What Is a System of Equations?
A system of equations is two (or more) equations that we want to be true at the same time. Each linear equation is a line on the coordinate plane.
A solution to the system is an ordered pair that makes every equation true. Graphically, it is the point where the lines cross.
One linear equation has a whole line of solutions. A system puts two of them side by side and asks which pair satisfies BOTH at once — which is why its picture is the place where two lines meet. Substitution replaces one variable by an expression the first equation says it equals; elimination adds multiples of the equations so that a variable cancels. Both are legal for the same reason: neither changes which pairs make the statements true. The answer is one point, no point (parallel lines), or every point on one line (the same line written twice), and the algebra says which: if every variable disappears, what is left is either false (no solution) or true (infinitely many).
Each equation alone is satisfied by a whole line of points: holds for , , and infinitely many more. Asking for both equations at once keeps only the points on both lines. Two different non-parallel lines share exactly one point, which is why a typical system has exactly one solution.
We usually write a system with a curly brace:
This means: find the one pair that fits both equations.
Is a solution of the system above?
- First equation: . ✓
- Second equation: . ✓
Both are true, so is the solution. If even one equation failed, it would not be a solution.
Checking a proposed solution in only one equation: satisfies , so "it is the solution." It fails (). A point on one line only is not a solution of the system.
Remember: A solution must satisfy all equations. Always plug your answer back into both equations to check.
Which of , , solves ?
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: , out. : , out. : and , both true. The solution is .
Method 1: Solving by Graphing
Graph both lines on the same axes. The intersection point is the solution, because that point lies on both lines at once.
Steps: (1) Put each equation in slope-intercept form if helpful. (2) Plot each line using its -intercept and slope (or a small table of points). (3) Read off the point where they cross. (4) Check by substituting.
A graph is a picture of every solution of one equation. Drawing two graphs on the same axes and looking for a shared point is therefore the definition of a system's solution made visible. Its weakness is precision: a crossing at cannot be read off a grid, so the algebraic methods are needed for exact answers.
Solve
Line 1: passes through and .
Line 2: passes through and .
Both tables contain the point , so the lines cross there.
Check: ✓ and ✓
Solution: .
Reading an intersection at a non-grid point as the nearest grid point. The lines and cross at ; reading "" off a rough sketch gives a point on neither line. Always check a graphed answer in both equations.
Tip: Graphing shows the answer at a glance, but it is only exact when the intersection lands on a clean grid point. For messy or fractional answers, use substitution or elimination instead.
Solve by graphing (use intercepts), then confirm algebraically.
Show the answerHide the answer
Line 1 passes through and ; line 2 through and . They share . Check: and .
Method 2: Solving by Substitution
Solve one equation for one variable, then substitute that expression into the other equation. Now you have one equation with one variable, which you can solve. Finally, back-substitute to find the other variable.
Best when: one variable is already isolated (like ) or is easy to isolate.
If is true at the solution, then and are the same number there, so either name can be used in the other equation. Replacing by removes one unknown and leaves an equation you already know how to solve. Nothing about the second equation changes except the name of one quantity.
Solve
Step 1. The first equation already gives .
Step 2. Substitute in place of in the second equation:
Step 3. Combine and solve for :
Step 4. Back-substitute into :
Check: ✓
Solution: .
Substituting back into the equation that was just used: finding from , then putting into again, which only returns . Use the isolated equation to get .
Solve
Step 1. Isolate in the first equation: .
Step 2. Substitute into the second: .
Step 3. Solve: .
Step 4. Then .
Check: ✓ Solution: .
Watch out: Use parentheses when you substitute, especially with a minus sign in front. Distribute carefully.
Solve .
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Substitute: , so , , . Back-substitute: . Check: . Solution .
Method 3: Solving by Elimination (Linear Combination)
Add or subtract the two equations so that one variable cancels. To make a variable cancel, its coefficients must be opposites (like and ). If they are not, multiply one or both equations by a number first.
Best when: the equations are in the form and no variable is isolated.
Adding two true equations gives another true equation, and every solution of the pair still satisfies it. When the -terms cancel, that new equation, , is the vertical line : a third line through the same crossing point. Elimination is a way of choosing a combination whose line is so simple it can be read off.
