Solutions of two-variable equations as ordered pairs, checking a candidate pair, the standard form that makes an equation linear, rewriting into function form to find solutions easily, building a table of values, and plotting it to obtain the straight line that is the graph of the equation.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 4 — Graphing Linear Equations and Functions
Graphing Linear Equations
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. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 210-215 — the lesson these objectives are drawn from
Warm-up
Lesson 1.4 checked whether a number was a solution. This lesson checks whether a pair is.
Discussion prompt
In Lesson 1.4 you checked whether 2 was a solution of 4x plus 1 equals 9. Now consider x minus 2y equals negative 3 — what would a candidate solution even look like, and why?
Hint: Count the letters in the equation.
Answer:
\[ x - 2y = -3 \quad \text{at } (1, 2): \; 1 - 4 = -3 \;\checkmark \]
There are two letters, so a candidate has to supply a value for each — a pair rather than a single number. And a pair of numbers is exactly what names a point on the plane from Lesson 4.1, which is why the solutions of a two-variable equation can be drawn.
Concept
A solution of an equation in two variables is an ordered pair that makes the equation true. Because every ordered pair is a point, the whole collection of solutions can be drawn — and for a linear equation that collection is a straight line.
solution of an equation — For an equation in two variables, an ordered pair that makes the equation true when its two numbers are substituted for the two variables.
The graph of an equation is the set of all points that are solutions of it.
Figure (svg): A solution of a two-variable equation shown as a single point on the plane
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 210-211
Section
Section 1
Concept
To check whether an ordered pair is a solution, substitute its first number for x and its second for y, simplify, and read off true or false. The routine is Lesson 1.4's with one extra substitution.
Both substitutions happen before any judgement, and a pair fails if the resulting statement is false by any amount.
Figure (svg): Two candidate pairs substituted into an equation, one giving a true statement and one false
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 210-210 — Example 1, Check Solutions of Linear Equations
Picture it
Both columns do the same work; only the verdict differs.
Figure (svg): Two candidate pairs substituted into an equation, one giving a true statement and one false
The failed check is as informative as the successful one. It rules out a point, which means that point is not on the graph.
Worked example
This is Example 1 from the textbook. Two candidates for one equation.
\[ \text{Is } (1, 2) \text{ or } (7, 3) \text{ a solution of } \; x - 2y = -3? \]
Substitute the first pair
Why: One for x and two for y.
\[ 1 - 2(2) = -3 \]
Simplify and judge
Why: One minus four is negative three, which matches the right side.
Substitute the second pair
Why: Seven for x and three for y.
\[ 7 - 2(3) = -3 \]
Simplify and judge
Why: Seven minus six is one, and one is not negative three.
Figure (svg): Two candidate pairs substituted into an equation, one giving a true statement and one false
\[ (1, 2) \;\checkmark \qquad (7, 3) \;\times \]
Verify: say what each verdict means about the graph
Why: The point (1, 2) lies on the graph of the equation and (7, 3) does not. Every check of a pair is really a question about whether a point is on the line, which is what makes checking useful once the graph is drawn.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 210-210
Sorting
Substitute both coordinates into x minus 2y equals negative 3.
Sort into buckets
Sort each pair by whether it is a solution.
Four of the six are solutions, and they include a negative coordinate and a decimal one. There are infinitely many more, which is exactly why the graph is the useful way to describe them.
Worked example
Guided Practice 1. Four pairs tested against one equation.
\[ \text{Which of } (3, 7), \; (-3, -7), \; \left(\tfrac{1}{2}, 0\right), \; \left(\tfrac{5}{2}, 6\right) \text{ solve } 2x - y = -1? \]
Test (3, 7)
Why: Two times three minus seven is six minus seven, which is negative one.
Test (-3, -7)
Why: Negative six minus negative seven is negative six plus seven, which is one — not negative one.
Test the pair with a half
Why: Two times a half minus zero is one, not negative one.
Test the pair with five halves
Why: Two times five halves is five, minus six is negative one.
