Systems whose lines do not cross exactly once: parallel lines giving no solution and coincident lines giving infinitely many. Includes recognising each case graphically from the slopes and intercepts, and algebraically from a false or always-true statement when the variables vanish.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 7 — Systems of Linear Equations and Inequalities
Special Types of Linear Systems
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-422 — the lesson these objectives are drawn from
Warm-up
Every system so far had exactly one solution. This lesson asks what happens when the two lines do not cross once.
Discussion prompt
Rewrite 2x plus y equals 5 and 2x plus y equals 1 in slope-intercept form. What do you notice, and what does it mean for the system?
Hint: Compare the slopes, then the intercepts.
Answer:
\[ y = -2x + 5 \qquad y = -2x + 1 \]
Both slopes are negative two and the intercepts differ, so the lines are parallel. Parallel lines never meet, so no pair of numbers satisfies both equations and the system has no solution.
Concept
Two lines in a plane can cross exactly once, never, or coincide entirely. So a linear system has exactly one solution, no solution, or infinitely many, and the slopes and intercepts decide which.
linear system — Two linear equations considered together. It has one solution when the slopes differ, none when the slopes match and the intercepts do not, and infinitely many when both match.
There is no fourth possibility for two straight lines.
Figure (svg): Three systems shown as intersecting, parallel and coincident lines
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 418-418
Section
Section 1
Concept
Rewriting both equations in slope-intercept form classifies the system in two comparisons. Different slopes give one solution; the same slope with different intercepts gives none; both the same gives infinitely many.
The comparison settles the question without solving anything.
Figure (svg): Three systems shown as intersecting, parallel and coincident lines
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 418-418 — the Number of Solutions of a Linear System summary
Picture it
Crossing, parallel, coincident.
Figure (svg): Three systems shown as intersecting, parallel and coincident lines
The third picture shows one line drawn twice, which is why it looks like a single line. Every point on it is a solution, so the system has infinitely many.
Worked example
Two comparisons settle each one.
\[ \text{Classify } \; y = 2x + 1, \; y = -x + 4; \quad y = -2x + 5, \; y = -2x + 1; \quad y = 2x - 3, \; y = 2x - 3. \]
Take the first
Why: The slopes are two and negative one, which differ.
Take the second
Why: Both slopes are negative two; the intercepts are five and one.
Take the third
Why: Both slopes and both intercepts match.
Note the order of the checks
Why: Slopes first; intercepts only matter if the slopes agree.
Figure (svg): Three systems shown as intersecting, parallel and coincident lines
\[ \text{one}, \quad \text{none}, \quad \text{infinitely many} \]
Verify: check that the classification needed no solving
Why: None of the three systems was solved; the slopes and intercepts settled all three. That is the advantage of classifying first, since two of the three would have produced confusing algebra if solved blindly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 418-418
Sorting
Compare slopes first, then intercepts.
Sort into buckets
Sort each system by its number of solutions.
The last pair is worth noticing: one half and 0.5 are the same number written differently, so those two equations are identical despite looking distinct.
Worked example
Standard form hides the slope.
\[ \text{Classify } \; 2x + y = 5 \; \text{ and } \; 2x + y = 1. \]
Rewrite the first
Why: Subtract 2x from each side.
\[ y = -2 x + 5 \]
Rewrite the second
Why: The same move.
\[ y = -2 x + 1 \]
Compare the slopes
Why: Both are negative two.
Compare the intercepts
Why: Five and one differ, so the lines are parallel.
Figure (svg): Two parallel lines with the same slope and different intercepts
\[ \text{no solution} \]
Verify: confirm with a test pair
Why: The pair (0, 5) satisfies the first equation and gives five rather than one in the second, and (0, 1) does the reverse. Every pair satisfying one fails the other, which is what parallel means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-417
Trap
\[ 2x - y = 3 \quad \text{and} \quad 4x - 2y = 6 \]
Say the systems differ, since the coefficients are 2 and 4
Why: The two equations look different, so they seem to describe different lines.