Solve
The and are already opposites. Add the equations:
Back-substitute into : .
Check: ✓ Solution: .
Subtracting only the first terms: computed as . Subtracting an equation changes the sign of every term: . Adding the opposite equation avoids the trap.
Solve
Step 1. Choose a variable to eliminate. Multiply the first equation by so the -terms match:
Step 2. Subtract the second equation from this new one:
Step 3. Back-substitute into : .
Check: ✓ Solution: .
Solve
To cancel , make both -coefficients . Multiply the first by and the second by :
Subtract: .
Back-substitute into : .
Check: ✓ Solution: .
Tips for elimination: Line up like variables in columns before you start. When you subtract, change the sign of every term in the equation you subtract. Double-check signs, since that is the most common mistake.
Solve .
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Multiply the first by : . Add the second: , so . Then gives . Check: . Solution .
How Many Solutions? Three Possibilities
Two lines can meet in exactly three ways:
- One solution: the lines cross once (different slopes). This is the usual case.
- No solution: the lines are parallel (same slope, different -intercept). They never meet.
- Infinitely many solutions: the two equations are really the same line (same slope and same intercept). Every point on the line works.
Two lines in a plane can only cross once, never meet, or coincide, and the slopes decide which. Different slopes: the lines separate at a steady rate in one direction and close in on the other, so they cross exactly once. Equal slopes: the gap between them never changes, so they never meet (different intercepts) or are the same line (same intercept).
No solution: Same slope , different intercepts parallel no solution.
Infinitely many: The second equation is just the first times , so they are the same line infinitely many solutions.
Reading at the end of elimination as "the solution is ." A true statement with no variables means the two equations describe the same line, so every point on it is a solution.
Algebra clue: If your work leads to a false statement like , there is no solution. If it leads to an always-true statement like , there are infinitely many solutions.
Without solving, how many solutions: (a) , ; (b) , ; (c) , ?
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(a) Both have slope ( and ), different intercepts: none. (b) The second is twice the first: the same line, infinitely many. (c) Slopes and differ: exactly one, .
Choosing the Best Method
- Graphing — great for a quick picture or when you want to see the answer; best with clean integer intersections.
- Substitution — best when a variable is already isolated or has a coefficient of (easy to solve for).
- Elimination — best when both equations are in form, especially if coefficients line up or match easily.
There is no single “right” method; pick whichever needs the least work for the problem in front of you.
All three methods find the same answer, so the choice is about avoiding fractions and sign errors. A variable already alone invites substitution; matching or opposite coefficients invite elimination; a quick sketch shows roughly where the answer is and catches an algebra slip.
Forcing substitution on , : isolating gives , and the fractions multiply the chances of an error. Elimination with multipliers and stays in whole numbers.
Choose a method for , solve it, and explain the choice.
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Both equations give , so set them equal (substitution): , , , and . Solution ; no rearranging was needed.
Systems of Linear Inequalities
A system of inequalities uses instead of . Instead of a single point, the solution is a whole region of the plane.
Steps: (1) Graph each boundary line. Use a solid line for or and a dashed line for or . (2) Shade the correct side of each line. (3) The solution is the overlap where the shadings cross. Any point in that overlap satisfies both inequalities.
Each linear inequality is satisfied by a whole half-plane: the boundary line splits the plane in two, and every point on one side passes while every point on the other fails (a point can only switch from passing to failing by crossing the line). The system keeps the points in every half-plane at once, their overlap. A test point decides which side is which.
Consider
Use a test point to decide which side to shade and to check membership.
Test : (true) and (true). Both hold, so is in the solution region.
Test : (true) but (false). One fails, so is not in the region.
The solution is the overlapping band above the line and below the line .
Shading "above" for every without solving for first. For , dividing by flips it to : the solutions are below the line. Test a point when in doubt.
Tip: is the easiest test point to plug in, as long as no boundary line passes through the origin. A point is a solution only if it satisfies every inequality.
For and , decide whether , and are solutions.
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: and , yes. : fails, no. : holds but fails, no (it is on the dashed boundary, which is excluded).
