The substitution method and the elimination method for solving a linear system exactly, how to choose between them, what the no-solution and infinitely-many cases look like algebraically, and modelling with a system solved by algebra.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Solve Linear Systems Algebraically
Objectives
Five outcomes. The first two are the methods; the rest is knowing which to reach for.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 160-165 — the lesson these objectives are drawn from
Warm-up
Lesson 3.1 solved systems by drawing. This lesson exists because of what drawing cannot do.
Discussion prompt
Two lines cross at the point with coordinates negative four thirds and negative two. Could you have read that off a hand-drawn graph? What would you have written down instead?
Hint: Try locating negative four thirds on a grid ruled in whole units.
Answer:
\[ \left(-\tfrac{4}{3}, -2\right) \approx (-1.33, -2) \]
You would have written something like negative one and a bit, which is not an answer. Graphing shows you what is happening — one crossing, none, or a whole line of them — and algebra gives you the exact numbers. This lesson supplies the algebra.
Concept
Both algebraic methods do the same thing by different routes: they turn a system in two variables into a single equation in one variable, which Lesson 1.3 already taught you to solve. Substitution replaces a variable; elimination cancels one.
substitution method — Solving one equation for one of its variables, then substituting that expression into the other equation to leave a single variable.
Neither method is more correct than the other. Both give exact answers, and the choice between them is about which produces cleaner arithmetic on the system in front of you.
Figure (svg): Two lines crossing at a point with fractional coordinates, showing why reading a graph is not enough
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 160-161
Section
Section 1
Concept
Pick whichever variable is easiest to isolate, solve one equation for it, and substitute that expression into the other equation. What is left has a single variable, so ordinary equation solving finishes it.
\[ x + 3y = 3 \;\Longrightarrow\; x = -3y + 3 \]
Substitute into the OTHER equation. Substituting back into the one you rearranged gives an identity and no information — the same trap as substituting a vertex in Lesson 2.7.
Figure (svg): The three steps of substitution: solve one equation for a variable, substitute into the other, then back-substitute
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 160-160 — The Substitution Method
Picture it
Example 1: the system 2x plus 5y equals negative 5, and x plus 3y equals 3.
Figure (svg): The three steps of substitution: solve one equation for a variable, substitute into the other, then back-substitute
Step three is the one people forget: finding y is only half the answer, and the value has to go back in to produce x.
Worked example
Example 1. The second equation has an x with coefficient one, which decides the plan.
\[ \begin{cases} 2x + 5y = -5 \\ x + 3y = 3 \end{cases} \]
Solve the second equation for x
Why: Its x has a coefficient of one, so isolating it costs one subtraction and creates no fractions.
\[ x = -3 y + 3 \]
Substitute that expression into the FIRST equation
Why: Every x in the first equation becomes the bracket, which leaves only y.
\[ 2(-3 y + 3) + 5 y = -5 \]
Solve for y
Why: Distributing gives negative 6y plus 6 plus 5y, so negative y plus 6 equals negative 5, and y is 11.
\[ y = 11 \]
Substitute back into the revised equation
Why: Negative three times eleven is negative thirty-three, plus three.
\[ x = -30 \]
Figure (svg): The solution to Worked example substitution, all three steps shown as a ladder of expressions, one row per algebraic move
\[ (-30, 11) \]
Verify: substitute into BOTH original equations
Why: First: 2 times -30 plus 5 times 11 is -60 plus 55, which is -5 — correct. Second: -30 plus 3 times 11 is -30 plus 33, which is 3 — correct. Both original equations hold, which is what makes this a solution of the system rather than of one equation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 160-160
Sorting
Look for a coefficient of 1 or negative 1.
Sort into buckets
Sort each system by which variable is easiest to isolate.
Scanning for a coefficient of one takes two seconds and decides both the method and which variable to isolate.
Worked example
Guided Practice 1. One of the four coefficients is a 1, and that decides everything.
\[ \begin{cases} 4x + 3y = -2 \\ x + 5y = -9 \end{cases} \]
Scan the four coefficients for a 1
Why: The x in the second equation has coefficient one; the others are 4, 3 and 5.
\[ \text{isolate } x\text{ in equation } 2 \]
Solve that equation for x
Why: Subtracting 5y from both sides.
\[ x = -5 y - 9 \]
Substitute into the first equation
Why: Four times the bracket, plus 3y, equals negative two.
\[ 4(-5 y - 9) + 3 y = -2 \]
Solve for y, then back-substitute
Why: Negative 20y minus 36 plus 3y is negative 17y minus 36, so negative 17y is 34 and y is negative 2. Then x is 10 minus 9.
