The substitution method: solving one equation for one variable, substituting that expression into the other equation to reduce the system to one variable, back-substituting to recover the second coordinate, and checking in both originals. Includes choosing which variable to isolate and why the method gives exact answers a graph cannot.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 7 — Systems of Linear Equations and Inequalities
Solving Linear Systems by Substitution
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 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-401 — the lesson these objectives are drawn from
Warm-up
Lesson 7.1 read solutions off graphs and could only estimate them. This lesson replaces the estimate with algebra.
Discussion prompt
In the system x plus y equals 1 and 2x minus y equals 2, solve the first equation for y. What happens if you put that expression into the second equation?
Hint: Count the letters afterwards.
Answer:
\[ y = -x + 1 \;\Longrightarrow\; 2x - (-x + 1) = 2 \;\Longrightarrow\; 3x - 1 = 2 \]
Only x is left, so the problem has become an ordinary Chapter 3 equation. Reducing two variables to one is what the whole method is for, and everything else is bookkeeping.
Concept
The substitution method solves one equation for one of its variables and puts the resulting expression into the other equation. That removes a variable, leaving a one-variable equation that can be solved directly.
substitution method — An algebraic method for solving a linear system: solve one equation for one variable, substitute that expression into the other equation, solve, and substitute back.
It gives exact answers whether or not the crossing falls on a grid point.
Figure (svg): A substitution turning a two-variable equation into a one-variable one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-396
Section
Section 1
Concept
Solve one of the equations for one of its variables, substitute the expression into the other equation and solve, substitute that value into the revised equation, and check the pair in both originals.
Each step is a Chapter 3 move; only the order is new.
Figure (svg): The four steps of the substitution method
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 397-397 — the Solving a Linear System by Substitution summary
Picture it
Two of reduction, one of recovery, one of proof.
Figure (svg): The four steps of the substitution method
The third step is where the second coordinate comes from, and it is the one most often skipped — an answer of x equals one is half a solution.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \; x + y = 1 \; \text{ and } \; 2x - y = 2. \]
Solve the first equation for y
Why: Subtract x from each side.
\[ y = -x + 1 \]
Substitute into the second
Why: Replace y with negative x plus one.
\[ 2 x - (-x + 1) = 2 \]
Solve for x
Why: Combining gives 3x minus one equals two.
\[ x = 1 \]
Substitute back into the revised equation
Why: y equals negative one plus one.
\[ y = 0 \]
Figure (svg): A substitution turning a two-variable equation into a one-variable one
\[ (1, 0) \]
Verify: check in both original equations
Why: One plus zero is one, and two minus zero is two. Both hold, so the pair is the solution. Checking in the originals rather than in the revised equation tests the rearrangement as well as the arithmetic.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-396
Sorting
Four steps, in a fixed order.
Sort into buckets
Sort each action by the step it belongs to.
Step four has two actions because there are two original equations, and step two has two because substituting and solving are separate moves. Only step three is a single action.
Worked example
Naming the purpose of each step makes the method memorable.
\[ \text{Describe what each of the four steps does.} \]
Step one
Why: Produces an expression for one variable in terms of the other.
Step two
Why: Removes that variable from the other equation, leaving one unknown.
Step three
Why: Uses the found value to recover the other coordinate.
Step four
Why: Confirms the pair against both original equations.
Figure (svg): The four steps of the substitution method
\[ \text{reduce} \to \text{solve} \to \text{recover} \to \text{confirm} \]
Verify: ask what would be missing without step three
Why: You would have one number and the question asks for a pair. The value of x names the vertical line the solution sits on, and the second coordinate is what locates it on that line — so step three is not a formality but half the answer.
Trap
\[ 3x - 1 = 2 \;\Longrightarrow\; x = 1 \]
Report the solution as x = 1
Why: A number has been found and the equation is solved, so the work feels complete.
A system's solution is an ordered pair. One coordinate names a vertical line, not a point, and the question asked where two lines cross.
\[ x = 1 \;\Longrightarrow\; y = -1 + 1 = 0 \;\Longrightarrow\; (1, 0) \]
Substitute the value back into the revised equation
Why: That is the third step, and it produces the second coordinate.
Writing the answer as an ordered pair from the start makes an unfinished solve visible immediately.
Faded example
Replace y with the expression.