The second is the first multiplied by two, so both rewrite to y equals 2x minus three. They are the same line and the system has infinitely many solutions.
\[ y = 2x - 3 \quad \text{and} \quad y = 2x - 3 \]
Rewrite both into slope-intercept form before comparing anything
Why: Only that form displays the slope and intercept directly.
This is the same warning Lesson 4.7 gave about parallelism: standard-form coefficients say nothing about the slope until the equation is rewritten.
Faded example
Slopes first, then intercepts.
Fill in the blanks
y = -2x + 5 and y = -2x + 1 have the same slope and different intercepts, so the system has no solution.
Why: Equal slopes make the lines parallel and different intercepts keep them apart, so nothing satisfies both equations. Had the intercepts also matched, the same two comparisons would have given infinitely many solutions instead.
Elimination
Two equations in slope-intercept form.
Eliminate the wrong options
What do you compare first?
Survives elimination: A
Why: Different slopes settle the question immediately: one solution, whatever the intercepts do. Only if the slopes agree does the intercept comparison decide between the two special cases, which is why the order matters.
Socratic
The classification claims to be complete.
Discussion prompt
Explain why two straight lines in a plane cannot meet in exactly two points, or in exactly seventeen. Then say what that means for the number of solutions a linear system can have.
Hint: Ask what two points determine.
Answer:
Two points determine exactly one line, as Lesson 4.2 established. So if two lines shared two points they would have to be the same line, and then they share every point rather than just two. Any finite number of shared points above one is therefore impossible.
That leaves exactly three possibilities: no shared points, one shared point, or all of them. Since a solution of the system is a shared point, the system has no solution, exactly one, or infinitely many — and the classification is complete rather than merely a list of the common cases.
Section
Section 2
Concept
When two equations have the same slope and different y-intercepts their graphs are parallel. Parallel lines never intersect, so the system has no solution.
\[ y = -2x + 5 \quad \text{and} \quad y = -2x + 1 \]
No pair of numbers can satisfy both equations.
Figure (svg): Two parallel lines with the same slope and different intercepts
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-417 — Example 1, A Linear System with No Solution
Picture it
The same tilt, four units apart.
Figure (svg): Two parallel lines with the same slope and different intercepts
The gap between them is the same everywhere, which is exactly what equal slopes guarantee. No amount of extending either line will bring them together.
Worked example
This is Example 1 from the textbook, by graphing.
\[ \text{Show that } \; 2x + y = 5 \; \text{ and } \; 2x + y = 1 \; \text{ has no solution.} \]
Rewrite both in slope-intercept form
Why: Subtract 2x from each.
\[ y = -2 x + 5\text{ and } y = -2 x + 1 \]
Compare the slopes
Why: Both are negative two.
Compare the intercepts
Why: Five and one differ.
Conclude
Why: The lines are parallel, so they never intersect.
Figure (svg): Two parallel lines with the same slope and different intercepts
\[ \text{no solution} \]
Verify: read the original equations aloud
Why: The first says 2x plus y is five and the second says the same expression is one. A quantity cannot be both five and one, so the impossibility is visible in the original equations without any rewriting at all.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-417
Elimination
Look for equal slopes with different intercepts.
Eliminate the wrong options
Which one?
Survives elimination: A
Why: Equal slopes with different intercepts is exactly the parallel case. Option B is worth noticing: equal intercepts do not make lines parallel, and in fact they force the lines to meet at that intercept.
Worked example
Guided Practice 1. The multiple is not exact.
\[ \text{Show that } \; x - 3y = 4 \; \text{ and } \; 2x - 6y = 4 \; \text{ has no solution.} \]
Multiply the first by 2
Why: Every term doubles.
\[ 2 x - 6 y = 8 \]
Compare with the second
Why: The left sides match and the right sides do not.
\[ 8\text{ against } 4 \]
Interpret
Why: The same expression cannot be both eight and four.
Conclude
Why: The lines are parallel and the system has no solution.
Figure (svg): The solution to Worked example a parallel pair in standard form shown as a ladder of expressions, one row per algebraic move
\[ 2x - 6y = 8 \text{ against } 2x - 6y = 4 \]
Verify: check by rewriting both
Why: The first gives y equals one third x minus four thirds and the second y equals one third x minus two thirds. Equal slopes and different intercepts, which agrees with the multiplication test — the two routes are two ways of seeing the same thing.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-417
Error analysis
The student was asked how many solutions a system has.