Applications: Word Problems
Many word problems hide a system. Steps: (1) Define two variables clearly. (2) Write one equation for each fact in the problem. (3) Solve with any method. (4) Answer in a full sentence and check that it makes sense.
Two unknowns need two independent facts, and word problems almost always supply one fact that counts (how many tickets) and one that measures (how many dollars). Each fact alone allows many combinations; the two together pin down one.
A theater sells $5 adult tickets and $3 child tickets. One night they sold tickets for a total of $38. How many of each?
Define: adult tickets, child tickets.
System:
Solve by substitution: , so , then .
Check: ✓ Answer: adult and child tickets.
Writing both facts as counts: and "somehow." Each price multiplies its own count: the money equation is . Check the answer against the story: adults and children make tickets and .
A chemist mixes a acid solution with a acid solution to make mL of a acid solution. How much of each?
Define: mL of solution, mL of solution.
System:
Solve: , so , then .
Check: ✓ Answer: mL of and mL of .
Tip: Distance–rate–time and current problems often become systems too. If a boat's speed is and the current is , then downstream speed is and upstream speed is . Two trips give you two equations.
A jar holds coins, all dimes and quarters, worth . How many of each?
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Let dimes and quarters. Count: . Value in cents: . Substitute : , so and , . Check: cents.
What a System Is Actually Asking
A single equation like has infinitely many solutions — pick any and follows. Its picture is a whole line of points. A system puts two such equations side by side and asks for the pairs that satisfy BOTH at once, which is why its picture is the place where two lines meet.
That is the whole idea, and it explains the shape of every method: you are not solving two separate equations, you are looking for the one input pair that keeps both statements true. Every legal step must preserve that pair — which is exactly what substitution and elimination are built to do.
One solution. The lines have different slopes and cross once. The algebra ends with a normal statement like .
No solution. The lines are parallel and distinct. The algebra ends with something FALSE and variable-free, such as .
Infinitely many solutions. The two equations describe the same line. The algebra ends with something TRUE and variable-free, such as , and the answer is the whole line, usually written or as .
Notice what decides the case: whether the variables all disappear, and then whether what is left is true.
The answer to a system is a POINT. Writing "" is half an answer; the solution is . Losing the second coordinate is the single most common way to lose marks on this topic.
Substitution, Step by Step
1. Pick the variable that is cheapest to isolate — one with a coefficient of or if there is one.
2. Solve that equation for it, giving an expression in the other variable.
3. Substitute that expression into the OTHER equation. (Into the same one and you get , which tells you nothing.)
4. Solve the resulting one-variable equation.
5. Back-substitute to get the second value, and state the answer as a point.
6. Check the point in BOTH original equations.
Solve and .
Step 1 — already isolated. The first equation hands us for free.
Step 2 — substitute into the other equation. .
Step 3 — solve. , so and .
Step 4 — back-substitute. .
Answer. .
Check. First: ✓. Second: ✓.
Solve and .
Step 1 — choose. in the first equation has coefficient , so isolating it costs nothing: .
Step 2 — substitute. .
Step 3 — solve. , so and .
Step 4 — back-substitute. .
Answer. .
Check. ✓ and ✓.
Replacing by is legitimate because the first equation SAYS those two things are the same number. You are not changing the second equation; you are writing one of its quantities under a different name.
Elimination, Step by Step
1. Line the equations up so like terms sit above like terms.
2. Choose a variable to kill. Multiply one or both equations by constants so its coefficients become opposites (or equal).
3. Add the equations (or subtract, if you made them equal). One variable disappears.
4. Solve for the survivor.
5. Back-substitute into either original equation, and state the point.
6. Check in both originals.
Solve and .
Step 1 — choose a target. Kill : the second equation has coefficient , so multiply it by .
Step 2 — scale. .
Step 3 — add. , so and .
Step 4 — back-substitute. , so .
Answer. .
Check. ✓ and ✓.
Solve and .
Step 1 — choose a target. Kill . The coefficients are and , whose least common multiple is .
Step 2 — scale both. Multiply the first by : . Multiply the second by : .
Step 3 — add. , so .
Step 4 — back-substitute. , so and .
Answer. .
Check. ✓ and ✓.
If and , then — you added equal amounts to the two sides of a true statement. The step is reversible (subtract the same equation again), so no solution is gained and none is lost.
