Compare a result of 0 equals 5, which is false for every x and y and therefore means NO solution. The variables vanish in both cases; only the truth of what is left differs.
Prediction
Commit before reasoning.
Predict first
Eliminating a variable leaves the statement 0 equals 0. What does the system have?
Correct: Infinitely many solutions.
\[ 0 = 0 \;\Rightarrow\; \text{infinitely many} \qquad 0 = 5 \;\Rightarrow\; \text{none} \]
Why: A statement with no variables left is either true or false for every possible pair. Zero equals zero is true always, so every point satisfying one equation satisfies the other, and the graphs coincide. Compare 0 equals 5, which is false always and therefore means no solution. The variables vanishing tells you the system is degenerate; whether the leftover statement is true tells you which kind.
Fill the middle
Example 2's check.
Fill in the blanks
8x - 6y = 16 \;\Longrightarrow\; -6y = -8x + 16 \;\Longrightarrow\; y = \frac{4}{3}x - \frac{8}{3}
Why: Negative eight over negative six reduces to four thirds, and sixteen over negative six reduces to negative eight thirds. That is exactly the slope-intercept form of the first equation, which is what makes the graphs identical. Note that both signs flipped because the divisor was negative, as Lesson 1.4 warned.
Counterexample
A classmate offers a rule about systems.
\[ \text{two different-looking equations always give two different lines} \]
Discussion prompt
Find two equations that look different and describe the same line, then say what operation turns one into the other and why that operation cannot change the solutions.
Hint: Multiply one equation through by something.
Answer:
\[ 4x - 3y = 8 \qquad \text{and} \qquad 8x - 6y = 16 \]
The second is the first multiplied through by 2. The multiplication property of equality from Lesson 1.3 says multiplying both sides of an equation by a nonzero constant produces an EQUIVALENT equation — one with exactly the same solutions.
So looking different and being different are not the same thing, and the fastest way to tell is to reduce both to slope-intercept form, where equivalent equations become literally identical.
Section
Section 3
Concept
If the two lines have the same slope but different intercepts, they never meet, so no ordered pair satisfies both equations. The system contradicts itself.
\[ \begin{cases} 2x + y = 4 \\ 2x + y = 1 \end{cases} \]
The two equations here claim that the same expression equals both 4 and 1, which nothing can. The parallel graphs are that contradiction drawn.
Figure (svg): Two equations with the same slope but different intercepts, graphed as parallel lines that never meet
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 154-154 — Solve a system with no solution
Picture it
Example 3: 2x plus y equals 4, and 2x plus y equals 1.
Figure (svg): Two equations with the same slope but different intercepts, graphed as parallel lines that never meet
Both lines have slope negative two, so they stay exactly three units apart for ever. There is no x at which they could meet.
Worked example
Example 3, with the slope-intercept check the textbook recommends.
\[ \begin{cases} 2x + y = 4 \\ 2x + y = 1 \end{cases} \quad \text{Solve and classify.} \]
Solve each for y
Why: Subtracting 2x from each gives negative two x plus four, and negative two x plus one.
\[ y = -2 x + 4, y = -2 x + 1 \]
Compare the slopes
Why: Both are negative two, so the lines are equally steep.
Compare the intercepts
Why: Four and one, which differ, so the lines are at different heights.
State the conclusion
Why: Equal slopes with different intercepts means parallel and distinct, so they never meet.
Figure (svg): The solution to Worked example no solution shown as a ladder of expressions, one row per algebraic move
\[ \text{no solution: the system is inconsistent} \]
Verify: show the contradiction directly
Why: The left sides are identical, so the system claims 2x plus y is both 4 and 1 at once. Subtracting the two equations gives 0 equals 3, which is false for every x and y — an algebraic version of the same fact the parallel graphs show.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 154-154
Comparison
Fill the blanks. Slopes and intercepts decide everything.