Fill in the blanks
y = -x + 1 \;\Longrightarrow\; 2x - (-x + 1) = 2 \;\Longrightarrow\; 3x - 1 = 2
Why: Subtracting the bracket changes both signs inside it, giving 2x plus x minus one, which is 3x minus one. The brackets are what make that sign change happen correctly.
Elimination
Substitution produced x equal to 1.
Eliminate the wrong options
What should be reported?
Survives elimination: A
Why: The solution is a single point, written as an ordered pair with x first. Option D is worth naming because the two numbers alone are ambiguous, and the convention from Lesson 4.1 is what removes the ambiguity.
Socratic
The step looks like bookkeeping and does real work.
Discussion prompt
Explain why replacing y with an expression in x turns a two-variable equation into a one-variable one. Then say why the resulting equation still has the same solution as the system.
Hint: Ask what the expression is asserting.
Answer:
The revised equation says that y equals that expression, so wherever a y appears you may write the expression instead without changing anything. After the replacement the second equation mentions only x, so it is a one-variable equation of the kind Chapter 3 solved.
The replacement is valid only for pairs satisfying the first equation, which is exactly the pairs the system is about. So the one-variable equation's solution is the x-coordinate of any point satisfying both — and since the two lines cross once, that is a single value. The method never adds or loses solutions, which is why the answer it produces is the answer.
Section
Section 2
Concept
Either variable in either equation may be isolated, and the answer is the same. Choosing one whose coefficient is one or negative one keeps the working free of fractions.
The textbook's Study Tip says to begin with the variable that is easier to isolate.
Figure (svg): Two columns on which variable to solve for first
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 397-397 — the Study Tip on choosing which variable to solve for first
Picture it
A coefficient of one costs nothing.
Figure (svg): Two columns on which variable to solve for first
Both columns reach the same answer. The right-hand one simply carries fractions through every subsequent line, which is more work and more risk for no benefit.
Worked example
This is Example 2 from the textbook, where the second equation is the easy one.
\[ \text{Solve } \; -2x + 2y = 3 \; \text{ and } \; x + 4y = 1. \]
Scan the coefficients
Why: The x in the second equation has a coefficient of one.
Solve the second equation for x
Why: Subtract 4y from each side.
\[ x = -4 y + 1 \]
Substitute into the first
Why: Brackets around the whole expression.
\[ -2(-4 y + 1) + 2 y = 3 \]
Solve for y, then back-substitute
Why: Ten y minus two is three, so y is a half; then x is negative one.
\[ (-1, \frac{1}{2}) \]
Figure (svg): An expression, not a number, substituted into the second equation
\[ \left(-1, \tfrac{1}{2}\right) \]
Verify: check in both originals
Why: For the first: negative two times negative one is two, plus two times a half is one, giving three. For the second: negative one plus two is one. Both hold, and the fractional coordinate is exactly the kind of answer a graph could not have produced.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 397-397
Sorting
Look for a coefficient of one or negative one.
Sort into buckets
Sort each system by the variable that is easiest to isolate.
Only one system here forces fractions, and it is exactly the kind of system the next lesson's method handles more comfortably. Scanning the coefficients is also how you choose between methods.
Worked example
Guided Practice 1 to 3. Naming the variable to isolate, with a reason.
\[ \text{Which variable would you solve for first in } \; 3x + y = 9, \; 2x + 4y = 8; \quad x - 3y = 11, \; 2x + 5y = 33; \quad x + 3y = 0, \; x - 2y = 10? \]
Take the first system
Why: The y in the first equation has a coefficient of one.
\[ y\text{ in Equation } 1 \]
Take the second
Why: The x in the first equation has a coefficient of one.
\[ x\text{ in Equation } 1 \]
Take the third
Why: Both equations have an x with coefficient one.
State the principle
Why: A coefficient of one means no division and no fractions.
Figure (svg): The solution to Worked example three from guided practice shown as a ladder of expressions, one row per algebraic move
\[ y, \quad x, \quad \text{either } x \]
Verify: try the awkward choice on the first system
Why: Isolating x in 3x plus y equals 9 gives x equals negative one third y plus three, and every later line carries thirds. The answer is the same and the arithmetic is three times as error-prone, which is the whole argument for scanning first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 397-397
Error analysis
The student substituted an expression into the other equation.
