Picture two walkers starting at different places on straight paths. A meeting address must belong to both paths at once. A system of linear equations asks you to obey several line rules together. With two lines having different directions, their point of intersection is their shared address. At that input, both rules give the same output. You can find it by setting their output formulas equal, solving for the input, then using that input to find the output. A drawing helps you see the meeting, while algebra gives exact coordinates. Parallel paths never meet. Two descriptions of the same path share every address, so that system has infinitely many solutions.
A shared point has the same input and output on both lines.
Reminder
Exact decimal coordinates.94 = 225100 = 2.25 and 114 = 275100 = 2.75. Multiply both fraction numbers by 25 to make hundredths.
Variables on both sides. 3t − 4 = 5 − t becomes 4t = 9 after adding t and 4.
Fraction subtraction. 5 − 94 = 204 − 94 = 114.
Substitution. h(94) means put 94 everywhere t occurs.
Coordinate order. The input t comes first and the common output comes second in the point.
Coincident and parallel lines. Equal slopes with different intercepts never meet; equal intercepts too mean the same line.
Undoing a coefficient. 4t = 9 gives t = 94 by dividing both sides by 4; 4 × 94 = 9 checks it.
Why it works. At a shared point, both equations must give the point's single output y. If y = h(t) and y = j(t), then h(t) = j(t). Solving that equality locates the shared input; substitution finds its height. Different slopes make one input satisfy it. Equal slopes cancel the input terms. Different intercepts then leave a false statement, a contradiction, meaning no solution. Equal intercepts leave a true statement such as 0 = 0, meaning every input on their common line works.
RuleRule: A solution to a system satisfies every equation. For y = f(x) and y = g(x), solve f(x) = g(x), find y, and check both. Two lines have one, no, or infinitely many shared points.
The same idea, five ways
Say it
Say: find an address that obeys both rules.
Write it
The point of intersection is a solution because it lies on both graphs.
In math
y = f(x) and y = g(x)
f(x) = g(x)
solution: (x, y)
no solution: no shared point
infinitely many solutions: the same line
Like
A meeting address must be on both people's paths.
See itA shared point has the same input and output on both lines.
The same idea, other ways
As a picture
Where the graphs cross, the input position and output height match. That is one address, not two separate answers.
A shared point has the same input and output on both lines.
As two machines
For input 2, one machine computes 2 × 2 + 1 = 5. The other computes −2 + 7 = 5. Matching outputs make (2, 5) a shared point.
The same numeric input produces the same output under both formulas.
As an equation
Setting 2x + 1 = −x + 7 says the outputs must agree. Adding x and subtracting 1 gives 3x = 6, so x = 2. Plugging 2 into both formulas gives 5.
The equality selects an input where both output expressions balance.
.1One solution
Two lines with different directions cross once. Their one shared address has an input and an output, so finding only the input is unfinished. A graph may give an approximate location unless the crossing coordinates are clearly marked.
Rule: Different slopes give one intersection for two nonvertical lines.
Rule: A vertical line and any nonvertical line also meet once.
Rule: Graph estimates can be rounded; substitution confirms an exact answer.
A shared point has the same input and output on both lines.
Reminder
Substitution. At x = 2, 2x + 1 = 5.
Both-side variables. 2x + 1 = −x + 7 becomes 3x = 6 after adding x and subtracting 1.
The same idea, five ways
Say it
Say: one shared address.
Write it
The two rules agree at exactly one point.
In math
f(x) = g(x) at one input
solution: (x, f(x))
Like
Two crossing streets have one crossing.
See itA shared point has the same input and output on both lines.
Worked exampleRead and check a crossing
You need the shared address of y = 2x + 1 and y = −x + 7.
A shared point has the same input and output on both lines.
Read the marked crossing (2, 5).The point lies where both plotted lines meet.
Set 2x + 1 = −x + 7. Add x and subtract 1: 3x = 6; divide by 3 to get x = 2.Equal outputs find the shared input exactly.