Figure (svg): The solution to Worked example four candidates from guided practice shown as a ladder of expressions, one row per algebraic move
\[ (3, 7) \;\checkmark \qquad \left(\tfrac{5}{2}, 6\right) \;\checkmark \]
Verify: notice that two very different pairs both work
Why: Three comma seven and five halves comma six are both solutions, and they look nothing alike. A two-variable equation has infinitely many solutions, which is precisely why they need a picture rather than a list.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 210-210
Trap
\[ \text{Is } (1, 2) \text{ a solution of } x - 2y = -3? \]
Put 2 in for x and 1 in for y
Why: The two numbers are both present, and nothing in the substitution itself announces which is which.
\[ 2 - 2(1) = 0 \neq -3 \quad \text{(wrong verdict)} \]
The pair really is a solution, and the check has rejected it. Every conclusion about the graph drawn from that verdict is now wrong.
\[ (1, 2): \; x = 1, \; y = 2 \;\Longrightarrow\; 1 - 2(2) = -3 \;\checkmark \]
Write down which value goes with which letter before substituting
Why: The first coordinate is always x, by the convention from Lesson 4.1.
Writing x equals one and y equals two on their own line costs a few seconds and makes the substitution mechanical rather than a memory test.
Faded example
The substitution is set up. Complete it and judge.
Fill in the blanks
x - 2y = -3 \text2 (1, 2): \quad 1 - 2(-3) = ___
Why: The y-coordinate of two goes into the 2y term, giving one minus four, which is negative three — matching the right side. Writing which coordinate goes where before substituting is what keeps the two from being swapped.
Elimination
Testing whether (4, 1) solves 3x minus y equals 11.
Eliminate the wrong options
Which substitution is right?
Survives elimination: A
Why: Four goes in for x and one for y, giving twelve minus one, which is eleven — so the pair is a solution. Option B is the common error and it produces a false verdict here, since three minus four is negative one rather than eleven.
Socratic
A one-variable linear equation had exactly one.
Discussion prompt
Explain why x minus 2y equals negative 3 has infinitely many solutions while 4x plus 1 equals 9 has only one. Then say how you would generate a solution of the two-variable equation on demand.
Hint: Count how much freedom each equation leaves you.
Answer:
A one-variable equation constrains its single unknown completely, so at most one value can satisfy it. A two-variable equation places one condition on two unknowns, which leaves one degree of freedom — you may choose either variable freely, and the equation then determines the other.
So to generate a solution, pick any value for x and solve the resulting one-variable equation for y. Choosing x equal to five gives five minus 2y equals negative three, so y is four and the pair (5, 4) is a solution. Every choice of x produces one, which is why there are infinitely many and why they form a line rather than a scatter of isolated points.
Section
Section 2
Concept
A linear equation in x and y is one that can be written as Ax plus By equals C, where A and B are not both zero. That form is exactly the condition guaranteeing a straight-line graph.
\[ Ax + By = C \]
No variable is raised to a power, no two variables are multiplied together, and no variable sits in a denominator.
Figure (svg): The standard form of a linear equation with its parts labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 210-210 — the definition of a linear equation in standard form
Picture it
Two coefficients, one constant, and one condition.
Figure (svg): The standard form of a linear equation with its parts labelled
The requirement that A and B are not both zero is what stops the form collapsing into a statement about nothing. If both were zero the equation would say C equals zero, which is not about x or y at all.
Worked example
Testing against the standard form is a matter of what the equation contains.
\[ \text{Which are linear: } \; 2x + 3y = 6, \quad y = x^2, \quad xy = 4, \quad y = 3x - 4? \]
Check the first
Why: It is already in the form Ax plus By equals C, with A two, B three and C six.
Check the second
Why: The x is squared, which the standard form does not allow.
Check the third
Why: The two variables are multiplied together, which the standard form does not allow.
Check the fourth
Why: It can be rewritten as negative 3x plus y equals negative four, which is the standard form.
Figure (svg): The solution to Worked example which equations are linear shown as a ladder of expressions, one row per algebraic move
\[ 2x + 3y = 6 \text{ and } y = 3x - 4 \text{ are linear} \]
Verify: test each with two solutions and see whether a line fits
Why: For the second equation, the pairs (1, 1), (2, 4) and (3, 9) are solutions and they do not lie on a straight line — the steps between the y-values are one and then five. The standard-form test and the straightness of the graph agree, which is the point of having the test.