Comparison matrix
| Slopes | Intercepts | Result |
|---|---|---|
| different | anything | exactly one solution |
| equal | different | no solution |
| equal | equal | infinitely many solutions |
| different | equal | one solution, at the shared intercept |
The last row is worth noticing: two lines with different slopes that share an intercept still meet exactly once — at that intercept. Different slopes always give exactly one solution, whatever the intercepts do.
Worked example
Guided Practice 6. One equation is in standard form and one in slope-intercept form.
\[ \begin{cases} -2x + y = 5 \\ y = 2x + 2 \end{cases} \quad \text{Solve and classify.} \]
Put the first into slope-intercept form
Why: Adding 2x to both sides gives y equals 2x plus 5.
\[ y = 2 x + 5 \]
Read both slopes
Why: Two and two — equal.
\[ \text{same slope } 2 \]
Read both intercepts
Why: Five and two — different.
\[ \text{intercepts } 5\text{ and } 2 \]
Classify
Why: Parallel and distinct, so no solution and the system is inconsistent.
Figure (svg): The solution to Worked example a disguised parallel pair shown as a ladder of expressions, one row per algebraic move
\[ \text{no solution: inconsistent} \]
Verify: substitute a candidate and watch it fail
Why: Try x equal to 0: the first equation needs y equal to 5 and the second needs y equal to 2. No single y can be both, and the same conflict appears at every x — the two required values always differ by exactly 3, which is the vertical gap between the parallel lines.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 155-155
Error analysis
A student graphs 3x minus 2y equals 10 and 3x minus 2y equals 2 and reads off an intersection.
Annotate
On: \( \text{the lines seem to converge at the right of the page, so the solution is about } (12, 13) \)
Whenever a graph suggests an intersection far off the drawn region, compare the two slopes algebraically before believing it. Equal slopes settle the question without any drawing.
Sorting
Compare slopes and intercepts without graphing.
Sort into buckets
Sort each system by its number of solutions.
Comparing slopes first is the fastest route: different slopes means one solution and you can stop there.
Prediction
Commit before reasoning.
Predict first
Subtracting one equation from another leaves the statement 0 equals 3. What does the system have?
Correct: No solution.
\[ 2x + y = 4 \;\text{ minus }\; 2x + y = 1 \;\Longrightarrow\; 0 = 3 \quad \text{false} \]
Why: The variables have vanished and what remains is false for every pair, so no pair can satisfy both original equations. The two zeros in the statement are not coordinates of anything — reading 0 equals 3 as a point is a common misreading, and substituting either candidate into the originals shows immediately that neither works. Compare 0 equals 0, which is true always and gives infinitely many solutions.
Explain it
A classmate finds it strange that a system can have no solution when each equation on its own has infinitely many.
Discussion prompt
In three sentences, explain how two equations that are each perfectly solvable can be unsolvable together. Give them a plain-English version of the contradiction.
Hint: Look at what the two equations claim about the same expression.
Answer:
Each equation on its own has infinitely many solutions — a whole line of them. A system asks for a pair satisfying both at once, and there is no reason two infinite sets have to overlap.
In plain English: 2x plus y equals 4 and 2x plus y equals 1 together claim that one number is both 4 and 1. Nothing can be, so the system asks for something impossible.
Section
Section 4
Concept
A system with at least one solution is consistent; one with none is inconsistent. A consistent system with exactly one solution is independent; one with infinitely many is dependent.
consistent and independent — A system with exactly one solution. Consistent means it has at least one; independent means it has no more than one.
Only three of the four combinations occur, because an inconsistent system has no solutions and so the second question does not arise.
Figure (svg): Three coordinate planes showing lines that cross once, lines that coincide, and lines that are parallel
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 154-154 — Classifying systems and Number of Solutions of a Linear System
Picture it
Crossing, coinciding, and parallel — with their names.
Figure (svg): Three coordinate planes showing lines that cross once, lines that coincide, and lines that are parallel
The vocabulary is worth learning because it appears in every later systems method: an inconsistent system stays inconsistent whether you solve it by graphing, substitution, elimination or matrices.