Annotate
On: \( \begin{aligned} 2x + y &= 5 \\ 2x + y &= 1 \\ &\text{the equations differ, so the lines cross once} \end{aligned} \)
The number of solutions depends on the slopes rather than on whether the equations look alike. Rewriting into slope-intercept form is the only reliable comparison.
Faded example
Scale one equation and compare.
Fill in the blanks
x - 3y = 4, \; \times 2: \quad 2x - 6y = 8, \text4 2x - 6y = ___
Why: The left sides now match and the right sides do not, so the same expression is being asked to equal two different numbers. That is impossible, so the lines are parallel and there is no solution.
Prediction
Two equations share the expression 2x + y.
Predict first
If 2x + y = 5 and 2x + y = 1, what follows?
Correct: No solution, since one expression cannot equal two values.
\[ 2x + y = 5 \;\text{ and }\; 2x + y = 1 \;\Longrightarrow\; 5 = 1, \text{ false} \]
Why: For a single pair of numbers the expression 2x plus y has one definite value, so it cannot be both five and one. Identical left sides with different right sides is the clearest form the parallel case can take, and it needs no rewriting at all to spot.
Socratic
The connection is worth stating rather than assuming.
Discussion prompt
Explain why two lines with the same slope and different intercepts can never meet. Then say what would have to be true for them to meet.
Hint: Try to solve for a crossing point.
Answer:
Setting mx plus b equal to mx plus c subtracts to give b equals c, which is false when the intercepts differ. So no value of x puts the two lines at the same height, and there is no crossing — the algebra reports the impossibility as a false statement, exactly as in Lesson 3.9.
They would meet only if the intercepts were equal, and then they would meet everywhere rather than once, since the two equations would be identical. So equal slopes leave only the two special cases, and which one you get is decided entirely by the intercepts.
Section
Section 3
Concept
When two equations have the same slope and the same intercept they describe the same line. Every point on it satisfies both, so the system has infinitely many solutions.
\[ 2x - y = 3 \;\xrightarrow{\times 2}\; 4x - 2y = 6 \]
One equation is the other multiplied through by a constant.
Figure (svg): Two equations describing the same line
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 418-418 — Example 2, A Linear System with Infinitely Many Solutions
Picture it
The second is the first, doubled.
Figure (svg): Two equations describing the same line
Only one line is visible because there is only one line. The two equations are two names for it, exactly as Lesson 5.4 showed standard form is not unique.
Worked example
This is Example 2 from the textbook, both methods.
\[ \text{Show that } \; 2x - y = 3 \; \text{ and } \; 4x - 2y = 6 \; \text{ has infinitely many solutions.} \]
Rewrite both in slope-intercept form
Why: Both give y equals 2x minus three.
Conclude graphically
Why: The equations represent the same line.
Try linear combinations instead
Why: Multiplying the first by two gives the second exactly.
Conclude algebraically
Why: Any solution of one is a solution of the other.
Figure (svg): Two equations describing the same line
\[ \text{infinitely many} \]
Verify: find three solutions
Why: The pairs (2, 1), (3, 3) and (0, -3) all satisfy both equations, and there are endlessly more. Producing several is a concrete way to see that the answer is a whole line rather than a point.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 418-418
Sorting
Check whether every term scales by the same factor.
Sort into buckets
Sort each pair by whether the two equations describe the same line.
The three parallel pairs differ from their partners by a single number on the right. That one number is what separates no solution from infinitely many.
Worked example
The multiplier need not be a whole number.
\[ \text{Classify } \; x + 2y = 4 \; \text{ and } \; 2x + 4y = 8. \]
Compare the coefficients
Why: Two, four and eight are double one, two and four.
\[ \text{multiplier } 2 \]
Check every term
Why: All three numbers are doubled, including the right side.
Conclude
Why: The second equation is the first multiplied by two.
State the answer
Why: The system has infinitely many solutions.