\[ y = -2, x = 1 \]
Figure (svg): The solution to Worked example choosing what to isolate shown as a ladder of expressions, one row per algebraic move
\[ (1, -2) \]
Verify: substitute into both originals
Why: First: 4 times 1 plus 3 times -2 is 4 minus 6, which is -2 — correct. Second: 1 plus 5 times -2 is 1 minus 10, which is -9 — correct. Note how choosing the coefficient of one kept every intermediate number a whole number; isolating y instead would have introduced thirds.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Trap
\[ \begin{cases} 2x + 5y = -5 \\ x + 3y = 3 \end{cases} \]
Solve the second for x, then substitute into the second
Why: The substitution goes back into the equation it came from.
\[ (-3y + 3) + 3y = 3 \;\Longrightarrow\; 3 = 3 \]
A true statement with no variables left, and no information gained. The first equation was never used.
\[ \begin{cases} 2x + 5y = -5 \\ x + 3y = 3 \end{cases} \]
Substitute the expression into the OTHER equation
Why: The two equations carry different information, and using both is the whole point of a system.
\[ 2(-3y + 3) + 5y = -5 \;\Longrightarrow\; y = 11 \]
Getting 3 equals 3 is not a sign of infinitely many solutions here — it is a sign that one equation was substituted into itself.
Fill the middle
Example 1, step two.
Fill in the blanks
2(-3y + 3) + 5y = -5 \;\Longrightarrow\; -6y + 6 + 5y = -5 \;\Longrightarrow\; y = 11
Why: Subtracting 3y from both sides of the second equation gives x equal to negative 3y plus 3, and that whole expression replaces x in the first equation. The brackets matter: without them the 2 would multiply only the first term. Distributing gives negative 6y plus 6 plus 5y, so negative y equals negative 11 and y is 11.
Error analysis
A student solves a system and reports only half an answer.
Annotate
On: \( \begin{aligned} x &= -5y - 9 \\ 4(-5y - 9) + 3y &= -2 \\ -17y - 36 &= -2 \\ y &= -2 \\ \text{answer} &: y = -2 \end{aligned} \)
Finish by writing the answer as an ordered pair. If your final line has one number in it, a step is missing.
Explain it to yourself
Replacing a variable with an expression feels like a sleight of hand until you say why it is legal.
\[ x = -3y + 3 \;\Longrightarrow\; 2x + 5y = -5 \text{ becomes } 2(-3y+3) + 5y = -5 \]
Discussion prompt
Explain why replacing x by that expression does not change the solutions. What is the first equation asserting about x, and what does that permit?
Hint: Ask what the second equation says about the relationship between x and y.
Answer:
The second equation asserts that for any solution, x and negative 3y plus 3 are the SAME NUMBER. So anywhere x appears in a solution, that expression can stand in its place without changing anything.
This is the substitution property of equality in action: equal quantities may replace one another. The method is not a trick — it is that property applied once.
Section
Section 2
Concept
Multiply one or both equations by constants so that one variable's coefficients differ only in sign. Adding the revised equations then makes that variable vanish, leaving a single equation in the other.
elimination method — Multiplying equations by constants so that one variable's coefficients are opposites, then adding to eliminate that variable.
\[ 3x - 7y = 10 \;\xrightarrow{\times(-2)}\; -6x + 14y = -20 \]
Multiplying an equation through by a nonzero constant is the multiplication property of equality from Lesson 1.3, so the revised equation has exactly the same solutions as the original.
Figure (svg): Two equations stacked with one multiplied so the x terms cancel when the equations are added
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161 — The Elimination Method
Picture it
Example 2: 3x minus 7y equals 10, and 6x minus 8y equals 8.
Figure (svg): Two equations stacked with one multiplied so the x terms cancel when the equations are added
The 3 and the 6 were already related by a factor of two, which is what made a single multiplication enough. When no such relation exists, both equations get multiplied.
Worked example
Example 2. The coefficients of x are 3 and 6, so one factor of negative two does it.
\[ \begin{cases} 3x - 7y = 10 \\ 6x - 8y = 8 \end{cases} \]
Look for a variable whose coefficients are easy multiples
Why: Three and six for x, against negative seven and negative eight for y. The x column is the easy one.
