A system of two linear equations in x and y shows up in this course constantly, so it's worth being fluent in every lens for looking at one: the picture it draws, and the two hand methods that pin down its solution exactly. None of this is new — it's the same system you solved in algebra — but Chapter 3 rebuilds it as an augmented matrix, so it pays to be sharp on what that matrix is standing in for before the notation changes.
Picture each equation as a straight line on a page. Two straight lines can only relate in three ways. They cross at one spot, they run side by side forever without touching, or they're secretly the same line drawn twice. A solution to the system is a point both lines agree on. So the geometry and the algebra tell the same story from two different angles. The algebra just tells it without needing a ruler.
A system \begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} has a solution set that is exactly one of three things, and the algebra always tells you which:
- One solution — the two equations, solved together, force a single pair (x, y).
- No solution — the equations reduce to a false numerical statement, such as 0 = 5.
- Infinitely many solutions — the equations reduce to a true statement with no variables left, such as 0 = 0, because the second equation was really the first one in disguise.
Every linear system in two variables lands in exactly one of these three cases — never a fourth.
Drag either line below and watch the marked point — that point is the solution to the system x + 2y = 8,\; 3x - y = 3, and it moves the instant either equation changes.
Two lines, one crossing point
Drag either line — the point where they cross is the solution to x + 2y = 8 and 3x - y = 3. Slide a line until it runs parallel to the other — what happens to the marked point?
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Check your understanding (counts toward your progress)
The lines x + y = 9 and x - y = 3 cross at one point. What is its x-coordinate?
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Why
Adding the equations eliminates y: (x+y)+(x-y) = 9+3 \Rightarrow 2x = 12 \Rightarrow x = 6. (Then y = 3, so the crossing point is (6, 3).)
Correct answer
Also accepted:
Why
Add or subtract the two equations to eliminate one variable, then solve for x.
Solving x - 4y = 2 together with 2x - 8y = 4 leads to the statement 0 = 0. What does that mean?
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0=0 is a true statement, not a false one like 0=5 — a false statement is what signals no solution.
0=0 carries no information about x or y specifically, so it can't be pinning down a single point like the origin.
0=0 is exactly what you should expect here — the second equation is just the first one multiplied by 2, so it never had new information to contribute.
Why
2x - 8y = 4 is exactly x - 4y = 2 multiplied by 2. The two equations carry identical information, so eliminating a variable leaves the always-true statement 0=0, and the solution set is the whole shared line.
Try dividing the second equation by a constant and compare it to the first -- are these actually two different lines, or the same line written twice?
Practice: Two lines, three outcomes (4 problems · doesn’t affect your progress)
Geometrically, the solution (x, y) of a 2\times2 linear system is... (For reference, the choices given are: the point where the two lines cross; the y-intercept of either line; the slope shared by both lines; the midpoint between the two lines' intercepts.)
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The y-intercept only satisfies one of the two equations, not both — the solution has to sit on both lines at once.
Slope describes a single line's steepness; it isn't a point, and it isn't what the system is asking for.
There's no reason the solution would land halfway between two unrelated intercepts — find where the lines actually meet instead.
Why
A solution to a system has to make every equation true simultaneously. Since each linear equation draws a line, a pair satisfying both equations is a point lying on both lines — that's their intersection.
the point where the two lines cross
The lines x + y = 9 and x - y = 3 cross at one point. What is its x-coordinate? (Work through it specifically for x + y = 9 and x - y = 3 and x.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Adding the equations eliminates y: (x+y)+(x-y) = 9+3 \Rightarrow 2x = 12 \Rightarrow x = 6. (Then y = 3, so the crossing point is (6, 3).)
Correct answer
Also accepted:
Why
Add or subtract the two equations to eliminate one variable, then solve for x.
The system 2x + y = 6 and 4x + 2y = 15 has how many solutions? (For reference, the choices given are: One; None; Infinitely many; Cannot be determined without graphing.)
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Check the slopes: the second equation is 2\times the first equation's left side, but not its right side — the lines are parallel, so they never cross once.
For infinitely many solutions the second equation would have to be an exact scaled copy of the entire first equation, including the constant. Doubling 2x+y=6 gives 4x+2y=12, not 15.
The algebra settles this without a picture: try eliminating a variable and see what statement is left.
Why
Doubling the first equation gives 4x + 2y = 12, which matches the second equation's left side but not its right side (15 \ne 12). The lines are parallel and distinct, so no point lies on both — no solution.