Worked example
Guided Practice 4, 5 and 6 together.
\[ \text{Classify: (a) } 2x+5y=6, \, 4x+10y=12; \; \text{(b) } 3x-2y=10, \, 3x-2y=2; \; \text{(c) } -2x+y=5, \, y=2x+2. \]
Classify the first
Why: Every term of the second is twice the first, so the graphs coincide and there are infinitely many solutions.
Classify the second
Why: Identical left sides with different right sides, so the lines are parallel and distinct.
Rewrite and classify the third
Why: The first becomes y equals 2x plus 5; the second is y equals 2x plus 2. Same slope, different intercepts.
Note the pattern
Why: Two of the three are inconsistent, and in both cases the tell was equal slopes with unequal intercepts.
Figure (svg): The solution to Worked example classify three systems shown as a ladder of expressions, one row per algebraic move
\[ \text{(a) consistent and dependent} \;\; \text{(b) inconsistent} \;\; \text{(c) inconsistent} \]
Verify: confirm each verdict a second way
Why: For (a), the point (3,0) satisfies both equations, and so does (-2,2) — two solutions is already enough to rule out exactly one. For (b) and (c), subtracting the equations gives 0 equals 8 and 0 equals 3 respectively, both false, which confirms no solution.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 155-155
Matching
Three graphs, three names.
Match the pairs
Why: The last two rows describe the same situation from the picture and from the algebra: coincident lines are exactly the case where eliminating a variable leaves a true statement with no variables. Recognising that the graphical and algebraic signatures match is what lets you classify a system by whichever route is quicker.
Worked example
Guided Practice 2 and 3, classified before being solved.
\[ \text{(a) } 4x-5y=-10, \, 2x-7y=4. \quad \text{(b) } 8x-y=8, \, 3x+2y=-16. \]
Find the slopes in the first system
Why: Four fifths and two sevenths, which are different.
\[ \frac{4}{5}\text{ and } \frac{2}{7} \]
Classify it
Why: Different slopes means the lines cross exactly once.
Find the slopes in the second system
Why: Eight and negative three halves, again different.
\[ 8\text{ and } -\frac{3}{2} \]
Classify it
Why: Different slopes again, so exactly one solution.
Figure (svg): The solution to Worked example classify from slopes alone shown as a ladder of expressions, one row per algebraic move
\[ \text{both have exactly one solution} \]
Verify: find the actual solutions
Why: The first system solves to (-5, -2): 4 times -5 minus 5 times -2 is -20 plus 10, which is -10, and 2 times -5 minus 7 times -2 is -10 plus 14, which is 4. The second solves to (0, -8): 8 times 0 minus -8 is 8, and 0 plus 2 times -8 is -16. Both check, confirming the classification made before any solving.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 153-153
Error analysis
A student classifies a system whose two equations are multiples of each other.
Annotate
On: \( \begin{cases} 2x + 5y = 6 \\ 4x + 10y = 12 \end{cases} \;\Longrightarrow\; \text{inconsistent, no solution} \)
When the variables disappear, read what is left and ask whether it is true. True means every point works; false means none does.
Definition probe
The two questions are separate: does it have a solution, and how many?
Sort into buckets
Sort each system by its classification.
Two truths and a lie
All three are about classifying systems.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: C
Why: The survivor is the false one. Dependent describes a consistent system with infinitely many solutions, while inconsistent means there are none, so the two cannot both apply. Only three combinations occur: consistent and independent, consistent and dependent, and inconsistent.
Estimation
The system 4x minus 5y equals negative 10 and 2x minus 7y equals 4.
Predict first
Before solving, how many solutions does it have?
Correct: Exactly one.
\[ \tfrac{4}{5} \neq \tfrac{2}{7} \;\Longrightarrow\; \text{one solution, at } (-5, -2) \]
Why: The slopes are four fifths and two sevenths, which are unequal, so the lines are not parallel and must cross exactly once. That verdict needs only the two slopes and takes about ten seconds — far less than solving. It is worth doing first, because knowing there is exactly one solution tells you that any method must produce a single ordered pair rather than a degenerate statement.
