Systems of two linear equations and their solutions as ordered pairs satisfying both. Includes reading a solution off the point where two graphs intersect, the graph-and-check method with its rewriting and verification steps, why a graphical answer is only an estimate, and modelling two growing quantities to find when they become equal.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 7 — Systems of Linear Equations and Inequalities
Graphing 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 7.1 Graphing Linear Systems §7.1, pp. 389-394 — the lesson these objectives are drawn from
Warm-up
Chapter 4 drew one line at a time. This chapter draws two and asks where they meet.
Discussion prompt
The pair (2, 1) satisfies 3x minus 2y equals 4. Does it also satisfy x plus 3y equals 5? Check both, and say what a pair satisfying both would mean geometrically.
Hint: Substitute into each equation separately.
Answer:
\[ 3(2) - 2(1) = 4 \;\checkmark \qquad 2 + 3(1) = 5 \;\checkmark \]
It satisfies both, so the point (2, 1) lies on both lines — it is where the two graphs cross. Finding such a point is the whole business of this chapter.
Concept
Two or more linear equations in the same variables form a system of linear equations. A solution is a pair of numbers making every equation true, and the point where the graphs cross is called the point of intersection.
system of linear equations — Two or more linear equations in the same variables, considered together. A solution is an ordered pair satisfying every equation in the system.
The solution is written as an ordered pair, exactly as a point is.
Figure (svg): A system written as two labelled equations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 389-389
Section
Section 1
Concept
A solution of a linear system in two variables is a pair of numbers that makes each equation a true statement. It is written as an ordered pair, and it is the point where the two graphs meet.
This is the word and from Lesson 6.4, applied to two equations.
Figure (svg): One ordered pair substituted into both equations of a system
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 389-389 — the definitions of a linear system and of a solution
Picture it
Two substitutions, two true statements.
Figure (svg): One ordered pair substituted into both equations of a system
Checking only one equation would confirm that the point is on one line, which almost every point of that line does. Both checks together are what pin down the crossing.
Worked example
This is the check accompanying Example 1 in the textbook.
\[ \text{Is } (2, 1) \text{ a solution of } \; 3x - 2y = 4 \; \text{ and } \; x + 3y = 5? \]
Substitute into the first equation
Why: Three times two minus two times one.
\[ 6 - 2 = 4 \]
Judge it
Why: Four equals four, so the first equation holds.
Substitute into the second
Why: Two plus three times one.
\[ 2 + 3 = 5 \]
Judge it
Why: Five equals five, so the second holds too.
Figure (svg): One ordered pair substituted into both equations of a system
\[ (2, 1) \text{ satisfies both equations} \]
Verify: test a point on only one line
Why: The pair (0, -2) satisfies the first equation, since zero minus negative four is four, and fails the second, since zero minus six is negative six rather than five. So it lies on one line and not the other, which is exactly what failing a system means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 389-389
Sorting
The system is 3x - 2y = 4 and x + 3y = 5.
Sort into buckets
Sort each pair by whether it solves the system.
Only one pair in the list works, which is typical: two lines that are not parallel cross exactly once, so a system usually has exactly one solution.
Worked example
Half a success is a failure for a system.
\[ \text{Is } (5, 0) \text{ a solution of } \; 3x - 2y = 4 \; \text{ and } \; x + 3y = 5? \]
Substitute into the first
Why: Fifteen minus zero is fifteen.
\[ 15\text{ is not } 4 \]
Judge it
Why: The first equation fails.
Substitute into the second
Why: Five plus zero is five.
\[ 5 = 5 \]
State the verdict
Why: One passes and one fails, so the pair is not a solution.
Figure (svg): The solution to Worked example a pair that fails one equation shown as a ladder of expressions, one row per algebraic move
\[ (5, 0): \; \text{Eq 1} \;\times, \; \text{Eq 2} \;\checkmark \]
Verify: locate the point relative to the two lines
Why: It lies on the second line and off the first, so it is one of the infinitely many points that solve one equation and not the system. Only the single crossing point solves both, which is why a system usually has exactly one solution.
Trap
\[ \text{Is } (5, 0) \text{ a solution of the system?} \]
Substitute into the second equation, find it works, and answer yes
Why: One true statement looks like confirmation, and checking is tiring.