None
Solving x - 4y = 2 together with 2x - 8y = 4 leads to the statement 0 = 0. What does that mean? (For reference, the choices given are: The two equations describe the same line, so every point on it is a solution; There is no solution; There is exactly one solution, at the origin; A mistake was made somewhere in the algebra.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
0=0 is a true statement, not a false one like 0=5 — a false statement is what signals no solution.
0=0 carries no information about x or y specifically, so it can't be pinning down a single point like the origin.
0=0 is exactly what you should expect here — the second equation is just the first one multiplied by 2, so it never had new information to contribute.
Why
2x - 8y = 4 is exactly x - 4y = 2 multiplied by 2. The two equations carry identical information, so eliminating a variable leaves the always-true statement 0=0, and the solution set is the whole shared line.
Try dividing the second equation by a constant and compare it to the first -- are these actually two different lines, or the same line written twice?
The two equations describe the same line, so every point on it is a solution
Two hand methods, and how to pick one
Substitution is running an errand for one equation. It hands you a note that says y is secretly 9 - 2x. You carry that note to the other equation and tape it in everywhere y appears. Elimination is more like tuning two radios to cancel each other's static. You turn the volume knob on each equation, multiplying through by a constant, until one variable's contribution is the exact opposite in both. Then you play them together and that variable vanishes. Neither method is "the right one." You just reach for whichever one skips the most arithmetic for the system in front of you.
Substitution solves one equation for one variable and replaces every occurrence of that variable in the other equation with the resulting expression, collapsing the system to one equation in one unknown. It is fastest when some equation already has a variable with coefficient 1 or -1, since solving for that variable costs no fractions.
Elimination scales the two equations by constants chosen so that one variable's coefficients become opposites, then adds the equations so that variable cancels. It is fastest when no variable in either equation already has coefficient \pm 1 — forcing a substitution there would introduce fractions immediately, while elimination avoids them until the very last division.
Worked example
Substitution: 2x + y = 11 and x - y = 1
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1no fractions needed — that's the tell to use substitution
The second equation already isolates a variable with coefficient 1: solve it for x, giving x = y + 1.
no fractions needed — that's the tell to use substitution
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2
Substitute y + 1 for x in the first equation: 2(y + 1) + y = 11
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3
Distribute and combine: 2y + 2 + y = 11 \;\Longrightarrow\; 3y = 9 \;\Longrightarrow\; y = 3.
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4
Back into x = y + 1: x = 3 + 1 = 4. Check both equations: 2(4) + 3 = 11 ✓ and 4 - 3 = 1 ✓. Solution: \boxed{(4, 3)}.
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Worked example
Elimination with two multipliers: 3x + 2y = 8 and 5x - 3y = 26
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1multiply every term, including the constant on the right
Neither variable has coefficient \pm 1, so substitution would drag fractions through every line — elimination instead. To cancel y, scale the first equation by 3 and the second by 2, matching the y-coefficients 6 and -6: 9x + 6y = 24 \qquad\qquad 10x - 6y = 52
multiply every term, including the constant on the right
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2
Add the two scaled equations — the \textcolor{#4f46e5}{y}-terms cancel: 9x + \textcolor{#4f46e5}{\cancel{6y}} + 10x - \textcolor{#4f46e5}{\cancel{6y}} = 24 + 52 \;\Longrightarrow\; 19x = 76
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3
Divide: x = 4. Substitute into the original first equation: 3(4) + 2y = 8 \;\Longrightarrow\; 12 + 2y = 8 \;\Longrightarrow\; y = -2
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4
Check in the second original equation: 5(4) - 3(-2) = 20 + 6 = 26 ✓. Solution: \boxed{(4, -2)}.
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A system can also refuse to cooperate. If elimination or substitution collapses to something like 0 = 7, stop — that's not an arithmetic mistake to hunt down, it's the system announcing the lines never meet. If it collapses to 0 = 0 instead, the second equation was a scaled copy of the first all along, and every point on the shared line solves the system.
Check your understanding (counts toward your progress)
For the system y = 3x - 5 and 2x + y = 10, which move avoids the most arithmetic?
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That works eventually, but the first equation is already solved for y — using that head start with substitution is far shorter than scaling both equations.
Same issue: elimination is a fine method in general, but it's extra setup here when y is already isolated for you.
A graph only gives an exact answer when the crossing point happens to fall on a grid line — algebra gives the exact solution every time.
Why
The first equation already reads y = 3x - 5 — that's substitution's ideal case, a variable handed to you pre-isolated. Reaching for elimination here means undoing that head start and redoing the work.
