The solution is where the two lines cross.
Graphing shows what is happening but rarely gives an exact answer — the crossing point here is (5, 0.5), which is hard to read off a sketch. That is why algebra is used instead.
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- Solving Systems of Linear Equations
Solving Systems of Linear Equations
Two equations, two unknowns. Get rid of one variable and the rest is easy.
Solve the system
3x + 2y = 16
x − 2y = 4
The y terms are already opposites, so add the two equations: 4x = 20, giving x = 5. Substituting into the second: 5 − 2y = 4, so 2y = 1 and y = 12. The solution is (5, 12). Check in the first: 15 + 1 = 16 ✓ A satisfies the second equation only — a solution must satisfy both, which is what makes a system a system.
Elimination: make one variable cancel.
Add or subtract the equations so one variable disappears.
3x + 2y = 16
equation 1
x − 2y = 4
equation 2 — the y terms are opposites
4x = 20
add them
x = 5, then y = 12
back-substitute
Same signs? Subtract. Opposite signs? Add. Here +2y and −2y are opposites, so adding removes them.
When nothing cancels yet, multiply first.
2x + 3y = 13
nothing matches
5x − 2y = 4
target: kill y
×2: 4x + 6y = 26
multiply the whole equation
×3: 15x − 6y = 12
now +6y and −6y
19x = 38, x = 2, y = 3
add and finish
Multiply every term, both sides
Doubling 2x + 3y = 13 means 4x + 6y = 26. Forgetting the 13 is the error that quietly ruins the whole system.
Substitution suits a variable already on its own.
y = 2x − 1 and 3x + y = 14
y is already isolated
3x + (2x − 1) = 14
substitute
5x = 15, x = 3
one variable left
y = 2(3) − 1 = 5
back into the first
Choose substitution when a coefficient is 1 or a variable is already alone. Otherwise elimination avoids fractions.
When both variables vanish, read what is left.
0 = 7
false
no solution — parallel lines
0 = 0
always true
infinitely many — the same line
x = 5
a number
exactly one solution
Losing both variables is not a mistake. It is the system telling you the two lines are parallel or identical — and which one depends entirely on whether the leftover statement is true.
Seeing "none" or "infinitely many" coming, and finding k.
Compare the two equations term by term: same ratio for x and y means parallel or identical lines.
Solve 6x − 4y = 10 and 3x − 2y = 5.
×2 the second: 6x − 4y = 10
multiply every term
subtract: 0 = 0
both variables vanish
0 = 0 is always true: the first equation is just the second doubled, so they are the same line and there are infinitely many solutions.
For which k does 3x + ky = 7 and 6x + 8y = 5 have no solution?
x terms: 3 and 6, ratio 1 : 2
the second is double
y terms must match the ratio: k : 8 = 1 : 2
same slope means parallel
k = 4
constants: 7 and 5 are not 1 : 2
so the lines do not coincide
Check: with k = 4, double the first: 6x + 8y = 14. Subtracting the second gives 0 = 9, false, so there is no solution ✓.
Parallel is not enough on its own
The k that matches the slopes gives no solution only if the constants do NOT follow the same ratio. Had the first equation been 3x + 4y = 2.5, the lines would be identical and there would be infinitely many solutions instead.
Variations: a fraction answer, negative answers, both solved for x.
Elimination or substitution as before — the answers just look different.
2x + 3y = 4 and 4x − 3y = 5
6x = 9, x = 32
add: the y terms cancel
3 + 3y = 4, y = 13
2(32) = 3
Check the second: 4(32) − 3(13) = 6 − 1 = 5 ✓. Fractions are allowed answers.
x + y = −1 and 3x − y = −11
4x = −12, x = −3
add
−3 + y = −1, y = 2
back-substitute
Check: 3(−3) − 2 = −11 ✓. The crossing point is in quadrant II.
x = 2y + 1 and x = −y + 7
2y + 1 = −y + 7
both equal x, so equal each other
3y = 6, y = 2
x = 2(2) + 1 = 5
back into either
Check the other: −2 + 7 = 5 ✓. When both equations are already "x = …", set the right sides equal.
On the exam
Exams ask for the solution pair or just one coordinate, what a leftover 0 = 5 or 0 = 0 means, or the value of k that gives no solution. The fastest route is substitution when a coefficient is 1 and elimination otherwise. For a 10-second check, put your pair into both original equations — satisfying one is not enough.
Now you try
Practice problem
Solve the system
4x + y = 11
2x − 3y = −5
From the first, y = 11 − 4x. Substituting: 2x − 3(11 − 4x) = −5, so 2x − 33 + 12x = −5, giving 14x = 28 and x = 2. Then y = 11 − 8 = 3, so (2, 3). Check both: 8 + 3 = 11 ✓ and 4 − 9 = −5 ✓ C satisfies the first equation only.
Hint: y has a coefficient of 1 in the first equation, so substitution is the tidier route. Watch the sign when you distribute the −3.