Every point of the second line satisfies that equation — infinitely many of them. Only the ones also satisfying the first are solutions of the system.
Substitute into both equations and require both to hold
Why: A system joins its equations with and, so both are conditions.
Stopping at the first failure is fine, but stopping at the first success is not.
Faded example
Two substitutions, both must hold.
Fill in the blanks
(2, 1): \quad 3(2) - 2(1) = 4 \;\checkmark \qquad 2 + 3(1) = 5 \;\checkmark
Why: Both substitutions produce the right-hand sides of their own equations, so the pair satisfies both and is the solution of the system. Either one failing would have been enough to reject it.
Elimination
Two equations, considered together.
Eliminate the wrong options
Which statement is correct?
Survives elimination: A
Why: A system joins its equations with and, so a solution must satisfy every one of them. Option C is the natural confusion with Lesson 6.5's or, and choosing it would replace a single point with two entire lines.
Socratic
The connection between the picture and the algebra deserves a reason.
Discussion prompt
Explain why the point where two lines cross is exactly the solution of the corresponding system. Then say how many solutions you would expect a system of two lines to have, and why.
Hint: Ask what being on a line means about an equation.
Answer:
Each line is the set of all points satisfying its own equation, as Lesson 4.2 established. So a point lies on the first line exactly when it satisfies the first equation, and on the second exactly when it satisfies the second. A point on both therefore satisfies both, and that is precisely a solution of the system.
Two straight lines that are not parallel cross exactly once, so you would expect exactly one solution. Parallel lines never meet, giving none, and two equations describing the same line meet everywhere, giving infinitely many — the three cases Lesson 7.5 takes up systematically.
Section
Section 2
Concept
If the two lines are already graphed, the solution can be estimated by reading the coordinates of their point of intersection. The estimate is then confirmed by substitution.
A reading from a graph is an estimate until the substitution checks out.
Figure (svg): Two lines crossing at a single point on the coordinate plane
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 389-389 — Example 1, Find the Point of Intersection
Picture it
The coordinates read off the picture.
Figure (svg): Two lines crossing at a single point on the coordinate plane
The crossing looks like (2, 1) and the substitution proves it. Had it been at (2.1, 0.9) the drawing would have looked much the same, which is why the check matters.
Worked example
This is Example 1 from the textbook.
\[ \text{Two graphed lines appear to cross at } (2, 1). \text{ Confirm this solves } \; 3x - 2y = 4, \; x + 3y = 5. \]
Read the crossing
Why: The lines appear to meet at two comma one.
\[ (2, 1) \]
Substitute into the first equation
Why: Six minus two is four.
\[ 4 = 4 \]
Substitute into the second
Why: Two plus three is five.
\[ 5 = 5 \]
State the conclusion
Why: Both hold, so the estimate is exact.
\[ (2, 1) \]
Figure (svg): Two lines crossing at a single point on the coordinate plane
\[ (2, 1) \]
Verify: ask what the check has added
Why: The graph suggested the answer and the substitution proved it. Without the check the answer would be a reading of a drawing, accurate to whatever the pencil managed — which for a crossing near a grid point is usually good and never certain.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 389-389
Prediction
You read a crossing at (3, 2) and the substitution gives 5 = 4.
Predict first
What should you conclude?
Correct: The crossing is not exactly at (3, 2).
The remedy is an algebraic method, which Lessons 7.2 and 7.3 supply.
Why: A failed substitution rejects that particular pair and says nothing about whether a solution exists. The lines evidently do cross, since a crossing was visible — it is simply not at the point that was read, most likely because it falls between grid lines. Concluding that there is no solution would confuse a misreading with a genuine feature of the system.
Worked example
Graphs cannot resolve fractional crossings.
\[ \text{Two lines cross near } (2, 1) \text{ but the check gives } 3.8 = 4. \text{ What has happened?} \]
Note the failure
Why: The first equation does not hold exactly.
Interpret it
Why: The true crossing is near, but not at, the read point.
Say what to do
Why: The graph has located it roughly; an algebraic method is needed for the exact value.
Look ahead
Why: Lessons 7.2 and 7.3 supply exact methods.