Choosing a Method
Both methods always work, so the choice is about arithmetic, not correctness.
Use substitution when a variable is already alone () or has a coefficient of or — isolating it costs nothing and creates no fractions.
Use elimination when every coefficient is or bigger, or when one variable already has opposite coefficients in the two equations. Substituting there would create fractions for no reason.
Use graphing only to SEE what is going on or to check an answer. A graph gives an approximate crossing point; it cannot certify .
Solve and .
By elimination. The coefficients are already opposites: adding gives , so , then gives . That is two lines of work.
By substitution. , then , so , , and . Also correct, but it took a distribution and a subtraction more.
Answer. — and the check is the same either way: ✓, ✓.
No Solution, and Infinitely Many
Solve and .
Step 1 — eliminate. Multiply the second by : . Add: .
Step 2 — read it. Every variable vanished and what is left is FALSE, so no pair can satisfy both equations.
Answer. No solution.
Why, in a picture. Both lines have slope , but the first passes through and the second through : parallel and distinct, so they never meet.
Solve and .
Step 1 — eliminate. Multiply the first by : . Add: .
Step 2 — read it. The variables vanished and what is left is TRUE, so the second equation was the first one in disguise (it is exactly twice it).
Answer. Every point on ; in parametric form for any real .
Check the logic. Two solutions of it: and — both satisfy as well.
Write both equations as . Compare the ratios , , .
If the first two differ, the lines cross: one solution.
If all three are equal, it is one line twice: infinitely many.
If the first two are equal but the third differs, the lines are parallel: no solution.
In the example above, but — parallel, as found.
Systems in Words
A word problem becomes a system as soon as you name the unknowns and find two independent statements about them. Almost always one statement COUNTS things (tickets, coins, litres) and the other MEASURES them (dollars, value, concentration) — that pairing is what makes the two equations independent rather than the same fact twice.
A cinema sold tickets and took . Adults pay and children . How many of each were sold?
Step 1 — name them. Let be the number of adult tickets and the number of child tickets.
Step 2 — the counting fact. .
Step 3 — the money fact. .
Step 4 — eliminate . Multiply the counting equation by : . Add: , so .
Step 5 — back-substitute. .
Answer. adult and child tickets.
Check. ✓ and ✓.
Why ? Multiplying the counting equation by the CHILD price turns it into "what the takings would have been if everyone paid child price". Subtracting that from the real takings leaves only the extra each adult paid — so the division by is counting adults.
A boat covers km downstream in hours and the same km back in hours. Find its speed in still water and the speed of the current.
Step 1 — name them. Let be the boat's speed in still water and the current, both in km/h.
Step 2 — turn distances into rates. Downstream the boat makes km/h; upstream km/h.
Step 3 — the two facts. Downstream the current helps: . Upstream it hinders: .
Step 4 — add and subtract. Adding: , so . Subtracting: , so .
Answer. km/h in still water, current km/h.
Check. ✓ and ✓ — and the current came out positive, as it must when the return trip is slower.
How many litres of a acid solution and of a solution make litres of a solution?
Step 1 — name them. litres of the weak one, litres of the strong one.
Step 2 — the volume fact. .
Step 3 — the acid fact. .
Step 4 — clear the decimals and eliminate. Multiply the acid equation by : . Multiply the volume equation by : . Add: , so and .
Answer. litres of the and litres of the .
Check. ✓, which is of ✓. Sanity: the target sits nearer than , so most of the mixture should be the weak solution — and it is.
When the Coefficients Are Letters
Elimination multiplies, adds and divides. None of those steps ever asked whether was or just the letter — so the same procedure solves exactly as it solves . What changes is the ending: instead of a number you get a FORMULA, and you have to say when that formula is allowed to exist.
Solve and for and .
Step 1 — add. , so .
Step 2 — subtract. , so .
Answer. , — the familiar "half the sum, half the difference".
Check. Their sum is ✓ and their difference is ✓, whatever and are.
Solve and for and .
Step 1 — eliminate . Multiply the first by and the second by , then subtract: .
Step 2 — divide. , and the mirror-image calculation gives .