Sorting
Check for powers, products of variables and variables in denominators.
Sort into buckets
Sort each equation by whether it is linear.
The three non-linear equations each fail in a different way, and each failure is visible in the equation without any graphing. That is what makes the standard form a usable test.
Worked example
Recognising a linear equation sometimes requires rearranging first.
\[ \text{Write } \; y = 3x - 4 \; \text{ in the standard form } Ax + By = C. \]
Move the variable term to the left
Why: Subtract 3x from both sides.
\[ -3 x + y = -4 \]
Identify the coefficients
Why: A is negative three, B is one, C is negative four.
\[ A = -3, B = 1, C = -4 \]
Check the condition
Why: A and B are not both zero, so the equation is linear.
Note that the form is not unique
Why: Multiplying through by negative one gives 3x minus y equals four, which is equally standard.
Figure (svg): The solution to Worked example put an equation into standard form shown as a ladder of expressions, one row per algebraic move
\[ -3x + y = -4 \;\Longleftrightarrow\; 3x - y = 4 \]
Verify: check a solution in both forms
Why: The pair (2, 2) satisfies y equals 3x minus 4, since six minus four is two. In the standard form, negative six plus two is negative four, and in the multiplied version six minus two is four. All three forms agree on the same solutions, which is what equivalent means.
Error analysis
The student judged four equations as linear or not. Two are wrong.
Annotate
On: \( \begin{aligned} 2x + 3y = 6 &: \; \text{linear} \\ y = x^2 &: \; \text{linear} \\ y = \tfrac{4}{x} &: \; \text{linear} \\ y = 3x - 4 &: \; \text{linear} \end{aligned} \)
The test is about what the equation contains rather than how it is arranged. Powers, products of variables and variables in denominators all disqualify it, whatever rearranging is done.
Elimination
Standard form is Ax plus By equals C.
Eliminate the wrong options
Which equation is NOT written in standard form?
Survives elimination: A
Why: Option A has a variable isolated on one side, which is function form rather than standard form. It is still a linear equation — it becomes negative 3x plus y equals negative four after one move — but as written it does not match the Ax plus By equals C pattern.
Faded example
Read A, B and C off the standard form.
Fill in the blanks
2x + 3y = 6: \quad A = 2, \; B = 3, \; C = 6
Why: The coefficients are read directly off the equation once it is in standard form. The condition that A and B are not both zero holds here, since both are non-zero, so the equation is linear and its graph is a straight line.
Socratic
The connection between the algebra and the geometry deserves a reason.
Discussion prompt
Give an informal reason why an equation with no powers and no products of variables should graph as a straight line, using the idea of a constant step from Lesson 1.8. Then say what goes wrong for y equals x squared.
Hint: Look at what happens to y when x increases by one.
Answer:
In function form a linear equation reads y equals a number times x plus a number, so increasing x by one always changes y by that same coefficient. A constant step is exactly what a straight line is, as the balloon table in Lesson 1.8 showed — every unit right moves you the same distance up.
For y equals x squared the steps are not constant: going from x equal to one to two changes y from one to four, a step of three, while going from two to three changes it from four to nine, a step of five. The steps grow, so the graph bends. The absence of powers in the standard form is precisely what keeps the step constant.
Section
Section 3
Concept
A two-variable equation is in function form when one of its variables is isolated on one side. Rewriting into that form turns finding a solution from solving an equation into evaluating an expression.
\[ -2x + y = 3 \;\Longrightarrow\; y = 2x + 3 \]
The rearrangement uses exactly the moves from Lesson 3.7, treating x as a known quantity.
Figure (svg): Two columns contrasting an equation in function form with one that is not
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 211-211 — the Function Form paragraph and Example 2
Picture it
Both describe the same solutions; one is easier to substitute into.
Figure (svg): Two columns contrasting an equation in function form with one that is not
Standard form is the definition and function form is the working form. Almost every graphing task begins by converting from the first to the second.