Figure (svg): Two columns contrasting what the graph gives with what the check gives
\[ \text{estimate near } (2, 1), \text{ exact value unknown} \]
Verify: say what the graph is still good for
Why: It shows roughly where the answer is, how many solutions there are, and whether the lines are parallel — none of which is wasted. The graph frames the problem and the algebra finishes it, which is why both appear in this chapter.
Error analysis
The student read a solution off a graph and reported it.
Annotate
On: \( \begin{aligned} &\text{the lines appear to cross at } (3, 2) \\ &\text{substituting into Equation 1 gives } 5 = 4 \\ &\text{so the solution is } (3, 2) \end{aligned} \)
The check exists to be believed. A graphical estimate that fails its check is information — it says the answer is nearby and not where you looked.
Faded example
Substitute into both.
Fill in the blanks
(2, 1) \text4 3x - 2y = 4: \; 5 = 4 \qquad \text___ x + 3y = 5: \; ___ = 5
Why: Both substitutions match their right-hand sides, so the estimate read from the graph is exact. Only after both checks pass is the graphical reading upgraded from a guess to an answer.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Reading the graph | Substituting | |
|---|---|---|
| What it gives | an estimate | a definite verdict |
| Handles fractional answers | poorly | exactly |
| Shows how many solutions there are | yes, at a glance | no, only tests one pair |
Each is good at what the other is not, which is why the method uses both. The graph frames the problem and the substitution settles it.
Socratic
The algebra will give the exact answer anyway.
Discussion prompt
Say what a graph of a system tells you that an algebraic solution does not. Then say when the graph is the only practical approach.
Hint: Think about what you see before solving anything.
Answer:
The graph shows immediately whether the lines cross once, never, or coincide entirely — the three cases of Lesson 7.5 — and roughly where the answer lies. It also shows how sensitive the crossing is: two nearly parallel lines meet at a point that moves a great deal if either line shifts slightly, which matters when the equations come from measured data.
It is the only practical approach when the equations are not linear, or when they come from graphed data with no formula at all — two experimental curves crossing on a chart. Then the crossing can be read and estimated even though no algebraic method exists, which is why the graphical idea outlives the linear case.
Section
Section 3
Concept
To solve a system by graphing, put each equation into a form that is easy to graph, draw both on one plane, estimate the crossing, and check it in both original equations.
Slope-intercept form is usually the easiest form to graph from.
Figure (svg): The four steps of the graph-and-check method
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 390-390 — the Solving a Linear System Using Graph-and-Check summary
Picture it
Two of preparation, one of reading, one of proof.
Figure (svg): The four steps of the graph-and-check method
The check uses the original equations rather than the rewritten ones, so it tests the rewriting as well as the reading.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; x + y = -2 \; \text{ and } \; 2x - 3y = -9 \; \text{ by graphing.} \]
Rewrite the first equation
Why: Subtract x from each side.
\[ y = -x - 2 \]
Rewrite the second
Why: Move the x-term and divide by negative three.
\[ y = (\frac{2}{3}) x + 3 \]
Graph both and estimate
Why: The lines appear to cross at negative three comma one.
\[ (-3, 1) \]
Check in both originals
Why: Negative three plus one is negative two; negative six minus three is negative nine.
Figure (svg): Two standard-form equations rewritten into slope-intercept form
\[ (-3, 1) \]
Verify: confirm the check used the original equations
Why: Substituting into x plus y equals negative two and 2x minus 3y equals negative nine tests both the reading and the two rewritings. Checking in the slope-intercept versions instead would miss an error made while rewriting.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 390-390
Translation
Isolate y in each equation.
Match the pairs
Why: Each rewriting is the Lesson 4.7 move: isolate y and divide every term by its coefficient. The second and third involve dividing by a negative and by two respectively, which is where signs are most often lost.
Worked example
Guided Practice 1 to 3. Three systems, three crossings.
\[ \text{Solve } \; x + y = 4, \; 2x - y = 5; \quad x - y = 5, \; 2x + 3y = 0; \quad x + y = 2, \; x - y = 4. \]
Take the first system
Why: Adding the equations suggests x is three, so y is one.
\[ (3, 1) \]
Take the second
Why: The pair three comma negative two satisfies both.
\[ (3, -2) \]
Take the third
Why: The pair three comma negative one satisfies both.
\[ (3, -1) \]
Check each
Why: Substituting into both equations of each system confirms all three.