Multiply the first equation by -2
Why: Every term, including the constant: negative 6x plus 14y equals negative 20.
\[ -6 x + 14 y = -20 \]
Add the two equations
Why: The x terms cancel, leaving 14y minus 8y, which is 6y, and negative 20 plus 8, which is negative 12.
\[ 6 y = -12 \]
Solve for y, then substitute back
Why: y is negative 2; putting that into the first original gives 3x plus 14 equals 10, so 3x is negative 4.
\[ y = -2, x = -\frac{4}{3} \]
Figure (svg): The solution to Worked example elimination with one multiplication shown as a ladder of expressions, one row per algebraic move
\[ \left(-\tfrac{4}{3}, -2\right) \]
Verify: substitute into both originals
Why: First: 3 times -4/3 minus 7 times -2 is -4 plus 14, which is 10 — correct. Second: 6 times -4/3 minus 8 times -2 is -8 plus 16, which is 8 — correct. Note the fractional x: this is exactly the solution no hand-drawn graph could have given.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Fill the middle
Guided Practice 2: eliminate x from 3x plus 3y equals negative 15 and 5x minus 9y equals 3.
Fill in the blanks
3x + 3y = -15 \;\xrightarrow9x + 9y = -45\; ___ \;\text___\; 5x - 9y = 3
Why: Multiplying every term by 3 gives 9x plus 9y equals negative 45. Adding that to the second equation cancels the y terms — 9y plus negative 9y is zero — leaving 14x equal to negative 42, so x is negative 3. Note that this multiplication targeted y rather than x, which is why the 3 was chosen: the y coefficients were 3 and negative 9.
Worked example
Guided Practice 3. Neither coefficient divides the other, so both equations are scaled.
\[ \begin{cases} 3x - 6y = 9 \\ -4x + 7y = -16 \end{cases} \]
Choose a variable and find a common multiple
Why: The x coefficients are 3 and negative 4, whose least common multiple is 12.
\[ \text{target } 12\text{ and } -12 \]
Multiply the first equation by 4
Why: Twelve x minus 24y equals 36.
\[ 12 x - 24 y = 36 \]
Multiply the second equation by 3
Why: Negative twelve x plus 21y equals negative 48.
\[ -12 x + 21 y = -48 \]
Add and solve
Why: The x terms cancel; negative 24y plus 21y is negative 3y, and 36 minus 48 is negative 12, so y is 4.
\[ y = 4 \]
Back-substitute
Why: Three x minus 24 equals 9, so 3x is 33 and x is 11.
\[ x = 11 \]
Figure (svg): The solution to Worked example multiplying both equations shown as a ladder of expressions, one row per algebraic move
\[ (11, 4) \]
Verify: substitute into both originals
Why: First: 33 minus 24 is 9 — correct. Second: negative 44 plus 28 is negative 16 — correct. Both check. Choosing the y column instead would have needed multipliers of 7 and 6, giving larger numbers, so scanning both columns before committing is worth the moment it takes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Trap
\[ 3x - 7y = 10 \quad \text{multiply by } -2 \]
Multiply the two variable terms and leave the constant alone
Why: The multiplier is applied to the terms being changed rather than to the whole equation.
\[ -6x + 14y = 10 \quad \text{(wrong)} \]
Adding this to the second equation gives 6y equals 18, so y would be 3 instead of negative 2 — and the check fails immediately.
\[ 3x - 7y = 10 \quad \text{multiply by } -2 \]
Multiply EVERY term on both sides
Why: The multiplication property of equality multiplies each side as a whole, so the constant is multiplied like everything else.
\[ -6x + 14y = -20 \]
This is the same slip as clearing fractions in Lesson 1.3, where the whole-number terms had to be multiplied too.
Discrimination
Look at whether one coefficient divides the other.
Sort into buckets
Sort each system by how many equations need multiplying.
Ranking
Solving a system by elimination.
Put in order
Why: Choosing the target variable first is what determines the multipliers, so it cannot come later. The addition only cancels if the multiplication has already made the coefficients opposites. And the back-substitution uses an original equation rather than a revised one, which is both simpler and a partial check — a revised equation would work too, but errors made during the multiplication would go undetected.
Prediction
Commit before computing.
Predict first
For 3x - 6y = 9 and -4x + 7y = -16, eliminating which variable needs smaller multipliers?
Correct: x, using multipliers of 4 and 3.
\[ \text{x column: lcm}(3,4) = 12 \qquad \text{y column: lcm}(6,7) = 42 \]
Why: The x coefficients are 3 and -4, whose least common multiple is 12, so the multipliers are 4 and 3. The y coefficients are -6 and 7, whose least common multiple is 42, needing multipliers of 7 and 6. Both routes give the same answer, but the x route keeps every number under 50 while the y route pushes past 100. Comparing the two least common multiples before starting is a ten-second decision that saves real arithmetic.