Worked example
This is Example 2 from the textbook. Rewrite first, then choose values.
\[ \text{Find three solutions of } \; -2x + y = 3. \]
Rewrite in function form
Why: Add 2x to each side to isolate y.
\[ y = 2 x + 3 \]
Choose the easiest value of x
Why: Zero is easiest, since the 2x term vanishes.
\[ x = 0\text{ gives } y = 3 \]
Choose two more values
Why: One gives five, and two gives seven.
\[ (1, 5)\text{ and } (2, 7) \]
Write the three pairs
Why: Each choice of x produced exactly one y.
\[ (0, 3), (1, 5), (2, 7) \]
Figure (svg): An equation rearranged into function form before a table is built
\[ (0, 3), \; (1, 5), \; (2, 7) \]
Verify: check one pair in the original equation
Why: For (1, 5): negative two times one plus five is negative two plus five, which is three — matching the right side. Checking in the original rather than the rewritten form tests the rearrangement as well as the substitution.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 211-211
Translation
Isolate y in each equation.
Match the pairs
Why: Each rearrangement moves the x term across and then divides by the coefficient of y. The second and third produce fractions because their y-coefficients do not divide the other terms evenly, and the fourth needs a sign change since its y-coefficient is negative one.
Worked example
The rearrangement needs two moves when y is multiplied by something.
\[ \text{Write } \; 2x + 3y = 6 \; \text{ in function form.} \]
Move the x term to the right
Why: Subtract 2x from both sides.
\[ 3 y = -2 x + 6 \]
Divide by the coefficient of y
Why: Divide every term on both sides by three.
\[ y = (-\frac{2}{3}) x + 2 \]
Check every term was divided
Why: Both the 2x and the six had to be divided, not just one of them.
Note what this buys
Why: Any x may now be chosen and y read off directly.
Figure (svg): The solution to Worked example rewrite when y has a coefficient shown as a ladder of expressions, one row per algebraic move
\[ 2x + 3y = 6 \;\Longrightarrow\; y = -\tfrac{2}{3}x + 2 \]
Verify: substitute a convenient value into both forms
Why: Taking x equal to three gives y equal to negative two plus two, which is zero. In the original, two times three plus three times zero is six, which matches. Choosing an x divisible by three keeps the check free of fractions.
Trap
\[ 3y = -2x + 6 \]
Divide the 3y by 3 and leave the right side alone
Why: The three is on the left, so the left is where the correction seems to be needed.
\[ y = -2x + 6 \quad \text{(wrong)} \]
The balance is broken. Substituting x equal to three now gives y equal to zero in one form and negative twelve plus six in the other.
\[ \tfrac{3y}{3} = \tfrac{-2x + 6}{3} \;\Longrightarrow\; y = -\tfrac{2}{3}x + 2 \]
Divide every term on both sides by the coefficient
Why: Dividing a side means dividing each of its terms, exactly as in Lesson 3.5.
A quick check on one pair catches this immediately, which is why every rearrangement is worth testing on a single convenient value of x.
Faded example
Divide every term by the coefficient of y.
Fill in the blanks
3y = -2x + 6 \;\rightarrow\; y = -2/3x + 2
Why: Both terms on the right are divided by three, giving negative two thirds x plus two. Dividing only one of them is the commonest error here, and substituting x equal to three catches it at once since the two versions then disagree.
Elimination
The equation is y equals negative two thirds x, plus 2.
Eliminate the wrong options
Which choice of x avoids fractions?
Survives elimination: A
Why: Choosing x as a multiple of the denominator makes the fraction cancel: three gives negative two plus two, which is zero. When a coefficient is a fraction, choosing inputs divisible by its denominator keeps the whole table free of fractions — and makes the points far easier to plot.
Socratic
Both forms describe the same solutions.
Discussion prompt
Explain what function form makes easy that standard form does not, using the equation 2x plus 3y equals 6. Then say when you would leave an equation in standard form instead.
Hint: Think about generating ten solutions from each form.