Section
Section 5
Concept
When two choices are described by two linear rules, the system's solution is where they agree. Each side of the intersection belongs to a different option, which is usually the real question.
\[ \begin{cases} y = 0.75x + 25 \\ y = 2x \end{cases} \]
As in Lesson 2.8, quantities that cannot be negative mean only the first quadrant is worth drawing.
Figure (svg): Two cost lines for bus fare options crossing at twenty rides, with only the first quadrant drawn
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 155-155 — the bus fare example
Picture it
Example 4: a 25 dollar pass at 0.75 a ride, against 2 dollars a ride with no pass.
Figure (svg): Two cost lines for bus fare options crossing at twenty rides, with only the first quadrant drawn
Twenty rides is where they cost the same. Below that the pay-per-ride option wins because the pass has not paid for itself; above it the pass does.
Worked example
Example 4. Two verbal models, then a system.
\[ \text{Option A: } \$25 \text{ pass plus } \$0.75 \text{ a ride. Option B: } \$2 \text{ a ride. When are they equal?} \]
Write the first verbal model
Why: Total cost equals cost per ride times rides plus the monthly fee, all in dollars.
\[ y = 0.75 x + 25 \]
Write the second
Why: Total cost equals cost per ride times rides, with no fee.
\[ y = 2 x \]
Graph both in the first quadrant only
Why: Negative rides and negative costs make no sense, so nothing outside it is worth drawing.
Read the intersection and check it
Why: The lines meet at (20, 40). Substituting: 0.75 times 20 plus 25 is 40, and 2 times 20 is also 40.
\[ (20, 40)\text{ checks} \]
Figure (svg): The solution to Worked example the bus fare comparison shown as a ladder of expressions, one row per algebraic move
\[ (20, 40): \text{ equal after } 20 \text{ rides} \]
Verify: test a value on each side of the crossing
Why: At 10 rides, Option A costs 32.50 and Option B costs 20, so B is cheaper. At 30 rides, A costs 47.50 and B costs 60, so A is cheaper. The crossing really does separate the two regimes, which is the information the question was actually after.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 155-155
Real world
Gym A charges a 60 dollar joining fee plus 20 dollars a month. Gym B charges 35 dollars a month with no fee.
Discussion prompt
Write the system, find where the costs are equal, and say which gym is cheaper for a six-month membership and which for a two-year one. How could you have answered the last part without solving anything?
Hint: Compare the two starting costs and the two rates.
Answer:
\[ \begin{cases} y = 20m + 60 \\ y = 35m \end{cases} \;\Longrightarrow\; 15m = 60 \;\Longrightarrow\; m = 4 \]
They cost the same at 4 months, at 140 dollars. So Gym B is cheaper below 4 months and Gym A above it: six months favours A, and two years favours A by a wide margin.
Without solving: A starts higher because of the fee but rises more slowly, so it must start behind and end ahead. The starting gap and the difference in rates settle the direction, exactly as they did for the phone plans in Lesson 1.3.
Worked example
Guided Practice 7. The pass price rises to 36 dollars.
\[ \text{With a } \$36 \text{ pass instead, when do the two options cost the same?} \]
Rewrite the first equation
Why: Only the constant changes; the per-ride rate is unchanged.
\[ y = 0.75 x + 36 \]
Set the two costs equal
Why: The intersection is where 0.75x plus 36 equals 2x.
\[ 0.75 x + 36 = 2 x \]
Solve for x
Why: Subtracting 0.75x gives 36 equals 1.25x, so x is 28.8.
\[ x = 28.8 \]
Interpret in the situation
Why: Rides come whole, so the pass becomes worthwhile from the 29th ride onward.
\[ 29\text{ rides} \]
Figure (svg): The solution to Worked example change one number shown as a ladder of expressions, one row per algebraic move
\[ x = 28.8 \;\Longrightarrow\; \text{the pass wins from } 29 \text{ rides} \]
Verify: compare the two answers and explain the shift
Why: A more expensive pass takes longer to pay for itself, so the crossing must move right — and it does, from 20 rides to 28.8. Checking at 29 rides: the pass costs 57.75 and pay-per-ride costs 58, so the pass is just ahead, confirming the rounding direction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 155-155
Error analysis
A student solves the bus problem and answers with the point.