Section
Section 3
Concept
Substitution shines when a variable already has a coefficient of one or an equation is already solved for a variable. Elimination shines when both equations are in standard form with no isolated variable.
Graphing from Lesson 3.1 remains useful for a third reason: it shows at a glance whether the system has one solution, none, or infinitely many, which the algebra only reveals at the end.
Figure (svg): Two columns comparing when substitution is the easier method and when elimination is
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 160-161
Picture it
The same system can be solved either way; the choice is about arithmetic.
Figure (svg): Two columns comparing when substitution is the easier method and when elimination is
If you cannot decide in five seconds, pick elimination. It never requires you to isolate anything, so it never creates fractions that were not already there.
Worked example
Guided Practice 1, solved twice, to see what each method costs.
\[ \begin{cases} 4x + 3y = -2 \\ x + 5y = -9 \end{cases} \quad \text{by substitution, then by elimination.} \]
Substitution route: isolate x in the second
Why: Its coefficient is one, so this costs one subtraction.
\[ x = -5 y - 9 \]
Substitute and solve
Why: Four times the bracket plus 3y gives negative 17y minus 36 equals negative 2, so y is negative 2 and x is 1.
\[ (1, -2) \]
Elimination route: multiply the second by -4
Why: Negative 4x minus 20y equals 36, so the x terms will cancel.
\[ -4 x - 20 y = 36 \]
Add and solve
Why: Three y minus 20y is negative 17y, and negative 2 plus 36 is 34, so y is negative 2 and x is 1.
\[ (1, -2) \]
Figure (svg): The solution to Worked example the same system, both ways shown as a ladder of expressions, one row per algebraic move
\[ (1, -2) \]
Verify: compare the two routes rather than just the answers
Why: Both reached negative 17y at the same point, which is not a coincidence: the two methods are doing the same elimination, one by replacing and one by adding. Substitution needed one rearrangement and elimination needed one multiplication, so on this system they cost about the same.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Sorting
Judge from the coefficients, not from the difficulty.
Sort into buckets
Sort each system by the method that keeps the arithmetic cleanest.
Both methods solve all five. The sorting is about arithmetic cost, and getting it right saves more time than solving quickly does.
Worked example
The same system solved by isolating the harder variable, to see the cost.
\[ \begin{cases} 3x + 3y = -15 \\ 5x - 9y = 3 \end{cases} \quad \text{by substitution, isolating } x \text{ in the second.} \]
Isolate x in the second equation
Why: Dividing by 5 introduces fifths straight away.
\[ x = \frac{9 y + 3}{5} \]
Substitute into the first
Why: Three times a fraction plus 3y equals negative 15.
\[ 3(9 y + 3) / 5 + 3 y = -15 \]
Clear the fraction and solve
Why: Multiplying through by 5 gives 27y plus 9 plus 15y equals negative 75, so 42y is negative 84 and y is negative 2.
\[ y = -2 \]
Compare with the elimination route
Why: Multiplying the first by 3 and adding gave 14x equal to negative 42 in two lines, with no fractions at all.
Figure (svg): The solution to Worked example a system that punishes the wrong choice shown as a ladder of expressions, one row per algebraic move
\[ (-3, -2) \]
Verify: substitute into both originals
Why: First: 3 times -3 plus 3 times -2 is -9 minus 6, which is -15 — correct. Second: 5 times -3 minus 9 times -2 is -15 plus 18, which is 3 — correct. The answer is the same by either route; only the number of lines and the presence of fractions differed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Error analysis
A student uses substitution on a system with no coefficient of one and makes an arithmetic slip in the fractions.
Annotate
On: \( 5x - 9y = 3 \;\Longrightarrow\; x = \frac{9y + 3}{5} \;\Longrightarrow\; 3\left(\frac{9y+3}{5}\right) + 3y = -15 \;\Longrightarrow\; \frac{27y + 3}{5} + 3y = -15 \)
Scan the four coefficients before choosing. Substitution with no coefficient of one is legal, tedious, and a reliable source of exactly this kind of slip.
Comparison
Fill the blanks. Neither is better in general.
Comparison matrix
| Feature | Substitution | Elimination |
|---|---|---|
| What it does to a variable | replaces it | cancels it |
| Best when | a coefficient is 1 or -1 | all coefficients are awkward |
| Risk it creates | fractions when nothing is isolated | forgetting to multiply the constant |
| Number of variables left after step 2 | one | one |
| Exactness of the answer | exact | exact |
The last two rows are identical, which is the real point: both methods reduce the system to one variable and both give exact answers. Everything else is convenience.
Explain it
A classmate always uses substitution because it was taught first.