Answer:
In standard form, choosing x equal to one leaves three y equals four, which is a small equation to solve — and doing that ten times means ten solves. In function form the same choice gives y directly by evaluating an expression, so ten solutions cost ten substitutions and no algebra at all.
Standard form is worth keeping when you want the intercepts, which Lesson 4.4 will show are read off it very easily, or when comparing two equations for a system in Chapter 7. The forms are tools rather than rivals, and which one to use depends on the question.
Section
Section 4
Concept
A table of values records several solutions at once. Choose values of x that include negatives, zero and positives, so that the graph's behaviour on both sides of the y-axis is visible.
Choosing inputs that avoid fractions makes the points far easier to plot accurately.
Figure (svg): A table of five solutions of y equals 3x minus 2, plotted and joined into a line
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 211-211 — the Study Tip on choosing values of x, and Example 3
Picture it
Five columns become five points, and the points fall on one line.
Figure (svg): A table of five solutions of y equals 3x minus 2, plotted and joined into a line
The five chosen inputs run from negative two to two, so the picture shows what the graph does on both sides of the vertical axis rather than only on one.
Worked example
This is Example 3 from the textbook. Five inputs spanning zero.
\[ \text{Make a table of values for } \; y = 3x - 2 \; \text{ at } x = -2, -1, 0, 1, 2. \]
Note that the equation is already in function form
Why: y is isolated, so no rearranging is needed.
\[ \text{already } y =... \]
Substitute the negative inputs
Why: Three times negative two minus two is negative eight; three times negative one minus two is negative five.
\[ -8\text{ and } -5 \]
Substitute zero
Why: Three times zero minus two is negative two.
\[ -2 \]
Substitute the positive inputs
Why: One gives one, and two gives four.
\[ 1\text{ and } 4 \]
Figure (svg): A table of five solutions of y equals 3x minus 2, plotted and joined into a line
\[ y = -8, -5, -2, 1, 4 \quad \text{for } x = -2, -1, 0, 1, 2 \]
Verify: check the step between consecutive outputs
Why: Each output is three more than the one before it, matching the coefficient of x. A constant step confirms every column at once, and any irregular gap would point straight at the offending substitution.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 211-211
Pattern
Each frame adds one output to the table for y equals 3x minus 2.
Step through it
What is the step between consecutive outputs, and where in the equation does that number appear?
The step is three every time, and three is the coefficient of x. Chapter 4 will call that number the slope, and it is what makes the graph straight.
Worked example
The choice of inputs is a decision worth making deliberately.
\[ \text{Tabulate } \; y = -\tfrac{2}{3}x + 2 \; \text{ with whole-number outputs.} \]
Look at the denominator of the coefficient
Why: It is three, so multiples of three will cancel it.
\[ \text{use multiples of } 3 \]
Choose inputs accordingly
Why: Negative three, zero, three and six.
\[ -3, 0, 3, 6 \]
Substitute each one
Why: Two plus two is four; zero plus two is two; negative two plus two is zero; negative four plus two is negative two.
\[ 4, 2, 0, -2 \]
Note the benefit
Why: Every output is a whole number and every point is easy to plot exactly.
Figure (svg): The solution to Worked example choose inputs that avoid fractions shown as a ladder of expressions, one row per algebraic move
\[ (-3, 4), \; (0, 2), \; (3, 0), \; (6, -2) \]
Verify: check the step against the coefficient
Why: Each input rises by three and each output falls by two, which is exactly the coefficient of negative two thirds. The constant step confirms the table, and the whole numbers make the plotting exact rather than approximate.
Trap
\[ y = 3x - 2 \text{ at } x = 1, 2, 3, 4 \]
Choose the first few counting numbers, since they are easiest
Why: Positive whole numbers are the most familiar inputs and the substitutions are simple.
The four points all sit on the right of the vertical axis, so the drawn portion of the line shows nothing about its behaviour on the left — and a plotting error there would go unnoticed.
\[ y = 3x - 2 \text{ at } x = -2, -1, 0, 1, 2 \]
Include negatives, zero and positives so the graph is seen on both sides
Why: The line extends in both directions, and a table spanning zero shows both.