Annotate
On: \( \begin{aligned} x &= -4y + 1 \\ -2x + 2y &= 3 \\ -2 \cdot -4y + 1 + 2y &= 3 \\ 8y + 1 + 2y &= 3 \end{aligned} \)
Writing the substituted expression in brackets before simplifying is the single habit that prevents this. It costs one pair of brackets and saves the whole solve.
Faded example
A coefficient of one needs no division.
Fill in the blanks
x + 4y = 1 \;\Longrightarrow\; x = -4y + 1
Why: Subtracting 4y isolates x with no division at all, so no fractions enter the working. Isolating y in the same equation would have required dividing by four and carried quarters through every later line.
Elimination
The system is 3x + y = 9 and 2x + 4y = 8.
Eliminate the wrong options
Which variable should you isolate?
Survives elimination: A
Why: The y in the first equation has a coefficient of one, so subtracting 3x isolates it with no division. All four choices reach the same answer, and only one of them does so without fractions.
Socratic
Four different starting points, one solution.
Discussion prompt
Explain why isolating any of the four variables leads to the same ordered pair. Then say what does change between the four routes.
Hint: Ask what the method is finding.
Answer:
All four routes are looking for the same thing: the pair satisfying both equations, which is the single point where the two lines cross. Each route is a valid sequence of steps that preserves the solution set, so all four must arrive at that point.
What changes is the arithmetic. A coefficient of one gives whole numbers throughout, while a coefficient of three gives thirds in every subsequent line, and one wrong third produces a wrong answer. So the choice affects the chance of getting there rather than where there is.
Section
Section 3
Concept
What gets substituted is an expression rather than a number, so it must be enclosed in brackets. Whatever multiplies the variable then multiplies every term of the expression.
\[ -2x + 2y = 3, \; x = -4y + 1 \;\Longrightarrow\; -2(-4y + 1) + 2y = 3 \]
Dropping the brackets changes which terms are multiplied.
Figure (svg): An expression, not a number, substituted into the second equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 397-397 — Example 2, where the substituted expression is bracketed
Picture it
One expression, multiplied whole.
Figure (svg): An expression, not a number, substituted into the second equation
This is the distributive property from Lesson 2.6 doing exactly what it always does. The only novelty is that the bracket arrived from another equation.
Worked example
The substitution and the expansion are separate steps.
\[ \text{Substitute } x = -4y + 1 \text{ into } \; -2x + 2y = 3 \; \text{ and simplify.} \]
Write the substitution with brackets
Why: The whole expression replaces x.
\[ -2(-4 y + 1) + 2 y = 3 \]
Distribute
Why: Negative two times negative 4y is 8y; negative two times one is negative two.
\[ 8 y - 2 + 2 y = 3 \]
Combine like terms
Why: 8y plus 2y is 10y.
\[ 10 y - 2 = 3 \]
Solve
Why: Add two, then divide by ten.
\[ y = \frac{1}{2} \]
Figure (svg): An expression, not a number, substituted into the second equation
\[ y = \tfrac{1}{2} \]
Verify: check the sign of each distributed term
Why: Negative two times a negative gives a positive 8y, and negative two times a positive gives a negative two. Both signs changed, which is what multiplying by a negative does — and getting only one of them is the standard slip.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 397-397
Translation
Brackets first, then distribute.
Match the pairs
Why: In each case the substituted expression is multiplied whole by whatever stood in front of the variable. The first two involve a negative multiplier, so both terms inside the bracket change sign — which is where the brackets earn their place.
Worked example
Example 1's substitution has a minus sign rather than a coefficient.
\[ \text{Substitute } y = -x + 1 \text{ into } \; 2x - y = 2. \]
Write it with brackets
Why: The whole expression is being subtracted.
\[ 2 x - (-x + 1) = 2 \]
Distribute the minus sign
Why: Both signs inside the bracket flip.
\[ 2 x + x - 1 = 2 \]
Combine like terms
Why: 2x plus x is 3x.
\[ 3 x - 1 = 2 \]
Solve
Why: Add one, divide by three.