Discussion prompt
In three sentences, tell them what to look at before choosing, why elimination is the safer default, and one system where substitution is clearly better.
Hint: The thing to look at takes two seconds.
Answer:
Scan the four coefficients for a 1 or a negative 1. If one is there, substitution costs a single subtraction; if not, isolating a variable creates fractions that will follow you through the whole solution.
Elimination is the safer default because it never requires isolating anything. But when an equation is already solved for a variable — like y equals 2x plus 2 — substitution is clearly better, since step one is already done.
Commit first
Answer, then rate your confidence honestly.
Predict first
Do substitution and elimination ever give different answers for the same system?
Correct: No — both are exact, so both give the same solution.
This is worth using deliberately. On an important problem, solve by one method and check by the other rather than re-reading the first solution.
Why: Every step of each method is one of the properties of equality from Lesson 1.3, so each produces an equivalent system at every stage. Equivalent systems have identical solution sets, which means the two routes cannot disagree. If they do, one of them has an arithmetic error — and solving the same system both ways is a genuinely strong check for exactly that reason.
Section
Section 4
Concept
Sometimes both variables disappear during substitution or elimination. What remains is a statement with no variables, and its truth decides the answer: true means infinitely many solutions, false means none.
\[ 0 = 0 \;\Rightarrow\; \text{infinitely many} \qquad 0 = 3 \;\Rightarrow\; \text{none} \]
These are the algebraic faces of the coincident and parallel lines from Lesson 3.1. Nothing new is happening; the same three outcomes are appearing in symbols instead of pictures.
Figure (svg): Two algebraic outcomes when the variables vanish: a true statement and a false one
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Picture it
The two ways a system can fail to have exactly one solution.
Figure (svg): Two algebraic outcomes when the variables vanish: a true statement and a false one
The instinct that a vanished variable means something went wrong is worth unlearning. It means the system is one of the two degenerate kinds, and that is a real answer.
Worked example
The parallel case from Lesson 3.1, now solved algebraically.
\[ \begin{cases} 2x + y = 4 \\ 2x + y = 1 \end{cases} \]
Multiply the second equation by -1
Why: Negative 2x minus y equals negative 1, ready to cancel with the first.
\[ -2 x - y = -1 \]
Add the two equations
Why: Both the x terms and the y terms cancel.
\[ 0 = 3 \]
Read the leftover statement
Why: Zero equals three is false, whatever x and y are.
State the conclusion
Why: No pair can satisfy both equations, so the system has no solution and is inconsistent.
Figure (svg): The solution to Worked example a system with no solution shown as a ladder of expressions, one row per algebraic move
\[ 0 = 3 \text{ is false} \;\Longrightarrow\; \text{no solution} \]
Verify: confirm with the graphical test from Lesson 3.1
Why: Solving each for y gives negative 2x plus 4 and negative 2x plus 1: same slope, different intercepts, so the lines are parallel and distinct. The algebra and the geometry agree, as they must.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Sorting
In each case all the variables have cancelled.
Sort into buckets
Sort each result by what the system has.
The rule is one question: is the leftover statement true? Not what the numbers are, but whether the two sides are equal.
Worked example
The coincident case, solved algebraically.
\[ \begin{cases} 4x - 3y = 8 \\ 8x - 6y = 16 \end{cases} \]
Multiply the first equation by -2
Why: Negative 8x plus 6y equals negative 16.
\[ -8 x + 6 y = -16 \]
Add the two equations
Why: Both variables cancel and so do the constants.
\[ 0 = 0 \]
Read the leftover statement
Why: Zero equals zero is true for every x and y.
State the conclusion
Why: Every pair satisfying one equation satisfies the other, so there are infinitely many solutions.
Figure (svg): The solution to Worked example a system with infinitely many shown as a ladder of expressions, one row per algebraic move
\[ 0 = 0 \text{ is true} \;\Longrightarrow\; \text{infinitely many solutions} \]
Verify: produce two different solutions explicitly
Why: The point (2,0) gives 8 in the first equation and 16 in the second — both correct. The point (5,4) gives 20 minus 12, which is 8, and 40 minus 24, which is 16 — also correct. Two distinct solutions already rules out exactly one, confirming the verdict.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 161-161
Error analysis
A student eliminates and, finding nothing left, starts over.
Annotate
On: \( -8x + 6y = -16 \;\text{ plus }\; 8x - 6y = 16 \;\Longrightarrow\; 0 = 0 \;\Longrightarrow\; \text{I must have made a mistake} \)
A vanished variable is one of the three possible outcomes, not a symptom of arithmetic trouble. Read what is left and decide whether it is true.