Annotate
On: \( \text{the solution is } (20, 40) \)
A system's solution is an ordered pair; a word problem's answer is usually one of its coordinates, with a unit attached. Reread the question before writing the final line.
Matching
Each pair of options becomes two equations.
Match the pairs
Why: Every one has the same shape: one option with a fixed cost and a lower rate, one with no fixed cost and a higher rate. The option with the fee always starts higher and always wins eventually, and the crossing point is the fee divided by the difference in the rates — which is worth noticing as a general result rather than four separate calculations.
Edge cases
The pattern of Section 5 assumes the two rates differ.
Discussion prompt
What would happen if both options charged the same per-ride rate but one also had a monthly fee? Write such a system, classify it, and say what it means in the situation.
Hint: Same rate means same slope.
Answer:
\[ \begin{cases} y = 2x + 25 \\ y = 2x \end{cases} \;\Longrightarrow\; \text{parallel, inconsistent} \]
The two costs are never equal: the pass option is 25 dollars dearer at every number of rides, for ever. The system is inconsistent, and its parallel graphs say exactly that.
So no solution is not a failure here — it is a genuine answer to the question. One option is simply always worse, and there is no crossover to find.
Commit first
Answer, then rate your confidence honestly.
Predict first
In the bus problem, why is only the first quadrant graphed?
Correct: Because negative rides and negative costs have no meaning.
Whenever a system models something physical, ask which part of the plane the situation actually permits. The algebra will happily answer questions the situation cannot.
Why: Both lines continue into the other quadrants perfectly well as mathematical objects — extending y equals 2x to the left gives negative costs for negative rides. The restriction comes from the situation, not from the algebra, and it is the same domain judgement that ended the paramotor model in Lesson 1.5 at eight minutes. The intersection being in the first quadrant is a convenient fact rather than the reason.
Comparison
Fill the blanks. Two numbers per equation decide the whole outcome.
Comparison matrix
| What you compare | What it tells you | Classification |
|---|---|---|
| Slopes differ | lines cross once | consistent and independent |
| Slopes equal, intercepts differ | lines are parallel | inconsistent |
| Slopes equal, intercepts equal | one line drawn twice | consistent and dependent |
| Eliminating leaves 0 = 0 | a true statement | consistent and dependent |
| Eliminating leaves 0 = 3 | a false statement | inconsistent |
The last two rows are the algebraic versions of the middle two. Whichever route you take, the same three outcomes appear.
Pattern
One routine solves any two-equation linear system by graphing.
Step one can settle the whole question in ten seconds when the slopes are equal, and step four is the difference between a reading and an answer.
OpenStax Algebra and Trigonometry 2e, §11.1 Systems of Linear Equations: Two Variables §11.1
Check
A solution must satisfy both equations.
Check your understanding
Which ordered pair is the solution of the system 8x - y = 8 and 3x + 2y = -16?
Answer: A
Why: Substituting gives 8 times 0 minus -8, which is 8, matching the first equation, and 3 times 0 plus 2 times -8, which is -16, matching the second. Both hold.
Check
Classify from slopes and intercepts.
Check your understanding
How many solutions does the system 3x - 2y = 10 and 3x - 2y = 2 have?
Answer: A
Why: The left sides are identical, so both equations have slope 3/2, but the intercepts are -5 and -1. Parallel distinct lines never meet, and subtracting gives 0 = 8, which is false.
Check
A modelling question. Answer what was asked.