Figure (svg): One equation shown to be a multiple of the other
\[ \text{infinitely many} \]
Verify: check what would change if the right side were 9
Why: Then the left sides would still be proportional and the right sides would not, so the lines would be parallel and the system would have no solution. One number decides which of the two special cases you are in, which is why every term has to be checked.
Trap
\[ 2x + 4y = 8 \quad \text{and} \quad 2x + 4y = 9 \]
Notice the left sides match and conclude the system has infinitely many solutions
Why: The interesting terms are identical, so the equations look the same.
The right sides differ, so the same expression is being asked to equal both eight and nine. The lines are parallel and the system has no solution — the opposite conclusion.
Check every term, including the right-hand side, for the same multiplier
Why: Matching variable terms alone give parallel lines; matching everything gives one line.
The right-hand side is where the two special cases are distinguished, which makes it the term that matters most.
Faded example
Every term, or the lines are parallel.
Fill in the blanks
2x - y = 3 multiplied by 2 gives 4x - 2y = 6, which is the second equation exactly.
Why: Doubling every term including the three on the right produces the second equation exactly, so the two describe the same line. Had the second equation ended in seven, the doubling would have failed on that one term and the lines would be parallel.
Elimination
Both equations describe the same line.
Eliminate the wrong options
What is the solution set?
Survives elimination: A
Why: The two equations impose the same condition, so any pair satisfying one satisfies the other, and there are infinitely many such pairs. The answer is usually written as all points on the line, with the line's equation given.
Socratic
Two equations, one condition.
Discussion prompt
Explain why a system whose second equation is a multiple of the first gives no more information than the first alone. Then say what that means about the count of facts and unknowns.
Hint: Ask what the second equation rules out.
Answer:
Multiplying an equation through by a non-zero number does not change which pairs satisfy it, as Lesson 5.4 established. So the second equation accepts exactly the pairs the first accepts and rejects exactly the same ones — it rules out nothing new, which is why the solution set does not shrink to a point.
So there are two unknowns and effectively only one fact, and one fact cannot pin down two unknowns. That matches Lesson 4.2's observation that a single two-variable equation has a whole line of solutions, and it is why the system's answer is a line rather than a point.
Section
Section 4
Concept
If you solve a special system without noticing, the variables cancel. A false statement means no solution; a statement true for all values means infinitely many.
This is the same reading of a false statement as in Lesson 3.9.
Figure (svg): The three algebraic outcomes and what each one means
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-418 — Method 2 in each of Examples 1 and 2
Picture it
A value, a falsehood, or a truism.
Figure (svg): The three algebraic outcomes and what each one means
The variables vanishing is a signal rather than a failure. Which special case you have is then decided by whether what remains is true or false.
Worked example
This is Example 1, Method 2, from the textbook.
\[ \text{Substitute } y = -2x + 1 \text{ into } \; 2x + y = 5. \]
Substitute
Why: Replace y with the expression.
\[ 2 x + (-2 x + 1) = 5 \]
Combine like terms
Why: The x-terms cancel.
\[ 1 = 5 \]
Read the result
Why: A statement with no variables in it, and false.
Conclude
Why: No pair of numbers satisfies both equations.
Figure (svg): A substitution producing a false statement
\[ 1 = 5: \text{ no solution} \]
Verify: compare with the graph
Why: The two lines are parallel, which agrees with the algebra. The false statement is the algebraic form of never meeting, so the two methods are reporting the same fact in different languages.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-417
Matching
What is left after the variables vanish.
Match the pairs
Why: A value for a variable means the lines crossed. A false statement means they never meet, and any always-true statement — zero equals zero, seven equals seven — means they coincide. The particular numbers do not matter; only whether the statement is true.
Worked example
This is Example 2, Method 2, from the textbook.
\[ \text{Combine } \; 2x - y = 3 \; \text{ and } \; 4x - 2y = 6. \]
Multiply the first by 2
Why: It becomes identical to the second.
\[ 4 x - 2 y = 6 \]
Subtract one from the other
Why: Every term cancels.
\[ 0 = 0 \]
Read the result
Why: A statement with no variables, and always true.