Matching
Each algebraic result corresponds to a picture from Lesson 3.1.
Match the pairs
Why: The last two rows describe the same situation from two angles: an equation that is a multiple of another is exactly the case that produces 0 equals 0 when eliminated. Recognising the algebraic signature saves you from graphing, and recognising the picture saves you from algebra — either route reaches the same three outcomes.
Prediction
Commit before solving.
Predict first
What will elimination produce for the system 2x + 5y = 6 and 4x + 10y = 13?
Correct: 0 = 1, so no solution.
\[ -4x - 10y = -12 \;\text{ plus }\; 4x + 10y = 13 \;\Longrightarrow\; 0 = 1 \]
Why: Multiplying the first by negative two gives negative 4x minus 10y equals negative 12. Adding leaves 0 equals 1, which is false, so the system has no solution. The coefficients scale by two but the constant does not — 13 is not twice 6 — which is precisely the parallel-lines signature from Lesson 3.1. Had the constant been 12, the same elimination would have given 0 equals 0 instead.
Edge cases
The system 2x plus 5y equals 6 and 4x plus 10y equals c.
Discussion prompt
For which value of c does this system have infinitely many solutions, and what does it have for every other value of c? Is there any value of c giving exactly one solution?
Hint: Compare the second equation with twice the first.
Answer:
Twice the first equation is 4x plus 10y equals 12, so c equal to 12 makes the two equations identical and gives infinitely many solutions.
For every other c the coefficients still match after scaling but the constants disagree, so the lines are parallel and there is no solution.
There is no value of c giving exactly one solution, because the coefficient ratio is fixed: 4 over 2 equals 10 over 5, so the two lines have equal slopes whatever c does. Only the constant is free, and it can only choose between coincident and parallel.
Section
Section 5
Concept
Real problems that describe two relationships between two unknowns become systems. Solving algebraically gives the exact values, which matters when the answer is not a whole number or when the graph would be hard to read.
\[ \begin{cases} x + y = 40 \\ 8x + 12y = 384 \end{cases} \]
As in Lesson 1.5, name the two unknowns explicitly before writing anything, and check the answer against the situation rather than only against the equations.
Figure (svg): Two lines crossing at a point with fractional coordinates, showing why reading a graph is not enough
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 160-161
Picture it
A crossing at negative four thirds cannot be read off a grid.
Figure (svg): Two lines crossing at a point with fractional coordinates, showing why reading a graph is not enough
In a modelling problem a fractional answer often signals something real — a quantity that must be rounded, or a combination that is not actually achievable.
Worked example
A ticket problem, built the way Lesson 1.5 built its models.
\[ \text{40 tickets sold for } \$384. \text{ Adult tickets cost } \$12 \text{ and student tickets } \$8. \text{ How many of each?} \]
Name the two unknowns
Why: Let s be the number of student tickets and a the number of adult tickets.
Write the two conditions as equations
Why: The counts total 40, and the money totals 384 dollars.
\[ s + a = 40\text{ and } 8 s + 12 a = 384 \]
Choose a method and solve
Why: The first equation has coefficients of one, so substitution is natural: s equals 40 minus a.
\[ 8(40 - a) + 12 a = 384 \]
Solve and back-substitute
Why: Three hundred and twenty minus 8a plus 12a is 320 plus 4a, so 4a is 64 and a is 16, giving s equal to 24.
\[ a = 16, s = 24 \]
Figure (svg): The solution to Worked example two unknowns, two facts shown as a ladder of expressions, one row per algebraic move
\[ 16 \text{ adult}, \; 24 \text{ student} \]
Verify: check both conditions against the situation
Why: Sixteen plus 24 is 40 tickets — correct. And 16 times 12 is 192, plus 24 times 8 which is 192, giving 384 dollars — correct. Both totals are recovered, and both counts are non-negative whole numbers, which the situation requires and the algebra does not guarantee.
Real world
A jar holds 30 coins, all 5-cent and 10-cent pieces, worth 2.30 dollars in total.
Discussion prompt
Name your variables, write the system, solve it exactly, and check both conditions. Then say which method you chose and why.
Hint: Work in cents to avoid decimals.
Answer:
\[ \begin{cases} n + d = 30 \\ 5n + 10d = 230 \end{cases} \]
\[ n = 30 - d \;\Longrightarrow\; 5(30-d) + 10d = 230 \;\Longrightarrow\; 5d = 80 \;\Longrightarrow\; d = 16 \]
Sixteen dimes and 14 nickels. Check: 30 coins, and 70 plus 160 cents is 230 cents. Substitution was the natural choice because the first equation has coefficients of one, so isolating cost a single subtraction.