Figure (svg): The solution to Worked example three from guided practice shown as a ladder of expressions, one row per algebraic move
\[ (3, 1), \quad (3, -2), \quad (3, -1) \]
Verify: check the second system carefully
Why: At (3, -2): three minus negative two is five, and six plus negative six is zero. Both hold. The negative y-coordinate is where a reading from a graph is easiest to get wrong by a unit, so it is the one worth substituting first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 390-390
Trap
\[ x + y = -2 \;\Longrightarrow\; y = -x - 2 \]
Check the answer by substituting into y = -x - 2
Why: That is the equation actually graphed, so it feels like the one to test against.
If the rewriting itself went wrong, the check would confirm the wrong line. Testing against the original equation is what catches an error made while rewriting.
Always check in the original equations as given
Why: The rewriting is a step in the method, so it needs testing too.
The textbook's method says this explicitly, and it costs nothing since the originals are still on the page.
Faded example
Move the x-term, then divide.
Fill in the blanks
2x - 3y = -9 \;\Longrightarrow\; -3y = -2x - 9 \;\Longrightarrow\; y = \tfrac33}x + ___
Why: Dividing every term by negative three gives a slope of two thirds and an intercept of three. Both signs flip during that division, which is the step to check by substituting one point back into the original.
Elimination
A student rewrites both equations, graphs them and reads a crossing.
Eliminate the wrong options
What still has to be done?
Survives elimination: A
Why: The fourth step is the check, and it is what turns an estimate into an answer. Skipping it is the difference between reporting where two pencil lines appeared to meet and reporting a solution.
Socratic
Standard form can be graphed too.
Discussion prompt
Say why the method's first step is to rewrite each equation, and name a case where you would graph directly from standard form instead.
Hint: Think about what each form makes easy.
Answer:
Slope-intercept form gives a point and a step, so each line can be drawn from two numbers with no computation. Standard form requires finding two points first, which is more work per line and more chances for an arithmetic slip.
Standard form is quicker when both intercepts are whole numbers, since Lesson 4.4's quick graph then gives two exact points in two one-step computations. For an equation like 3x plus 4y equals 12 that is faster than rewriting, so the first step is really choose a convenient form rather than always use slope-intercept.
Section
Section 4
Concept
A crossing read from a hand-drawn graph is an estimate. It is reliable when the crossing falls on a grid point and unreliable when it falls between them, which is why the method ends with a substitution.
Lessons 7.2 and 7.3 supply exact methods for the awkward cases.
Figure (svg): Two columns contrasting what the graph gives with what the check gives
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 390-394 — the graph-and-check method's reliance on a final substitution
Picture it
Speed against certainty.
Figure (svg): Two columns contrasting what the graph gives with what the check gives
Neither column is dispensable. The graph is what tells you roughly where to look, and the substitution is what makes the answer defensible.
Worked example
Not every system has a convenient answer.
\[ \text{The system } \; 2x + y = 5, \; x - y = 1 \; \text{ crosses where?} \]
Rewrite both
Why: y equals negative 2x plus five, and y equals x minus one.
Estimate from a graph
Why: The crossing looks like it is near two comma one.
\[ \text{near } (2, 1) \]
Check the estimate
Why: Four plus one is five, and two minus one is one.
State the answer
Why: The estimate was exact this time.
\[ (2, 1) \]
Figure (svg): The solution to Worked example a crossing between grid lines shown as a ladder of expressions, one row per algebraic move
\[ (2, 1) \]
Verify: consider what a nearby system would do
Why: Changing the first equation to 2x plus y equals 6 moves the crossing to (7/3, 4/3), which no hand-drawn graph would resolve. The method's reliability depends on the numbers rather than on the drawing, which is the limitation worth knowing.
Elimination
Some systems resist a graphical answer.
Eliminate the wrong options
In which case would you not solve by graphing?
Survives elimination: A
Why: A fractional crossing cannot be read accurately from a drawing, so the estimate will fail its check and an algebraic method is needed. Knowing which method suits which situation is more useful than preferring one of them generally.
Worked example
The picture answers a question the substitution cannot.
\[ \text{How many solutions does a system of two lines have?} \]
Draw two crossing lines
Why: They meet once.
Draw two parallel lines
Why: They never meet.