Zero is worth including for its own reason: it gives the point where the graph crosses the vertical axis, which Lesson 4.4 will name and use.
Faded example
Two outputs are missing. Substitute and supply them.
Fill in the blanks
y = 3x - 2: \quad x = -2 \rightarrow -8, \; x = -1 \rightarrow -5, \; x = 0 \rightarrow -2, \; x = 1 \rightarrow 1, \; x = 2 \rightarrow 4
Why: At negative two the output is negative six minus two, which is negative eight, and at two it is six minus two, which is four. Both can also be found by continuing the constant step of three in each direction, and getting the same answer two ways is a genuine check.
Elimination
You are tabulating y equals negative two thirds x, plus 2.
Eliminate the wrong options
Which inputs give whole-number outputs and span the y-axis?
Survives elimination: A
Why: Multiples of three cancel the denominator, giving whole outputs, and including negative three shows the graph on both sides of the vertical axis. Option D satisfies one requirement and not the other, which is worth noticing — the two considerations are independent and both matter.
Socratic
Zero is one of the easiest inputs and also one of the most informative.
Discussion prompt
Give two reasons for always including x equal to zero in a table of values. Then say what the corresponding y-value tells you about the equation in function form.
Hint: One reason is about arithmetic and one is about the graph.
Answer:
First, it is the easiest substitution: the term containing x vanishes, so the output is just the constant. Second, it gives the point where the graph crosses the vertical axis, which is a landmark you can check by eye once the line is drawn.
In function form the y-value at zero is exactly the constant term. For y equals 3x minus 2 it is negative two, and the line does cross the vertical axis at negative two. Lesson 4.4 will call that number the y-intercept and use it to graph a line without any table at all.
Section
Section 5
Concept
The graph of an equation is the set of all points that are solutions. For a linear equation those points form a straight line, so plotting a few and drawing a line through them displays every solution at once.
Two points fix a line and a third one checks it, which is why a table should have at least three rows.
Figure (svg): Two points determining a line, with a third point used as a check
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 211-211 — the Graphs of Linear Equations paragraph and Example 3
Picture it
The third point is the one that catches an error.
Figure (svg): Two points determining a line, with a third point used as a check
Any two points can be joined by a line, whether or not either is correct. Only a third point can tell you that the line is right.
Worked example
The table from the previous section, plotted and joined.
\[ \text{Graph } \; y = 3x - 2 \; \text{ using the table of values.} \]
Plot the five pairs
Why: From negative two comma negative eight up to two comma four.
Check they are collinear
Why: Each point is one right and three up from the previous one, so they line up.
Draw one straight line through them
Why: Extend it past the outermost points, since the solutions continue in both directions.
State what the line represents
Why: Every point on it is a solution, and every solution is on it.
Figure (svg): A table of five solutions of y equals 3x minus 2, plotted and joined into a line
\[ \text{the line through } (-2, -8) \text{ and } (2, 4) \]
Verify: test a point the table did not use
Why: The pair (3, 7) should be on the line, and substituting gives nine minus two, which is seven — so it is. Testing a point beyond the table confirms that the drawn line really does carry solutions the table never listed.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 211-211
Elimination
Two points are plotted and joined, and a third computed solution misses the line.
Eliminate the wrong options
What is the most likely explanation?
Survives elimination: A
Why: Three solutions of a linear equation must be collinear, so a miss means one of them is wrong. Rechecking each substitution finds it quickly, and that is precisely the job the third point exists to do — two points can never disagree with each other.
Worked example
Two points were enough to draw the line. The third is what verifies it.
\[ \text{Graph } \; y = x + 2 \; \text{ using } (-2, 0) \text{ and } (2, 4), \text{ then check with } x = 0. \]
Plot the two given points and draw a line
Why: Two points determine exactly one line.
Compute a third solution
Why: At x equal to zero, y is two.
\[ (0, 2) \]
Check whether it lies on the drawn line
Why: The line passes through zero comma two, so it does.
Say what a miss would have meant
Why: If the third point were off the line, one of the three pairs is wrong and the table must be rechecked.