Worked example
The same shape of problem, with numbers that do not divide evenly.
\[ \text{35 items cost } \$300 \text{ in total, at } \$7 \text{ and } \$11 \text{ each. How many of each?} \]
Name and write the equations
Why: Let x be the number at 7 dollars and y the number at 11.
\[ x + y = 35\text{ and } 7 x + 11 y = 300 \]
Substitute using the simpler equation
Why: x equals 35 minus y.
\[ 7(35 - y) + 11 y = 300 \]
Solve
Why: Two hundred and forty-five minus 7y plus 11y is 245 plus 4y, so 4y is 55 and y is 13.75.
\[ y = 13.75 \]
Interpret the result
Why: A count of items cannot be fractional, so no combination of whole items gives exactly these totals.
Figure (svg): The solution to Worked example an answer that is not whole shown as a ladder of expressions, one row per algebraic move
\[ y = 13.75 \;\Longrightarrow\; \text{no whole-number solution exists} \]
Verify: confirm by testing the nearest whole numbers
Why: With 14 at 11 dollars and 21 at 7, the total is 154 plus 147, which is 301. With 13 and 22, it is 143 plus 154, which is 297. Neither hits 300, so the fractional answer really is telling you something about the situation rather than about the arithmetic.
Error analysis
A student models the ticket problem with a single equation.
Annotate
On: \( 8s + 12a = 384 \quad \text{with } s \text{ and } a \text{ both unknown} \)
Count the unknowns and count the equations. Two unknowns need two independent facts, and a problem that supplies two has supplied them for a reason.
Matching
Each situation gives two facts about two unknowns.
Match the pairs
Why: Every one supplies a count condition and a value condition, or a sum and a difference. The last two are the easiest of all to eliminate: adding the two equations cancels the second variable immediately, with no multiplication at all — which is why sum-and-difference problems are a standard first example of elimination.
Estimation
Forty tickets sold for 384 dollars, at 8 and 12 dollars each.
Predict first
Roughly what fraction of the tickets were the dearer kind?
Correct: About 40 percent.
\[ \frac{384}{40} = 9.60 \qquad \frac{9.60 - 8}{12 - 8} = 0.40 \]
Why: The average ticket price is 384 divided by 40, which is 9.60 dollars. That sits between 8 and 12, and closer to 8, so most tickets were the cheaper kind. Precisely, 9.60 is 1.60 above 8 out of a 4-dollar range, which is 40 percent — and the exact answer is 16 adult tickets out of 40, which is exactly 40 percent. Computing the average price first is a fast way to sanity-check any mixture problem.
Socratic
One question, and nothing else on this slide.
\[ y = 13.75 \quad \text{items} \]
Discussion prompt
A mixture problem produces an answer of 13.75 items. The algebra is correct. What are the possible explanations, and how would you decide between them? Is reporting 14 ever the right response?
Hint: Consider both the model and the data it was built from.
Answer:
Either the stated totals are inconsistent with whole items — no combination gives exactly those numbers — or one of the given figures is slightly wrong, or the quantity genuinely can be fractional, as with kilograms of two kinds of flour.
Deciding between them means going back to the situation. If the items are countable, the honest answer is that no whole-number solution exists, and rounding to 14 would give a total of 301 rather than 300.
Rounding is right only when the model is an approximation of something continuous. Reporting 14 as if it solved the stated problem hides the fact that the problem, as stated, has no solution.
Comparison
Fill the blanks. Graphing, substitution and elimination agree on everything.
Comparison matrix
| Outcome | Graphing shows | Algebra produces |
|---|---|---|
| One solution | lines crossing once | a variable equals a number |
| Infinitely many | lines coinciding | a true statement like 0 = 0 |
| No solution | parallel distinct lines | a false statement like 0 = 3 |
| Exact fractional answer | unreadable from a drawing | exact, by either method |
The last row is why this lesson exists. Graphing tells you which of the three cases you are in; algebra tells you the numbers.
Pattern
One routine solves any two-equation linear system algebraically.
Steps four and five are where finished-looking work loses marks: half an answer, or an unchecked one, or a vanished variable mistaken for an error.
OpenStax Algebra and Trigonometry 2e, §11.1 Systems of Linear Equations: Two Variables §11.1
Check
Substitution. Look for the coefficient of 1.
Check your understanding
Solve the system 2x + 5y = -5 and x + 3y = 3.
Answer: A
Why: Solving the second for x gives x = -3y + 3. Substituting into the first gives -6y + 6 + 5y = -5, so y = 11 and x = -30. Both originals check.