Draw two identical lines
Why: They meet everywhere.
Note what settled it
Why: A glance at the graph, not a substitution.
Figure (svg): A point tested against each of two lines separately
\[ \text{one, none, or infinitely many} \]
Verify: ask what a single substitution could have told you
Why: Testing one pair says whether that pair is a solution and nothing about how many exist. Counting solutions is a question about the whole system, which the graph answers immediately and a substitution never does — which is why Lesson 7.5 begins from the picture.
Trap
The crossing looks like it is at about (2.3, 1.4).
Report (2.3, 1.4) as the solution
Why: The graph seemed to show that much detail, and decimals look precise.
A hand-drawn line is not accurate to a tenth of a unit, so those decimals are invented. The substitution would reject them, and an algebraic method is what the situation calls for.
Report a grid-point reading only when the check confirms it, and use algebra otherwise
Why: The graph locates the answer; it does not measure it.
Saying the crossing is near (2, 1) and solving algebraically for the exact value is both honest and quick.
Faded example
The graph estimates; the check decides.
Fill in the blanks
A crossing read from a graph is an estimate, and the substitution is what confirms it exactly.
Why: The textbook's method is called graph-and-check precisely because the graph alone does not finish the job. The substitution is what upgrades a reading into an answer.
Hypothesis
Predict before you decide.
Predict first
Two lines cross somewhere near (2, 1). How accurately can a careful hand-drawn graph locate the crossing?
Correct: To about the nearest half unit, sometimes better at a grid point.
This is the same precision limit as reading a slope in Lesson 5.1, and it has the same remedy: use exact coordinates rather than the picture.
Why: Pencil width, ruler placement and point plotting each contribute error, so a fraction of a unit is realistic. A crossing that lands exactly on a lattice point can be identified confidently because there is only one candidate nearby, which is why the method works well on textbook systems and poorly on arbitrary ones.
Socratic
Graphing already solves systems.
Discussion prompt
Say what limitation of the graphing method the rest of Chapter 7 exists to remove. Then say what graphing will still be used for after the algebraic methods arrive.
Hint: Think about the answers graphing cannot produce.
Answer:
Graphing cannot produce an exact answer when the crossing is not at a grid point, and most real systems do not have integer solutions. Lessons 7.2 and 7.3 give algebraic methods — substitution and linear combinations — that produce exact coordinates whatever the numbers are.
The graph remains the way to see how many solutions there are and to sanity-check an algebraic answer. Lesson 7.5 uses it to classify systems into one solution, none and infinitely many, and Lesson 7.6 needs it for inequalities, where the answer is a region and cannot be written as a pair at all.
Section
Section 5
Concept
Two quantities changing at steady rates give two linear models. The point where their graphs meet is the moment they are equal, which is often the question being asked.
One quantity usually starts ahead and the other gains faster.
Figure (svg): Two growing quantities modelled as lines meeting at a crossover month
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 391-391 — Example 3, Write and Solve a Real-Life Linear System
Picture it
One ahead, one gaining.
Figure (svg): Two growing quantities modelled as lines meeting at a crossover month
The science site's two-hundred-visit head start is exactly cancelled by the math site's extra twenty-five visits a month, after eight months. Both coordinates of the crossing mean something.
Worked example
This is Example 3 from the textbook.
\[ \text{Science: } 400 \text{ visits, } +25 \text{ a month. Math: } 200 \text{ visits, } +50 \text{ a month. When are they equal?} \]
Write the science model
Why: Four hundred plus twenty-five per month.
\[ V = 400 + 25 t \]
Write the math model
Why: Two hundred plus fifty per month.
\[ V = 200 + 50 t \]
Graph both and estimate
Why: The lines appear to cross at eight comma six hundred.
\[ (8, 600) \]
Check both models
Why: Four hundred plus two hundred is six hundred; two hundred plus four hundred is six hundred.
Figure (svg): Two growing quantities modelled as lines meeting at a crossover month
\[ (8, 600) \]
Verify: interpret the two coordinates separately
Why: Eight is a number of months and six hundred a number of daily visits, so the answer is both when and at what level. Reporting only the eight would leave out half the information the crossing carries.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 391-391
Faded example
Starting value plus rate times time.