\[ x = 1 \]
Figure (svg): A substitution turning a two-variable equation into a one-variable one
\[ x = 1 \]
Verify: check what happens without the brackets
Why: Writing 2x minus negative x plus one would give 2x plus x plus one, so the constant keeps the wrong sign and the answer comes out as one third instead of one. The brackets are what tell the minus sign to act on both terms.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-396
Trap
\[ x = -4y + 1 \text{ into } -2x + 2y = 3 \]
Write -2 times -4y + 1 + 2y = 3 without brackets
Why: The expression is written straight into the place the variable occupied.
\[ 8y + 1 + 2y = 3 \quad \text{(wrong)} \]
The one was never multiplied by negative two, so the constant is wrong by three and the answer for y comes out as a fifth rather than a half.
\[ -2(-4y + 1) + 2y = 3 \;\Longrightarrow\; 8y - 2 + 2y = 3 \]
Write brackets around the substituted expression before simplifying
Why: Whatever multiplied the variable multiplies the whole expression.
Checking the answer in both originals catches this, since the wrong value fails the first equation.
Faded example
Both terms inside the bracket.
Fill in the blanks
-2(-4y + 1) = 8y - 2
Why: Negative two times negative 4y gives positive 8y, and negative two times positive one gives negative two. Both terms were multiplied, which is exactly what the brackets required.
Elimination
Substituting y = 2x - 3 into 4x - 3y = 1.
Eliminate the wrong options
Which line is right?
Survives elimination: A
Why: The brackets enclose the whole expression so that the negative three multiplies both of its terms, giving 4x minus 6x plus nine. Options B and C are the same error written two ways, which is a reminder that the bracket has to close after the whole expression.
Socratic
It is only a pair of marks.
Discussion prompt
Explain what the brackets are recording when an expression is substituted, and why omitting them changes the equation rather than just the notation. Then say what earlier lesson the expansion comes from.
Hint: Ask what the coefficient was multiplying before the substitution.
Answer:
Before the substitution, the coefficient multiplied the single quantity y. After it, that quantity is an expression with several terms, and the coefficient still multiplies the quantity — which now means all of it. The brackets record that the expression is one object, and without them the coefficient attaches to only the first term, which is a different equation with a different solution.
The expansion is the distributive property from Lesson 2.6, unchanged. What is new is only that the bracket arrived by substitution rather than being written in the problem, and that is why it is easy to forget it should be there at all.
Section
Section 4
Concept
Once one variable is found, substitute it into the revised equation from the first step to get the other. Then check the pair in both original equations.
Checking against the revised equation would confirm an error made while producing it.
Figure (svg): A found value substituted back into the revised equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-397 — steps 3 and 4 of the summary
Picture it
The revised equation is used again.
Figure (svg): A found value substituted back into the revised equation
Keeping the revised equation labelled makes this step trivial. Rederiving it, or substituting into the wrong equation, is where the extra work creeps in.
Worked example
The third step of Example 1.
\[ \text{Given } x = 1 \text{ and the revised equation } y = -x + 1, \text{ find } y. \]
Take the revised equation
Why: It already has y isolated.
\[ y = -x + 1 \]
Substitute the found value
Why: Negative one plus one.
\[ y = -1 + 1 \]
Simplify
Why: The result is zero.
\[ y = 0 \]
Write the ordered pair
Why: x first, then y.
\[ (1, 0) \]
Figure (svg): A found value substituted back into the revised equation
\[ (1, 0) \]
Verify: substitute into the other original equation instead
Why: Putting x equal to one into 2x minus y equals two gives two minus y equals two, so y is zero — the same answer by a different route. Either original equation recovers the second coordinate, and agreeing is a free check.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-396
Faded example
Use the revised equation.
Fill in the blanks
x = 1, \; y = -x + 1 \;\Longrightarrow\; y = -1 + 1 = 0
Why: Substituting one for x gives negative one plus one, which is zero, so the solution is the pair (1, 0). The revised equation from step one is what makes this a single line of work.
Worked example
The fourth step, on the fractional answer from Example 2.
\[ \text{Check } \left(-1, \tfrac{1}{2}\right) \text{ in } \; -2x + 2y = 3 \; \text{ and } \; x + 4y = 1. \]
Substitute into the first
Why: Negative two times negative one is two; two times a half is one.
\[ 2 + 1 = 3 \]
Judge it
Why: Three equals three.
Substitute into the second
Why: Negative one plus four times a half is negative one plus two.
\[ 1 = 1 \]
Judge it
Why: One equals one, so both hold.