Figure (svg): Two points determining a line, with a third point used as a check
\[ (0, 2) \text{ lies on the line through } (-2, 0) \text{ and } (2, 4) \]
Verify: consider what the check can and cannot catch
Why: The check catches an arithmetic error in any one of the three substitutions, since a single wrong point will not lie on the line through the other two. It cannot catch an error in the original equation, which is why the equation itself is worth rereading if the graph looks unexpected.
Trap
Five points are plotted and the student joins them dot to dot with short segments.
Connect each point to the next, as in a scatter or a dot-to-dot drawing
Why: Joining adjacent points is the natural instinct once several are on the page.
A linear equation's graph is one straight line, not a chain of segments. If the segments bend, one of the points is plotted wrongly and the bend is the evidence.
Check that the points are collinear, then draw a single straight line through all of them and beyond.
Use a ruler and let the line extend past the outermost points
Why: The solutions continue forever in both directions, so the drawing should indicate that.
A bend in a dot-to-dot join is not a feature of the equation — it is a plotting error announcing itself, and it is worth investigating rather than drawing.
Prediction
The line is the graph of y equals 3x minus 2.
Predict first
Is the point (4, 10) on the line?
Correct: Yes, since 3 times 4 minus 2 is 10.
\[ y = 3(4) - 2 = 10 \;\Longrightarrow\; (4, 10) \text{ is on the line} \]
Why: Substituting four gives twelve minus two, which is ten, so the pair satisfies the equation and lies on the line. The table listed only five of infinitely many solutions, so a pair's absence from the table says nothing — and substituting is faster and more reliable than extending a drawing.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| A table of values | The graph | |
|---|---|---|
| How many solutions it shows | as many as it has columns | all of them at once |
| Best at | exact values at chosen inputs | showing the shape and the trend |
| How to test a new pair | substitute into the equation | see whether the point is on the line |
The two carry the same information in different forms, exactly as in Lesson 1.8. Neither replaces the other, and moving between them is the skill.
Socratic
The table stopped at x equal to two, and the line does not.
Discussion prompt
Explain why the drawn line should extend beyond the outermost plotted points, and say what would justify stopping it at a particular place instead.
Hint: Ask which pairs are solutions.
Answer:
Every choice of x gives a solution, including values far outside the table, so the solution set continues forever in both directions. Stopping the line at the last plotted point would suggest that solutions stop there too, which is false — arrows or an extended line record that the graph goes on.
What would justify stopping is a restriction from the situation, exactly as the balloon's five-minute burn restricted its domain in Lesson 1.8. If x counts something that cannot be negative, or the model only holds up to some value, the graph should stop there and the restriction should be stated alongside the equation.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| One variable | Two variables | |
|---|---|---|
| A solution is | a number | an ordered pair |
| How many solutions | usually exactly one | infinitely many |
| How to display them all | state the single number | draw the graph |
The jump from one variable to two is the jump from an answer to a picture, and it is what the rest of this chapter is built on.
Pattern
Whether you are checking a pair or drawing a whole graph, the same five moves cover it.
Step four's constant-step check verifies every row at once, and it is faster than rechecking the substitutions one at a time.
OpenStax Elementary Algebra 2e, §4.2 Graph Linear Equations in Two Variables §4.2
Check
Checking a pair. Substitute both coordinates.
Check your understanding
Is (2, 5) a solution of 4x minus y equals 3?
Answer: A
Why: Substituting two for x gives eight, and subtracting the y-value of five gives three, which matches the right side. The pair is a solution, so the point lies on the graph of the equation.
Check
Function form. Divide every term.
Check your understanding
Write 4x plus 2y equals 10 in function form.
Answer: A
Why: Subtracting 4x gives 2y equals negative 4x plus 10, and dividing every term by two gives y equals negative 2x plus 5. Substituting x equal to one gives y equal to three, and four plus six is ten, confirming it.
Check
A table of values. Check the step.
Check your understanding
For y equals 2x plus 1, what are the outputs at x equal to -1, 0 and 1?
Answer: A
Why: Substituting gives negative two plus one, zero plus one, and two plus one, which are negative one, one and three. The step between consecutive outputs is two, matching the coefficient of x — a constant step, as a linear equation requires.