Step 3 — state the condition. Dividing is only allowed when . That quantity is the determinant of the system, and it is the same number that decides the three cases from earlier: when it is zero the two lines are parallel or identical, and there is no single point to report.
Check with numbers. Put : the formula gives and , and indeed ✓, ✓.
An answer in letters is not finished until you have said which values are excluded. ", provided " is a complete answer; the fraction on its own is not.
Going Deeper: Advanced Systems
A three-variable system has three equations in , , and . A solution is an ordered triple that satisfies all three. Geometrically each equation is a plane in space, and a single solution is the point where all three planes meet.
Elimination strategy: pick one variable to remove. Combine equations in two different pairs to eliminate that same variable, leaving a system in the other two variables. Solve that smaller system, then back-substitute to recover the third variable.
Solve
Step 1. Eliminate from two pairs. Add equation (1) and equation (3):
Add equation (2) and equation (3):
Step 2. Back-substitute into :
Step 3. Find from equation (1): .
Check (all three): ✓, ✓, ✓
Solution: .
Sometimes a coefficient is a letter (a parameter) and we ask: for which values does the system have one, none, or infinitely many solutions? Compare the ratios of the coefficients. For
- One solution when (different slopes).
- Infinitely many when (same line).
- No solution when (parallel lines).
For which value of does fail to have a unique solution?
The -ratio is . The -ratio is . These match when , i.e. . So:
- If : exactly one solution for any .
- If and : the second equation is just the first, so infinitely many solutions.
- If and : parallel lines, so no solution.
A system is symmetric if swapping and leaves it unchanged (for example, using , , or ). Let (the sum) and (the product). A key identity converts squares into these:
Once you know and , the two numbers and are exactly the roots of
Solve
Step 1. Let . Use the identity :
Step 2. Now and are the roots of , which factors as , giving or .
Step 3. So . The solutions are and .
Check: ✓ and ✓
When a system of linear inequalities defines a feasible region, we often want to maximize (or minimize) a linear objective function such as . The Corner-Point Theorem says the maximum and minimum of a linear objective always occur at a vertex (corner) of the feasible region.
Steps: (1) Graph the constraints and find the feasible region. (2) Find each corner point (each is the intersection of two boundary lines). (3) Evaluate the objective at every corner. (4) The largest value is the maximum; the smallest is the minimum.
Maximize subject to
Find the corners. The boundary lines are and , together with the axes.
- — origin.
- — where meets .
- — where meets .
- — solve and : subtract to get , then .
Evaluate at each corner:
Maximum: at the corner .
Not every system is two lines. When one equation is a parabola (like ) and the other is a line, substitution still works, but the result is a quadratic. A line and a parabola can cross in two points, one point (tangent), or none, depending on the discriminant of that quadratic.
Solve
Step 1. Substitute into :
Step 2. Factor: , so or .
Step 3. Find each from : if , ; if , .
Check: gives ✓ and ✓
Two solutions: and .
For a system
the determinant of the coefficients is
If there is exactly one solution, given by Cramer's Rule:
Each numerator replaces one coefficient column with the constants . If , the system has no unique solution (either none or infinitely many).
Solve using determinants.
Here and .
Main determinant:
Solve for :
Solve for :
Solution: , matching the elimination result earlier.
Big picture: Every method here rests on the same idea — reduce a hard system to a simpler one. Three-variable elimination shrinks to a ; symmetric systems shrink to a single quadratic in ; Cramer's Rule packages the answer as a ratio of determinants. Whenever a determinant is (or ratios all match), watch for the “no solution / infinitely many” special cases.
Why It Works
Elimination is legitimate because adding equal quantities to both sides of an equation preserves it: if and then . Any solution of the original pair satisfies the new equation, and since the step is reversible (subtract again), no solutions are gained. Substitution is the same principle: replacing a quantity by something known to equal it changes nothing about the solution set.
Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.
Formulas, Proofs & Tips
What it means. A solution is a point on both lines, so the answer count is how many times they meet.
Example. and : adding gives , so , .
Why it works. Elimination adds a multiple of one equation to the other. Since both sides are equal quantities, the new equation still holds for every common solution — but one variable is gone.
Tip. If the variables all vanish and leave a false statement (), there is no solution; a true one () means infinitely many.