Figure (svg): A solution pair checked in both original equations
\[ \left(-1, \tfrac{1}{2}\right) \text{ confirmed} \]
Verify: notice that a graph could not have produced this
Why: The y-coordinate is a half, which no hand-drawn crossing would resolve. The algebra produced it exactly and the check confirmed it exactly, which is the advantage substitution has over Lesson 7.1's method.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 397-397
Trap
\[ x + y = 1 \;\Longrightarrow\; y = -x + 1 \]
Check the answer by substituting into y = -x + 1
Why: That is the equation the second coordinate came from, so it is the one to hand.
The pair was constructed to satisfy that equation, so the check passes automatically and confirms nothing. If the rearrangement itself went wrong, this check would miss it entirely.
Check in both original equations as given
Why: Neither was used to produce the second coordinate directly, so both can genuinely fail.
The second original equation is the strongest test, since the pair was never built to satisfy it.
Elimination
The solution came from isolating y in the first equation.
Eliminate the wrong options
Which substitution tests the most?
Survives elimination: A
Why: The second original equation played no part in producing the answer, so it is the only one that can fail independently. Option C is not wrong — checking both is the instruction — and it is the weaker of the two.
Prediction
The pair satisfies the first equation and fails the second.
Predict first
What has most likely happened?
Correct: An arithmetic error somewhere in the solving.
The place to look first is the substitution step, where a dropped bracket changes the equation.
Why: A pair satisfying one equation and not the other lies on one line and off the other, which means the crossing was computed wrongly. Systems with no solution announce themselves differently — the variables cancel and leave a false statement, which Lesson 7.5 takes up. Isolating a different variable never changes the answer, only the arithmetic.
Socratic
Two numbers, in a fixed order.
Discussion prompt
Explain why a system's answer is written as an ordered pair rather than as two separate statements about x and y. Then say what would be lost by writing the two values in the other order.
Hint: Ask what the answer describes.
Answer:
The answer is a single point of the plane, and Lesson 4.1's convention names a point by an ordered pair. Writing it that way records that the two numbers belong together as the coordinates of one location, rather than being two unrelated facts.
Reversing the order names a different point. The pair (1, 0) is on the horizontal axis and (0, 1) is on the vertical one, and only one of them solves the system. The convention that x comes first is what makes a bare pair of numbers unambiguous, which is why it is worth keeping even when the two values are obvious.
Section
Section 5
Concept
Substitution produces the exact coordinates of the crossing whether or not it falls on a grid point. That is the limitation of Lesson 7.1's graphing method, removed.
Example 2's answer has a coordinate of one half, which no drawing would resolve.
Figure (svg): A fractional solution that a graph could not have read
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-401 — the lesson's opening remark that systems can be solved without graphs
Picture it
Exactly a half, found algebraically.
Figure (svg): A fractional solution that a graph could not have read
Reading this crossing off a graph would have given about negative one comma a half, with no way to know whether the half was exact. The algebra says it is.
Worked example
Example 2's solution has a fractional coordinate.
\[ \text{Why could a graph not have produced } \left(-1, \tfrac{1}{2}\right)? \]
Look at the coordinates
Why: One is a whole number and one is a half.
Consider a hand-drawn reading
Why: A half can be estimated and not confirmed.
Consider a nearby value
Why: A crossing at 0.48 would look identical.
State the advantage
Why: Substitution gives the exact value.
Figure (svg): A fractional solution that a graph could not have read
\[ \text{graph: about } \tfrac{1}{2}; \quad \text{algebra: exactly } \tfrac{1}{2} \]
Verify: ask what the graph is still contributing
Why: It shows that the lines do cross, and roughly where, so it confirms the algebraic answer is in the right region. A sketch is a fast sanity check on a computed answer even when it cannot produce one.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Graphing (7.1) | Substitution (7.2) | |
|---|---|---|
| What the answer is | an estimate | exact |
| Handles fractional crossings | poorly | exactly |
| Shows how many solutions exist | at a glance | not directly |
Each method has the strength the other lacks, which is why both are worth having. A sketch alongside an algebraic solve catches errors neither would catch alone.