Real world
A taxi charges 3 dollars to get in plus 2 dollars a mile, so the fare y for m miles satisfies y equals 2m plus 3.
Discussion prompt
Build a table for zero to four miles and describe the graph. Then say which part of the plane the graph should actually occupy, and what the point where it meets the vertical axis means in the situation.
Hint: Think about whether a negative number of miles makes sense.
Answer:
\[ m = 0, 1, 2, 3, 4 \;\Longrightarrow\; y = 3, 5, 7, 9, 11 \]
The graph is a straight line rising two dollars for every mile. Only the part with m at least zero belongs in the picture, since a negative number of miles is meaningless — so the graph is a ray starting on the vertical axis rather than a full line.
The point where it meets the vertical axis is zero comma three, which is the fare for a journey of no miles: the three dollar boarding charge. That is the constant term of the equation appearing as a landmark on the graph, and Lesson 4.4 will make systematic use of it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
How many solutions does the equation x minus 2y equals negative 3 have?
Correct: Infinitely many, one for every choice of x.
\[ x = 1 \rightarrow (1, 2), \quad x = 3 \rightarrow (3, 3), \quad x = -3 \rightarrow (-3, 0) \]
Three of infinitely many, and all three lie on one line.
Why: One equation constraining two unknowns leaves one degree of freedom, so any value may be chosen for x and the equation then determines y. Every such choice gives a solution, and together they form the straight line that is the graph. The instinct that an equation has one answer comes from Chapter 3, where every equation had a single variable — and that is exactly what changes here.
Explain it
They can solve one-variable equations and plot points, and have never seen the two ideas combined.
Discussion prompt
In no more than four sentences, explain what it means for an ordered pair to solve an equation and why such an equation has infinitely many solutions. Then tell them the quickest way to find one.
Hint: Start from how many letters the equation has.
Answer:
A usable answer: an equation with two letters needs two numbers to test it, so a candidate answer is a pair rather than a single number. One equation cannot pin down two unknowns, so you are free to pick either letter's value and the equation then decides the other. Every pick gives a different solution, which is why there are endlessly many.
The quickest way to find one is to set x to zero, because whatever multiplies x then disappears and the remaining equation has one letter in it. That single substitution gives a solution in a few seconds, and it happens to be the point where the graph crosses the vertical axis.
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: Substitution order is fixed by writing x equals and y equals on their own line before touching the equation. Function form is fixed by dividing every term when y has a coefficient, and checking one pair afterwards. Choosing inputs is fixed by spanning zero and preferring multiples of any denominator present. Recognising linearity is fixed by checking for powers, products of variables and variables underneath — three things to look for and nothing 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
At the top of a page write one linear equation in standard form and rearrange it into function form, showing every move. Underneath, build a table of at least five values with inputs spanning zero, and write the constant step between the outputs beside it. To the right, draw a coordinate plane, plot the five points, and draw one straight line through them extending past the outermost points. Circle the point where your line crosses the vertical axis and write beside it the number from your equation that it matches. Near the bottom, write two equations that are not linear and beside each the reason it fails the standard-form test. Finally, in the margin, write one ordered pair that is not in your table, and show by substitution whether it is on your line.
The circled crossing point should match the constant term of your function-form equation. If it does not, either the rearrangement or the plotting has gone wrong, and the table's constant step will tell you which.
Recap
Five things, and the first is the shift the whole chapter turns on.
| If the question says | Your first move is |
|---|---|
| Is this pair a solution | Write x equals and y equals, then substitute |
| Find three solutions | Rewrite in function form, then pick values of x |
| Use a table to graph | Choose inputs spanning zero, avoiding fractions |
| Is this equation linear | Look for powers, products and denominators |
| Draw the graph | Plot, check collinear, then one straight line |
Lesson 4.3 looks at the two special cases the table method handles awkwardly: equations in which only one variable appears, whose graphs turn out to be horizontal and vertical lines.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.2 Graphing Linear Equations §4.2, pp. 210-215 — everything on these slides traces back here
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