Fill in the blanks
\text25: V = 400 + 200t \qquad \text___: V = ___ + 50t
Why: Each model has a starting value as its intercept and a monthly increase as its slope, exactly as in Lesson 5.5. Setting the two expressions equal is what the system does, and the crossing is where the two quantities coincide.
Worked example
Guided Practice 4. The Spanish club against the science club.
\[ \text{Spanish: } 500 \text{ visits, } +20 \text{ a month. When does it match the science site?} \]
Write the two models
Why: Five hundred plus twenty t, against four hundred plus twenty-five t.
Compare the head start and the rate
Why: Spanish starts a hundred ahead and gains five less a month.
\[ \text{closing at } 5 \]
Find where they meet
Why: A hundred divided by five is twenty months.
\[ t = 20 \]
Find the level
Why: Five hundred plus four hundred is nine hundred.
\[ (20, 900) \]
Figure (svg): The solution to Worked example a third site shown as a ladder of expressions, one row per algebraic move
\[ (20, 900) \]
Verify: check the crossing in both models
Why: At twenty months Spanish gives five hundred plus four hundred, which is nine hundred, and science gives four hundred plus five hundred, also nine hundred. The head start of a hundred divided by the rate gap of five gives twenty directly, which is a useful shortcut worth noticing.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 391-391
Trap
When will the two sites have the same number of daily visits?
Answer: after 8 months
Why: The question asked when, so a time seems to be the whole answer.
That is correct as far as it goes and leaves out the level. The crossing is the pair (8, 600), and the six hundred is what the sites will each be getting — usually the more interesting half.
After 8 months, when each site will have 600 daily visits.
Report both coordinates with their units
Why: A solution of a system is an ordered pair, and each coordinate answers something.
Checking the units of each coordinate — months and daily visits — makes it obvious that two different facts are being reported.
Prediction
Science starts at 400 gaining 25; math starts at 200 gaining 50.
Predict first
What happens after the crossing at 8 months?
Correct: Math is ahead, since it gains faster.
\[ t = 12: \; \text{science } 700, \; \text{math } 800 \]
Why: The math site's steeper slope means it adds more visits each month, so once it has caught up it keeps pulling away. The head start decides who leads before the crossing and the rate decides who leads after — which is what makes the crossing the interesting point rather than either endpoint.
Elimination
The models cross at (8, 600).
Eliminate the wrong options
Which interpretation is right?
Survives elimination: A
Why: The first coordinate is the number of months and the second the number of daily visits, so the crossing says when they become equal and at what level. Attaching units to each coordinate rejects the other three options immediately.
Socratic
Two growing quantities need not ever meet.
Discussion prompt
Describe a pair of real quantities whose models would never cross, and say what that looks like both algebraically and on a graph. Then say what it would mean if two models coincided entirely.
Hint: Think about the rates rather than the starting values.
Answer:
If one site starts ahead and grows at the same rate as the other, it stays exactly that far ahead forever. The two models have the same slope and different intercepts, so their lines are parallel and never meet — the system has no solution, which is a true and useful answer.
Two coinciding models would mean the same slope and the same intercept, so the two sites had identical visit counts at every moment — the same line written twice. Then every month is a solution, and the answer is infinitely many. Lesson 7.5 treats all three cases together.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| One linear equation | A system of two | |
|---|---|---|
| The solutions form | a line | usually a single point |
| A solution is | an ordered pair on that line | an ordered pair on both lines |
| How to check | substitute into the one equation | substitute into both equations |
Adding a second equation cuts the solution set from a whole line down to one point, in the same way that adding a second condition narrowed a set in Lesson 6.4.
Pattern
Whether the equations arrive in standard form or already rewritten, the same five moves cover it.
Step four uses the original equations rather than the rewritten ones, so it tests the rewriting as well as the reading.
OpenStax Elementary Algebra 2e, §5.1 Solve Systems of Equations by Graphing §5.1
Check
Both equations must hold.
Check your understanding
Is (1, 3) a solution of x + y = 4 and 2x - y = 5?
Answer: A
Why: One plus three is four, so the first equation holds. But two minus three is negative one rather than five, so the second fails and the pair is not a solution of the system.
Check
Rewrite before graphing.
Check your understanding
In slope-intercept form, what is 2x - 3y = -9?