Plug in: 2 × 2 + 1 = 5 and −2 + 7 = 5.Both original rules must return the found output.
Answer
(2, 5)
Check Their slopes 2 and −1 differ, so there is only one crossing; the point passes both substitution checks.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: the system's answer is x = 2, with no output.
The question asks for the shared point, which needs two coordinates.
✓ Instead: Substitute 2 to find y = 5, then report (2, 5).
Tips and tricks
Tip: Read across for the input and up for the output at the crossing.
.2No solution
Distinct parallel lines are separated paths. Their equations cannot share an address. When you try to equate their outputs, their matching input terms cancel and leave a false statement, called a contradiction.
Rule: Equal slopes and different intercepts give no solution.
Rule: A contradiction such as 0 = 3 says no input can make the equations true together.
Distinct parallel lines never meet, so this system has no solution.
Reminder
Subtract on both sides. x + 1 = x + 4 becomes 1 = 4 after subtracting x.
The same idea, five ways
Say it
Say: no address obeys both.
Write it
The system has no solution because the graphs never meet.
In math
no solution
0 = 3 is a contradiction
Like
A point cannot be on both separated train rails.
See itDistinct parallel lines never meet, so this system has no solution.
Worked exampleRecognize a contradiction
You need the shared point, if any, of y = x + 1 and y = x + 4.
Distinct parallel lines never meet, so this system has no solution.
Set x + 1 = x + 4 and subtract x: 1 = 4.The matching slopes cancel, leaving their unequal intercepts.
Subtract 1: 0 = 3. Report no solution.A false statement cannot hold for any input; there is no value to plug back in.
Answer
No solution.
Check At every input, the second output is 3 larger. For example, input 0 gives outputs 1 and 4, and the same gap remains everywhere.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: 0 = 3 means x = 0.
There is no x left in the statement, and 0 does not equal 3.
✓ Instead: No input works. Substituting x = 0 gives unequal outputs 1 and 4.
Tips and tricks
Tip: When the variable disappears, read the remaining statement instead of inventing a variable value.
.3Infinitely many solutions
Two coincident line equations are two names for one path. Every point on that path obeys both. The system has infinitely many solutions, meaning there is no last solution point, but points away from the line are still not solutions.
Rule: Coincident lines have infinitely many solutions.
Rule: A true equality such as 0 = 0 after valid simplification means every allowed input on the common line works.
A shared point has the same input and output on both lines.
Reminder
Coincident lines. Same slope and same intercept mean the same line.
The same idea, five ways
Say it
Say: every point on the shared line works.
Write it
Infinitely many solutions belong to the common line, not to the entire plane.
In math
y = x + 1 and 2y = 2x + 2
solutions: (x, x + 1) for every real x
Like
Two maps show the exact same road.
See itA shared point has the same input and output on both lines.
Worked exampleRecognize the same line twice
You need all shared points of y = x + 1 and 2y = 2x + 2.
A shared point has the same input and output on both lines.
Divide the second equation by 2 to get y = x + 1.Every term divides by the same nonzero number.
Equate outputs: x + 1 = x + 1. Subtract x and 1 to obtain 0 = 0.This is true for every input, so algebra imposes no further restriction.
Write the solutions as (x, x + 1) for every real x.The output must still lie on the common line.
Check x = 2: y = 3, and 2 × 3 = 2 × 2 + 2 = 6.A sample point confirms both original equations.
Answer
Infinitely many solutions: every point (x, x + 1).
Check The equations have the same slope 1 and intercept 1, confirming coincidence from their graphs.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: infinitely many solutions means every point in the plane works.
The output still must equal x + 1. The point (0, 0) fails that equation.
✓ Instead: Only points on y = x + 1 work; at input 0 the output must be 1.
Tips and tricks
Tip: Keep the common line equation when stating infinitely many solutions.