Worked example
The method does not care whether the answer is tidy.
\[ \text{Solve } \; y = 2x + 1 \; \text{ and } \; 3x + y = 6. \]
Note the first equation is already solved for y
Why: No rearranging needed.
\[ y = 2 x + 1 \]
Substitute into the second
Why: 3x plus the bracket equals six.
\[ 3 x + (2 x + 1) = 6 \]
Solve for x
Why: 5x plus one is six, so x is one.
\[ x = 1 \]
Back-substitute
Why: y equals two plus one.
\[ (1, 3) \]
Figure (svg): The solution to Worked example solving a system with awkward numbers shown as a ladder of expressions, one row per algebraic move
\[ (1, 3) \]
Verify: check in both originals
Why: Two times one plus one is three, and three plus three is six. Both hold. Notice that the first equation was already in the form step one produces, so the method skipped straight to step two — which is common when one equation is given in function form.
Trap
\[ y = \tfrac{1}{2} \]
Report the solution as (-1, 0.5) rounded to (-1, 1)
Why: Whole numbers look like proper answers, and a half seems close enough to one.
Substituting (-1, 1) into the first equation gives two plus two, which is four rather than three. The rounded pair is not a solution at all.
\[ \left(-1, \tfrac{1}{2}\right) \text{, exactly} \]
Keep fractional coordinates as fractions
Why: The exact value is what the method produced and what the check confirms.
Rounding is appropriate only when a real situation demands whole units, and then it should be stated as an approximation.
Faded example
Step one is sometimes already done.
Fill in the blanks
y = 2x + 1 \text2x + 1 3x + y = 6: \quad 3x + (5) = 6 \;\Longrightarrow\; ___x + 1 = 6
Why: The first equation was already solved for y, so the method starts at step two. Combining 3x and 2x gives 5x, and solving gives x equal to one, then y equal to three.
Hypothesis
Predict before you decide.
Predict first
For which system would substitution be least convenient?
Correct: 5x + 3y = 7 and 4x - 2y = 9.
\[ 5x + 3y = 7 \;\Longrightarrow\; x = \tfrac{7 - 3y}{5} \quad \text{fractions from the start} \]
Why: None of its four coefficients is one or negative one, so isolating any variable introduces fractions that persist through every later line. The other three each have a coefficient of one somewhere, or an equation already solved for a variable. This is exactly the shape Lesson 7.3's method is designed for.
Socratic
The chapter continues past this lesson.
Discussion prompt
Say what kind of system substitution handles awkwardly, and what a better method would need to do differently. Then say what question about a system substitution still cannot answer.
Hint: Think about coefficients and about counting.
Answer:
It handles awkwardly any system with no coefficient of one, since isolating a variable then forces fractions through the whole solve. A better method would remove a variable without first isolating one — which is exactly what Lesson 7.3 does by adding or subtracting the two equations so that one variable cancels.
It still cannot say how many solutions a system has before you attempt it. Substitution on a parallel pair produces a false statement and on a coincident pair produces a true one, which are answers of a kind — but recognising and interpreting them is a separate skill, and Lesson 7.5 is where it is developed.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Graph-and-check (7.1) | Substitution (7.2) | |
|---|---|---|
| Where the answer comes from | reading a picture | algebra |
| Precision | limited by the drawing | exact |
| Final step | check in both originals | check in both originals |
The last row is the same in both columns, which is worth noticing. Whatever produced a candidate answer, the substitution into both originals is what confirms it.
Pattern
Whether one equation arrives ready or both need rearranging, the same five moves cover it.
Step one costs a glance and decides whether the rest of the solve carries fractions. It is the only step that is optional and the one most worth doing.
OpenStax Elementary Algebra 2e, §5.2 Solving Systems of Equations by Substitution §5.2
Check
Scan for a coefficient of one.
Check your understanding
In the system 4x + y = 10 and 3x - 2y = 2, which variable is easiest to isolate?
Answer: A
Why: Its coefficient is one, so subtracting 4x isolates it with no division and the working stays free of fractions. Every other choice requires dividing by two, three or four.
Check
Brackets around the whole expression.
Check your understanding
Substituting y = 3x - 2 into 5x - 2y = 4 gives which equation?
Answer: A
Why: The whole expression replaces y and is enclosed in brackets, so the negative two multiplies both of its terms. Expanding gives 5x minus 6x plus four, which is negative x plus four.