Conclude
Why: Every solution of one equation solves the system.
Figure (svg): A combination producing a statement that is always true
\[ 0 = 0: \text{ infinitely many} \]
Verify: say what an always-true statement means
Why: It imposes no condition on x or y at all, so nothing has been ruled out beyond what the first equation already required. The remaining condition is that single equation, whose solutions form a line.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 418-418
Trap
\[ 2x + (-2x + 1) = 5 \;\Longrightarrow\; 1 = 5 \]
Assume a mistake was made and start the whole solve again
Why: Every previous system produced a value for x, so a line with no x in it looks wrong.
The cancellation is the answer. Redoing the work will produce the same false statement, since it reflects a property of the system rather than a slip.
The variables cancelled and 1 = 5 is false, so the system has no solution.
Read what is left: false means none, always true means infinitely many
Why: The vanishing of the variables is the signal that a special case has appeared.
One recheck is reasonable; a second one is a sign that the signal has not been recognised for what it is.
Faded example
True or false decides which case.
Fill in the blanks
If the variables cancel and 1 = 5 remains, the system has no solution; if 0 = 0 remains, it has infinitely many solutions.
Why: A false statement means no pair can satisfy both equations, and an always-true statement means the second equation added no condition. Both are answers rather than obstacles, and both are recognised the same way.
Prediction
A solve produced the statement 0 = 0.
Predict first
What would the graph show?
Correct: One line, drawn twice.
\[ 2x - y = 3 \;\Longleftrightarrow\; 4x - 2y = 6 \]
Why: An always-true statement means the two equations impose the same condition, so their graphs are the same line and every point on it is a solution. Parallel lines would have produced a false statement instead, and crossing lines a value for a variable. The three algebraic signals and the three pictures correspond exactly.
Socratic
In an ordinary system they do not.
Discussion prompt
Explain why the variables disappear when a system is parallel or coincident, but not when the lines cross. Then say what the surviving statement is telling you.
Hint: Ask what the two equations say about the same expression.
Answer:
Parallel and coincident equations have proportional variable parts, so scaling one to match the other makes the variable terms identical — and subtracting or substituting then cancels them exactly. Crossing lines have variable parts that are not proportional, so no scaling makes them cancel and a variable always survives.
What is left compares the two right-hand sides after the scaling. If they agree, the equations were the same condition and the statement is true; if they disagree, the same expression was asked to take two values and the statement is false. So the surviving statement is a verdict on the right-hand sides, which is exactly where the two special cases differ.
Section
Section 5
Concept
Rewriting both equations into slope-intercept form and comparing the slopes and intercepts classifies a system in seconds, before any solving is attempted.
The classification also tells you what to expect from the algebra.
Figure (svg): Two columns on how the slopes and intercepts classify a system
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 418-422 — the summary and its use of slope-intercept form
Picture it
Slopes first, intercepts second.
Figure (svg): Two columns on how the slopes and intercepts classify a system
Knowing which case you are in before starting means a vanished variable is expected rather than alarming, which is worth more than the seconds the comparison costs.
Worked example
The classification determines the next step.
\[ \text{Classify and handle } \; 3x - y = 2, \; 6x - 2y = 5; \quad x + y = 4, \; 2x - y = 5. \]
Rewrite the first system
Why: y equals 3x minus two, and y equals 3x minus two and a half.
Compare intercepts
Why: They differ, so the lines are parallel.
Rewrite the second
Why: y equals negative x plus four, and y equals 2x minus five.
Act on it
Why: One solution, so solve: adding gives 3x equals nine.
\[ (3, 1) \]
Figure (svg): Two columns on how the slopes and intercepts classify a system
\[ \text{none}; \quad (3, 1) \]
Verify: confirm the second answer
Why: Three plus one is four and six minus one is five, so both equations hold. The first system was never solved, because the classification showed there was nothing to find.
Sorting
Classify first, then act.
Sort into buckets
Sort each system by the right next step.
Four of these six need no algebra at all once classified. The two comparisons cost about ten seconds and settle two thirds of the work here.