Check
Elimination. Multiply every term.
Check your understanding
Solve the system 3x - 7y = 10 and 6x - 8y = 8.
Answer: A
Why: Multiplying the first equation by -2 gives -6x + 14y = -20. Adding gives 6y = -12, so y = -2, and back-substituting gives 3x = -4, so x = -4/3.
Check
A degenerate case. Read the leftover statement.
Check your understanding
Eliminating a variable from a system leaves the statement 0 = 0. What does the system have?
Answer: A
Why: Zero equals zero is true for every pair of values, so every point satisfying one equation satisfies the other. The two equations describe the same line, and the system is consistent and dependent.
Real world
A chemist has two solutions, one 20 percent acid and one 50 percent, and needs 12 litres of a 30 percent mixture.
Discussion prompt
Set this up as a system, solve it algebraically, and check both conditions. Then explain why substitution is the natural method here, and what a negative answer would have told you.
Hint: One equation counts litres and the other counts litres of acid.
Answer:
\[ \begin{cases} x + y = 12 \\ 0.20x + 0.50y = 0.30(12) = 3.6 \end{cases} \]
\[ x = 12 - y \;\Longrightarrow\; 0.2(12-y) + 0.5y = 3.6 \;\Longrightarrow\; 0.3y = 1.2 \;\Longrightarrow\; y = 4 \]
Four litres of the 50 percent solution and 8 of the 20 percent. Check: 12 litres total, and 1.6 plus 2.0 is 3.6 litres of acid, which is 30 percent of 12. Substitution is natural because the first equation has coefficients of one.
A negative answer would have meant the target concentration lies outside the range of the two solutions — you cannot mix 20 and 50 percent to get 60 percent, and the algebra would say so by demanding a negative volume.
Commit first
Answer, then rate your confidence honestly.
Predict first
When you multiply one equation of a system by a constant, does the system's solution change?
Correct: No, provided the constant is not zero.
\[ 3x - 7y = 10 \;\xrightarrow{\times(-2)}\; -6x + 14y = -20 \quad \text{same solutions} \]
Why: Multiplying an equation by a nonzero constant gives an equivalent equation, with exactly the same solutions — the multiplication property of equality from Lesson 1.3. So the system as a whole is unchanged, which is what makes elimination legitimate. Multiplying by zero would replace the equation with 0 equals 0, destroying its information and turning a system with one solution into one with infinitely many, which is why the nonzero condition is stated.
Explain it
They can solve a single equation but freeze when there are two of them.
Discussion prompt
In four sentences or fewer, explain what both algebraic methods are really doing, describe one of them concretely, and give them the check that catches almost every error.
Hint: The two methods share one goal.
Answer:
Both methods do the same thing: they get rid of one variable so that what is left is a single equation you already know how to solve. Substitution does it by replacing a variable with an expression from the other equation; elimination does it by adding the equations so one variable cancels.
Either way, once you have one value, put it back to find the other and write the answer as an ordered pair. Then substitute that pair into BOTH original equations — an answer that satisfies only one of them is not a solution of the system.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For choosing, scan the four coefficients for a 1 and pick substitution only if you find one. For multiplying, write the revised equation out in full rather than in your head. For the second variable, make the last line of every solution an ordered pair. For the degenerate cases, read the leftover statement and ask whether it is true. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Divide a page in two. On the left, write out a system of your own with a coefficient of one somewhere and solve it by substitution, labelling each of the three steps. On the right, write a system with no coefficient of one and solve it by elimination, showing the multiplication of every term including the constant. Under both, substitute each answer into BOTH original equations and tick them off. Across the bottom, write the two degenerate outcomes — a true statement and a false one — with a small sketch of the corresponding lines beside each, and write one sentence saying which of the three Lesson 3.1 classifications each represents. Finally, in a margin, write the two-second test you use to choose a method.
If your margin note is longer than one line, shorten it: the test is just whether any coefficient is 1 or negative 1.
Recap
Five things, and the first two are the same idea reached by different roads.
| If you see | Then |
|---|---|
| A coefficient of 1 or -1 | Substitution is cheap |
| All coefficients awkward | Use elimination |
| One equation solved for a variable | Substitute it straight in |
| 0 = 0 after the variables cancel | Infinitely many solutions |
| 0 = a nonzero number | No solution |
Lesson 3.3 returns to graphing, but with inequalities: two shaded half-planes at once, and the region where both hold.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.2 Solve Linear Systems Algebraically §3.2, pp. 160-165 — everything on these slides traces back here
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