Strategy: step by step
1. Graph the two lines to locate or estimate their shared point.
2. If both outputs are formulas in the same input, set those formulas equal.
3. Solve for the shared input. Watch for a contradiction or an equality true for every input.
4. Substitute a found input into one original formula to obtain the output.
5. Substitute the point into both original equations, then report both coordinates.
Strategy
Strategy: solve a two-line system
1
Does one line have equation x = c?
YesSubstitute c into the other nonvertical line. If both lines are vertical, compare their fixed inputs.
NoEquate their output formulas.
↓
2
Does the input variable disappear?
YesA false equality means no solution; a true equality means the lines coincide.
NoSolve for the input, substitute to find the output, and check both.
Put each nonvertical equation in output form.
Set the formulas equal and solve.
If a variable value remains, find its output and check both equations.
If the variable disappears, decide whether the remaining equality is false or always true.
Use the graph to interpret the result.
Worked exampleThe verified fractional intersection
You need the address where h(t) = 3t − 4 and j(t) = 5 − t have the same output.
A shared point has the same input and output on both lines.
Set 3t − 4 = 5 − t.At a shared input, both outputs must be equal.
Add t to both sides: 4t − 4 = 5.Collecting input terms makes one coefficient to undo.
Add 4 to both sides: 4t = 9.This isolates the input's multiplied amount.
Divide by 4: t = 94.Division finds the one input that can make the outputs agree.
Find its output in j: j(94) = 5 − 94 = 204 − 94 = 114.Rewrite 5 in quarters to subtract equal-sized pieces.
Plug the same input into h: h(94) = 274 − 164 = 114.The original second rule must give the same output; this verifies the solved input and point.
Answer
(94, 114)
Check The exact point is (2.25, 2.75). Both numbers terminate exactly. The graph places the crossing between inputs 2 and 3 at that height, confirming the location.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: a whole-number crossing
You need the shared point of y = 2x + 1 and y = −x + 7.
A shared point has the same input and output on both lines.
Set 2x + 1 = −x + 7.A shared input must make the two output formulas agree.
Add x to both sides: 3x + 1 = 7.This gathers the input terms on one side.
Subtract 1 from both sides: 3x = 6.This removes the constant beside the multiplied input.
Divide by 3: x = 2.Division finds the one input that can satisfy the equal-output condition.
Substitute into both original formulas: 2 × 2 + 1 = 5, and −2 + 7 = 5.The found input must give the same output under each rule, so the complete point is (2, 5).
Answer
(2, 5)
Check The graph crosses at the marked point (2, 5), confirming the exact calculation.
Rung 2Rung 2: a fractional crossing
You need the shared point of h(t) = 3t − 4 and j(t) = 5 − t.
A shared point has the same input and output on both lines.
Set 3t − 4 = 5 − t.This selects a common input where both outputs agree.
Add t to both sides: 4t − 4 = 5.This collects all input terms on the left.
Add 4 to both sides: 4t = 9.This clears the constant beside the multiplied input.
Divide both sides by 4: t = 94.This finds the input that can make the original outputs equal.
j(94) = 5 − 94 = 204 − 94 = 114.The common input still needs a height; rewriting 5 in quarters permits the subtraction.
h(94) = 3 × 94 − 4 = 274 − 164 = 114.The same height under the other original rule verifies both coordinates.
Answer
(94, 114)
Check Both exact coordinates match the graph location (2.25, 2.75).
Rung 3Rung 3: no crossing
You need any shared point of y = 3x − 1 and y = 3x − 4.
Distinct parallel lines never meet, so this system has no solution.
Set 3x − 1 = 3x − 4.This asks whether any input can produce the same output in both rules.
Subtract 3x from both sides: −1 = −4.Matching slopes cancel the variable terms, leaving only the different intercepts.
Add 4 to both sides: 3 = 0.The resulting equality is false; it cannot be made true by choosing an input.
Report no solution. At x = 0 the original outputs are −1 and −4, and their gap stays 3 for every x.There is no found input to substitute; the constant gap confirms that no shared point exists.
Answer
No solution.