Worked example
Exercise 31 concerns finding the weight of a jewellery bead.
\[ \text{A real problem produces a system with no solution. What does that mean?} \]
Recall what no solution means
Why: No pair of values satisfies both conditions.
Read it back into the situation
Why: The two measurements cannot both be right as described.
Consider infinitely many instead
Why: The two facts say the same thing.
State the practical response
Why: Either recheck the data or collect a genuinely new fact.
Figure (svg): The solution to Worked example what a real problem's special case means shown as a ladder of expressions, one row per algebraic move
\[ \text{inconsistent}, \text{ or not independent} \]
Verify: say which is the more common in practice
Why: Infinitely many is more common in a well-collected problem, because it usually means the second measurement repeated information the first already gave. No solution usually points at a measurement error or a misread condition, since a real situation does actually have an answer.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 422-422
Trap
\[ 3x - y = 2 \quad \text{and} \quad 6x - 2y = 5 \]
Start substituting immediately
Why: Every previous system had a solution, so the reflex is to solve.
The variables will cancel and leave four equals five, and several minutes will have been spent discovering something two comparisons would have shown at once.
Rewrite both and compare the slopes before starting
Why: It costs two lines and tells you what to expect.
The classification is not a replacement for solving; it is the step that says whether solving is worth doing.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Slopes and intercepts | The lines | Solutions |
|---|---|---|
| different slopes | cross once | exactly one |
| same slope, different intercepts | are parallel | none |
| same slope, same intercept | coincide | infinitely many |
Reading the table downwards, the lines get closer together until they merge. That is the whole classification, and there is no fourth row.
Hypothesis
Predict before you decide.
Predict first
A measured situation is modelled as a system, and the system turns out to have no solution. What is the most likely explanation?
Correct: One of the measurements or conditions is wrong.
Infinitely many solutions has the opposite cause: the second fact repeated the first rather than adding anything.
Why: A real situation that happened does have an answer, so a model saying otherwise is describing the data rather than the world. The usual cause is a measurement error or a misread condition, and the right response is to check the facts rather than the algebra. Rechecking the algebra once is sensible; concluding that the situation is impossible rarely is.
Socratic
Two unknowns and two facts, in principle.
Discussion prompt
Say what no solution and infinitely many solutions each reveal about the two facts a problem supplied. Then say what you would do about each in practice.
Hint: Ask whether the two facts agree, and whether they differ.
Answer:
No solution means the two facts contradict each other: they cannot both be true of any pair of values. Infinitely many means the two facts are not independent — the second says the same thing as the first, so effectively only one fact was given and two unknowns cannot be determined from it.
For a contradiction, recheck the measurements and the wording, since a real situation does have an answer. For a repeated fact, look for a genuinely new piece of information — a different measurement, another condition — because no amount of algebra can extract two numbers from one equation. Both diagnoses point at the data rather than at the method.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| One solution | No solution | Infinitely many | |
|---|---|---|---|
| Slopes | different | same | same |
| Intercepts | any | different | same |
| Algebraic signal | a value for a variable | a false statement | an always-true statement |
The first row separates the ordinary case from the two special ones, and the second row separates those two from each other. Two comparisons cover every possibility.
Pattern
Whether you classify first or discover the case mid-solve, the same five moves cover it.
Steps one to three take about ten seconds and can replace an entire solve, which makes them worth doing even when you expect an ordinary system.
OpenStax Elementary Algebra 2e, §5.1 Solve Systems of Equations by Graphing §5.1
Check
Compare slopes, then intercepts.
Check your understanding
How many solutions does the system y = 3x - 4 and y = 3x + 2 have?
Answer: A
Why: Both slopes are three and the intercepts are negative four and two, so the lines are parallel and never meet. Setting the two expressions equal gives negative four equals two, which is false.
Check
Check every term.
Check your understanding
How many solutions does the system x + 3y = 6 and 2x + 6y = 12 have?
Answer: A
Why: Every term of the second equation is twice the corresponding term of the first, including the right-hand side, so the two describe the same line and every point on it is a solution.
Check
Read the surviving statement.
Check your understanding
A solve ends with the statement 0 = 7. What does that mean?