Check
The answer is a pair.
Check your understanding
Substitution gives x = 4 with revised equation y = -2x + 9. What is the solution?
Answer: A
Why: Substituting four gives negative eight plus nine, which is one, so the solution is the pair (4, 1). Both coordinates are needed, and x comes first.
Real world
This is Exercise 29's situation. A club orders softballs and bats. It orders 24 items in total, softballs cost 5 dollars each and bats 20 dollars each, and the bill comes to 285 dollars.
Discussion prompt
Write a system for the number of each item, solve it by substitution, and interpret the answer. Then say which variable you isolated and why.
Hint: One equation counts items and the other counts dollars.
Answer:
\[ s + b = 24 \qquad 5s + 20b = 285 \]
\[ s = 24 - b \;\Longrightarrow\; 5(24 - b) + 20b = 285 \;\Longrightarrow\; 120 + 15b = 285 \]
So b is eleven and s is thirteen: thirteen softballs and eleven bats. Checking both: thirteen plus eleven is twenty-four items, and sixty-five plus two hundred and twenty is two hundred and eighty-five dollars.
The variable to isolate was s in the first equation, because its coefficient is one — the counting equation almost always has coefficients of one, which is why it is the natural place to start. Isolating b in the cost equation would have meant dividing by twenty and carrying twentieths through the whole solve for the same answer.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Substituting x = -4y + 1 into -2x + 2y = 3, what is the constant term after expanding?
Correct: -2, since the -2 multiplies the whole expression.
\[ -2(-4y + 1) + 2y = 8y - 2 + 2y = 10y - 2 \]
\[ 10y - 2 = 3 \;\Longrightarrow\; y = \tfrac{1}{2} \]
Why: The brackets mean the negative two multiplies both terms, so the constant one becomes negative two. The first option is what happens when the brackets are dropped, and it leaves the answer for y as a fifth instead of a half — an error invisible in the algebra and immediately visible in the check. The distributive property from Lesson 2.6 is doing exactly its usual job here.
Explain it
They can solve one equation with one letter and freeze at two equations with two.
Discussion prompt
In no more than four sentences, explain what substitution does and why it works. Then tell them the two things that go wrong most often.
Hint: Turn two letters into one.
Answer:
A usable answer: take one of the equations and get one letter on its own, so you have a statement like y equals something with x in it. Then wherever that letter appears in the other equation, write the something instead — now only one letter is left, and you can solve it the way you already know.
Two things go wrong. First, people stop once they have one number: put it back into your rearranged equation to get the other one, because the answer is a pair. Second, put brackets round the expression when you substitute, or whatever multiplies the letter will only reach the first bit of it.
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 choice is fixed by scanning all four coefficients for a one before writing anything. Brackets are fixed by writing them before simplifying, every time. Back-substitution is fixed by writing the answer as an empty ordered pair at the start, so the missing coordinate is visible. The checks are fixed by keeping both originals on the page and testing against them rather than against the revised equation. 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 which exactly one coefficient is a one, and circle that coefficient. Solve the system by substitution in four clearly numbered steps, labelling the revised equation and drawing brackets around the substituted expression before you expand it. Write the answer as an ordered pair and check it underneath in both original equations, one check per line. To the right, solve the same system again by isolating a different variable, and box both answers to show they match while noting which route produced fractions. In the lower half, sketch the two lines on a small coordinate plane and mark the crossing, writing one sentence on whether a graph alone could have given the same answer. Finally, in the margin, write the four steps of the method in your own words.
Your two boxed answers must be identical. If they differ, substitute both candidate pairs into the second original equation — the one that fails is the route where a bracket or a sign was lost.
Recap
Five things, and the third is the one that turns a number into an answer.
| If the question says | Your first move is |
|---|---|
| Solve the system by substitution | Scan for a coefficient of 1 |
| One equation is already y = something | Skip to step two and substitute |
| You have found x | Back-substitute into the revised equation |
| The answer has a fraction in it | Keep it exact; do not round |
| Check your answer | Use both original equations |
Lesson 7.3 handles the systems substitution finds awkward. When no coefficient is one, adding or subtracting the two equations can make a variable cancel outright, with no isolating and no fractions.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.2 Solving Linear Systems by Substitution §7.2, pp. 396-401 — everything on these slides traces back here
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