Answer: A
Why: Subtracting 2x gives negative 3y equals negative 2x minus nine, and dividing every term by negative three gives a slope of two thirds and an intercept of three. Substituting x equal to zero into the original confirms y equals three.
Check
The picture answers a counting question.
Check your understanding
What does a graph of a system tell you that a single substitution does not?
Answer: A
Why: Two lines crossing once, never or coinciding are visible at a glance, and no single substitution can establish how many solutions exist. That is the question Lesson 7.5 takes up.
Real world
Two gyms compete for members. One charges 60 dollars to join plus 25 a month; the other charges no joining fee but 35 a month.
Discussion prompt
Write a system for the total cost at each gym, find where the two models cross, and say what that point means. Then say which gym is cheaper before and after that point.
Hint: Each total cost is a linear model in the number of months.
Answer:
\[ C = 25m + 60 \qquad C = 35m \]
\[ \text{equal when } 25m + 60 = 35m \;\Longrightarrow\; m = 6, \; C = 210 \]
The two models cross at six months and two hundred and ten dollars: after half a year the two gyms have cost exactly the same. Both coordinates matter, since the six answers when and the two hundred and ten says how much has been spent by then.
Before six months the second gym is cheaper, because the first's joining fee has not yet been offset. After six months the first is cheaper, because its lower monthly rate keeps gaining. The crossing is exactly where the head start is used up — the same structure as the two web sites, with a fee in place of a visit count.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A pair satisfies the first equation of a system. Is it a solution?
Correct: Not necessarily; it must satisfy the second as well.
\[ (5, 0): \; \text{Eq 2 holds}, \; \text{Eq 1 gives } 15 \neq 4 \]
\[ (2, 1): \; \text{both hold} \]
Why: Every point of the first line satisfies the first equation, and there are infinitely many of them. Only the ones also on the second line solve the system, and usually that is a single point. The first option would replace one point with an entire line, which is the difference between solving one equation and solving a system — and it is the same and-versus-or distinction as in Lesson 6.4.
Explain it
They can graph a line and have never seen two drawn together.
Discussion prompt
In no more than four sentences, explain what a system is and why its solution is where the two lines cross. Then tell them the step people skip.
Hint: Each line is its own equation's solutions.
Answer:
A usable answer: a system is two equations you have to satisfy at the same time. Each equation's solutions make up a line, so a pair working for both has to be on both lines — which is the point where they cross. Draw both lines and read off where they meet.
The step people skip is checking. Put your answer back into both of the original equations, because a crossing read off a hand-drawn graph is a guess until both substitutions come out true.
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: Both checks are fixed by writing the two originals side by side and substituting into each. Rewriting is fixed by dividing every term and testing one point against the original. Reading a crossing is fixed by treating it as an estimate and letting the check decide. Interpretation is fixed by attaching units to each coordinate before writing a sentence. 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
At the top of a page write a system of two equations in standard form and rewrite both into slope-intercept form beside them, showing every step. Draw one coordinate plane and graph both lines on it, marking and labelling the point where they cross. Underneath, substitute that point into both of the original equations, writing each check on its own line and marking it true or false. To the right, draw a second small plane and sketch two parallel lines and two coinciding lines, writing beside each how many solutions the corresponding system has. In the lower half, invent a real situation with two quantities changing at different steady rates, write the two models, find their crossing, and write one sentence interpreting both coordinates with their units. Finally, in the margin, write one sentence saying why the check uses the original equations rather than the rewritten ones.
Both of your checks should come out true. If one fails, the crossing was misread or one of the rewritings lost a sign — and testing a single point of that line against its original equation will tell you which.
Recap
Five things, and the last step of the method is the one that gets skipped.
| If the question says | Your first move is |
|---|---|
| Is this pair a solution of the system | Substitute into both equations |
| Solve by graphing | Rewrite both into slope-intercept form |
| The lines seem to cross here | Check the coordinates in both originals |
| When will the two be equal | Write a model for each, then find the crossing |
| The check fails | Reread the graph or solve algebraically |
Lesson 7.2 removes the guesswork. Instead of reading a crossing off a picture, it solves one equation for a variable and substitutes into the other, producing exact coordinates whatever the numbers are.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.1 Graphing Linear Systems §7.1, pp. 389-394 — everything on these slides traces back here
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