Look for an equation that's already solved for one variable -- substituting that in avoids extra algebra.
Before substituting, solve 2x + y = 9 for y.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Isolating y means moving the 2x term to the other side: y = 9 - 2x. That expression is what gets carried into the other equation.
Correct answer
Also accepted:
Why
Find the value of x that satisfies both 4x + 3y = 7 and 5x - 2y = -20.
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Why
Scale the first equation by 2 and the second by 3 to match the y-coefficients: 8x+6y=14 and 15x-6y=-60. Adding gives 23x = -46, so x = -2. (Then y = 5.)
Correct answer
Also accepted:
Why
Multiply the equations by constants that make the coefficients of one variable match (or become opposites), then add or subtract to eliminate it.
Practice: Solving a 2x2 system by hand (5 problems · doesn’t affect your progress)
Rearranging 2x + y = 9, what does y equal?
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Isolating y means moving the 2x term to the other side: y = 9 - 2x. That expression is what gets carried into the other equation.
Correct answer
Also accepted:
Why
What value of x solves the pair x + y = 7 and x - y = 1?
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Why
Adding the two equations cancels y: 2x = 8, so x = 4. (Then y = 3.)
Correct answer
Also accepted:
Why
Determine x from the system 2x + 3y = 16 and x - y = 3.
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Why
Solve the second equation for x: x = y + 3. Substitute: 2(y+3) + 3y = 16 \Rightarrow 5y + 6 = 16 \Rightarrow y = 2, so x = 2 + 3 = 5.
Correct answer
Also accepted:
Why
Solve the simpler equation x-y=3 for one variable, then substitute that expression into the other equation.
Given the system 4x + 3y = 7 and 5x - 2y = -20, compute x.
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Why
Scale the first equation by 2 and the second by 3 to match the y-coefficients: 8x+6y=14 and 15x-6y=-60. Adding gives 23x = -46, so x = -2. (Then y = 5.)
Correct answer
Also accepted:
Why
Multiply the equations by constants that make the coefficients of one variable match (or become opposites), then add or subtract to eliminate it.
Two numbers have a sum of 22 and a difference of 6. What is the larger number? (For reference, the choices given are: 14; 8; 16; 28.)
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That's the smaller number — 8 = 22 - 14, and 14 - 8 = 6 checks out, but the question asks for the larger one.
16 would need a difference of 16 - 6 = 10 from the other number to sum to 22, not 6 — recheck the system x+y=22,\, x-y=6.
28 is already bigger than the sum of both numbers combined, so it can't be one of them.
Why
Let x, y be the numbers with x > y: x + y = 22 and x - y = 6. Adding gives 2x = 28, so x = 14 (and y = 8).
Let the two numbers be x and y, write the sum and difference as two equations, then add them to eliminate one variable.
14
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
A system's two lines turn out to be parallel and distinct. How many solutions does the system have?
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Exactly one solution needs the lines to actually cross — parallel lines never meet.
Infinitely many requires the lines to be the same line, not just parallel. 'Distinct' rules that out here.
Two straight lines can never cross exactly twice — that shape isn't geometrically possible.
Why
Parallel and distinct lines run side by side forever without touching, so no point satisfies both equations at once: no solution. This is the inconsistent case.
From College Algebra
Solve the system \;2x + y = 9,\;\; x - y = 3,\; where the y-coefficients are unit values. Enter the value of x.
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Why
Add the two equations: the y's cancel, leaving 3x = 12, so x = 4 (and y = 1). Check: 2(4)+1=9 ✓ and 4-1=3 ✓.
Correct answer
Also accepted:
Why
Where this shows up: matching supply to demand
Where this goes next
Linear algebra
Everything above generalizes rather than getting replaced. A system of three equations in three unknowns is still solved by the same eliminate and back-substitute idea — it just takes more bookkeeping to track which variable each step removes. Chapter 3 makes that bookkeeping automatic: the coefficients get lifted into an augmented matrix, and "eliminate a variable" becomes "apply a row operation." The row operations you'll meet there — swap two rows, scale a row, add a multiple of one row to another — are exactly substitution and elimination's moves, just renamed and applied to rows of numbers instead of to x's and y's directly.
That's also why the three-outcome classification above matters beyond this one section: one solution, no solution, or infinitely many will resurface as rank once matrices are on the table, and "the lines are parallel" will become "the rows are inconsistent" — same fact, new vocabulary.
What's next
Section 7 reviews the matrix tools — row operations and the 2\times2 determinant — that this section's elimination moves are quietly a preview of, before the chapter closes with how to read and write a short proof.