Answer: A
Why: The variables have cancelled and what remains is false, so no pair of numbers can satisfy both equations. Graphically, the two lines are parallel.
Real world
This is Exercise 31's situation. You try to find the weight of a bead by weighing necklaces. One necklace with 5 beads and a clasp weighs 21 grams; another with 10 beads and 2 clasps weighs 42 grams.
Discussion prompt
Write the system, classify it, and say what the outcome means about the two measurements. Then say what extra measurement would settle the bead's weight.
Hint: Compare the second equation with the first.
Answer:
\[ 5b + c = 21 \qquad 10b + 2c = 42 \]
The second equation is exactly twice the first, so the two describe the same line and the system has infinitely many solutions. The second necklace was simply two of the first, so weighing it added no information at all.
Practically, that means the bead's weight cannot be found from these two measurements: a bead of two grams with an eleven-gram clasp fits, and so does a bead of three grams with a six-gram clasp. What would settle it is a genuinely independent measurement — weighing a clasp on its own, or a necklace with a different ratio of beads to clasps, such as three beads and two clasps.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Solving a system, the variables cancel and you are left with 4 = 5. What is the answer?
Correct: The system has no solution.
\[ 2x + y = 5, \; 2x + y = 4 \;\Longrightarrow\; 5 = 4, \text{ false} \]
\[ \text{compare } 0 = 0: \text{ infinitely many} \]
Why: A false statement with no variables in it means no pair of numbers can satisfy both equations, so the two lines are parallel. The first option is the natural reaction, since every ordinary system produces a value — and redoing the work will simply produce the same false statement, because it reflects a property of the system rather than a slip. An always-true statement would have meant infinitely many instead.
Explain it
They solved a system, everything cancelled, and they think they have gone wrong.
Discussion prompt
In no more than four sentences, explain what a vanished variable means and how to tell which of the two cases they have. Then tell them how to see it before it happens.
Hint: Read what is left.
Answer:
A usable answer: nothing has gone wrong — the cancelling is the answer. Look at the statement you are left with: if it is false, like one equals five, the two lines are parallel and there is no solution; if it is always true, like zero equals zero, the two equations describe the same line and every point on it works.
To see it coming, rewrite both equations as y equals something before you start. If the slopes are different you will get one answer; if the slopes match, look at the intercepts — different means no solution, the same means infinitely many, and you never have to solve at all.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The two cases are told apart by whether the surviving statement is true or false, and graphically by the intercepts. A vanished variable is fixed by expecting it: it is a signal, not a slip. The multiplier check is fixed by including the right-hand side, which is where the two cases differ. Classifying first is fixed by rewriting both equations before touching anything else. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Draw three small coordinate planes across the top of a page and on them draw two crossing lines, two parallel lines and one line drawn twice, labelling each with the number of solutions and with the slope-and-intercept condition that produces it. Underneath each, write a system of two equations in standard form that gives that case, and rewrite both equations of each system into slope-intercept form beside it. In the middle of the page, take your parallel system and solve it by substitution, carrying on until the variables cancel, and box the false statement you get. Do the same for your coincident system and box the always-true statement. In the lower half, write two systems that differ only in one number on a right-hand side, one giving no solution and one giving infinitely many, and circle the number that changes. Finally, in the margin, write what each special case would mean if the system came from real measurements.
Your two boxed statements should be the two kinds of signal: one false and one always true. If a variable survives in either, the system you wrote is not the special case you intended — check that the variable terms are proportional.
Recap
Five things, and the second is the one that looks like a mistake.
| If the question says | Your first move is |
|---|---|
| How many solutions | Rewrite both and compare slopes |
| The slopes match | Compare the intercepts |
| The variables cancelled | Read whether what is left is true |
| One equation looks like a multiple | Check the right-hand side too |
| A real problem gives no solution | Recheck the data, not the algebra |
Lesson 7.6 closes the chapter by replacing the equations with inequalities. Two half-planes overlapping give a region rather than a point, which is the two-variable version of the compound and inequalities from Lesson 6.4.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.5 Special Types of Linear Systems §7.5, pp. 417-422 — everything on these slides traces back here
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