Check Their equal slopes and different intercepts make them parallel; the gap stays 3 at every input.
Rung 4Rung 4: every point on one line
You need the shared points of y = 2x + 1 and 2y = 4x + 2.
A shared point has the same input and output on both lines.
Divide the second equation by 2: y = 2x + 1.This reveals it is the first equation again.
Equate outputs and subtract both terms: 2x + 1 = 2x + 1 gives 0 = 0.Every input makes this equality true.
State points (x, 2x + 1) for every real x. For x = 1, y = 3 and both originals give 3 = 3 and 6 = 6.All answers lie on the shared line; the sample substitution checks the description.
Answer
Infinitely many solutions: (x, 2x + 1) for every real x.
Check Same slope 2 and intercept 1 confirm that the graph is one line drawn twice.
Rung 5Rung 5: one fixed input
You need the point satisfying both x = −3 and y = 4x + 10.
x = −3: fixed input
y = 4(−3) + 10 = −2
Shared point: (−3, −2)
Both equations are true
Use the fixed input in the other rule instead of equating an input with an output.The nonvertical rule gives height −2 at input −3.
Read x = −3 from the vertical equation.This equation already supplies the input coordinate of every possible shared point.
Substitute that input into the other rule: y = 4(−3) + 10 = −12 + 10 = −2.The nonvertical equation determines the output at that fixed input.
Write the point (−3, −2).A system solution needs the given input and calculated output together.
Check the first equation: x = −3. Check the second: −2 = 4(−3) + 10.Both original equations must be true at the same address.
Answer
(−3, −2)
Check The vertical line holds the across position at −3, while the other line has only one height there, −2. That is why these lines have one shared point.
Rung 6Rung 6: two vertical conditions
You need shared points for two systems. First use x = −3 and x = 5. Then use x = −3 and 2x = −6.
x = −3 and x = 5: separated lines
x = −3 and 2x = −6: same line
Common-line points: (−3, y)
Two vertical equations are compared by their fixed input values.
In the first system, the same point would need its first coordinate to be both −3 and 5.A coordinate cannot have two different values at the same point.
Therefore the first system has no solution.The separated vertical lines share no input position and no point.
For the second system, divide 2x = −6 by 2 to obtain x = −3.Balanced division shows that both equations impose exactly the same input condition.
Every point (−3, y) is a solution to the second system, with y any real number.Neither original equation restricts the output coordinate.
For example, (−3, 2) gives x = −3 and 2x = 2(−3) = −6.Substitution checks a sample point, while the absence of y from either equation explains all the others.
Answer
First system: no solution.
Second system: infinitely many solutions, every point (−3, y) for real y.
Check The first pair is two distinct vertical lines. The second pair is one vertical line written twice, so its infinite solutions stay on that line.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: set h(t) = 0 and j(t) = 0 to find their intersection.
That finds each line's separate x-intercept, not where their outputs agree. A crossing can occur above or below the x-axis.
✓ Instead: Set h(t) = j(t). Their common output here is 114, not 0.
✗ Not this: Counterexample: an apparent graph crossing gives exact coordinates even when the grid is coarse.
A drawing can show location without enough precision to distinguish nearby fractions.
✓ Instead: Treat an unmarked reading as an estimate, then solve algebraically and check both formulas.
✗ Not this: For x = −3 and y = 4x + 10, set −3 = 4x + 10.
The left number is an input condition, while the expression on the right gives an output. They are not two output formulas.
✓ Instead: Use x = −3 inside the output formula: y = −12 + 10 = −2. The shared point is (−3, −2).
Tips and tricks
Tip: Memory cue: shared input, shared output. Write both coordinates.
Tip: A contradiction means no solution; 0 = 0 means return to the common line.
Understand, then rebuild it when needed: equal outputs give the system equation. You do not need a separate intersection formula.
Trap. Finding the shared input and stopping. A point needs its output too. Also check both original equations; a point that fits only one is not a system solution.