Check your understanding
Option A costs 25 dollars plus 0.75 per ride; Option B costs 2 dollars per ride. After how many rides are the costs equal?
Answer: A
Why: Setting 0.75x + 25 equal to 2x gives 25 = 1.25x, so x = 20. At 20 rides both options cost 40 dollars.
Real world
Two printing quotes: Shop A charges a 90 dollar setup plus 0.15 per page; Shop B charges 0.40 per page with no setup.
Discussion prompt
Write the system, find the break-even number of pages, and say which shop to use for a 200-page job and which for a 1000-page job. Then say what would have to change about the two quotes for there to be no break-even point at all.
Hint: Compare the two rates as well as the two fixed costs.
Answer:
\[ \begin{cases} y = 0.15x + 90 \\ y = 0.40x \end{cases} \;\Longrightarrow\; 0.25x = 90 \;\Longrightarrow\; x = 360 \]
They cost the same at 360 pages, at 144 dollars. Below that Shop B is cheaper, so a 200-page job goes to B; above it Shop A is, so a 1000-page job goes to A.
There would be no break-even if the two per-page rates were equal, since then the lines would be parallel: the shop with the setup fee would be dearer at every size, for ever. That is the inconsistent case, and it is a perfectly meaningful answer to a real question.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can a system of two linear equations in two variables have exactly two solutions?
Correct: No — two distinct lines meet at most once.
The count of solutions is a structural fact about lines, not an accident of the numbers. Any method you use must produce one of the three outcomes.
Why: Two points determine a line, so if two lines shared two points they would be the same line, and then they would share infinitely many. That is why the only possibilities are one, none, or infinitely many. Exactly two never occurs for linear systems, which is worth knowing because it does occur for the nonlinear systems of Lesson 9.7 — a line can cut a circle twice.
Explain it
They can graph a line but do not see why two of them matter.
Discussion prompt
In four sentences or fewer, explain what a system asks for, why the answer is where the graphs cross, and the one check they must do before believing a point read off a graph.
Hint: The check is a substitution, and it is into both equations.
Answer:
A system asks for the pair of values that makes both equations true at the same time. Since each line is the set of all points satisfying its own equation, a point on both lines satisfies both equations — which is why the answer is the crossing point.
Before believing it, substitute the point into BOTH original equations. Reading a graph is an estimate, and a point that satisfies one equation but not the other is simply on one of the lines.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For checking, substitute into the second equation first — it is the one people skip. For the degenerate cases, read what is left after the variables vanish and ask whether it is true. For the vocabulary, consistent answers whether any solution exists and independent answers whether there is only one. For modelling, write each option's total cost as a rate times a quantity plus a fixed amount. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Across the top of a page draw three coordinate planes showing lines that cross once, lines that coincide and lines that are parallel, writing under each the number of solutions and the two classification words. In the middle, take a system of your own with different slopes, graph both lines, mark the intersection, and then write out the algebraic check in BOTH equations, showing every substitution. Below that, write one system whose second equation is a multiple of the first and one whose equations differ only in their constants, and beside each write what happens when you eliminate a variable — the true statement or the false one. At the bottom of the page, invent two payment options with different fixed costs and different rates, write the system, find the crossing, and write one sentence saying which option wins on each side of it.
If your last sentence names only the crossing point and not which option wins on each side, add that: the crossing is rarely the answer anyone actually wants.
Recap
Five things, and the second is what turns a reading into an answer.
| If you see | Then |
|---|---|
| Different slopes | Exactly one solution |
| Equal slopes, different intercepts | No solution; inconsistent |
| One equation a multiple of the other | Infinitely many; dependent |
| 0 = 0 after eliminating | Infinitely many |
| 0 = a nonzero number | No solution |
Lesson 3.2 replaces the graph with algebra, so that a solution like negative four thirds can be found exactly rather than estimated.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.1 Solve Linear Systems by Graphing §3.1, pp. 153-159 — everything on these slides traces back here
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