Systems of Linear Equations — Print
Two conditions at once. The answer is the single pair of numbers that satisfies both.
What the practice questions cover
These questions use 3 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
Two shops hire out the same bike. Shop A asks $20 to join and $6 a day. Shop B asks $50 to join but only $3 a day.
Which is cheaper? The honest answer is: it depends how long you need it.
For a weekend, A wins easily — you are barely paying off B's joining fee.
For a month, B wins: those cheaper days add up faster than the $30 extra you paid at the door.
Somewhere between the two, the lines cross. On day 10 both shops charge $80.
That crossing point is the solution of the system — the one pair of numbers that satisfies both rules at once. Before it one shop is cheaper, after it the other, and the question "which is cheaper?" has no answer without it.
What you are looking for
Graphically, each equation is a line. The solution is where they cross — the one point on both lines.
Algebraically, it is the pair of numbers that makes both equations true. Always check it in both.
Parallel lines never cross, so that system has no solution. Identical lines cross everywhere, so it has infinitely many.
Same slope tells you the answer early
Same slope, different intercept: no solution. Same slope, same intercept: infinitely many. Different slopes: exactly one.
Solving by substitution
Best when one equation already has a lone variable, like y = 2x + 1.
Put that whole expression into the other equation in place of y, using brackets. Now there is one variable, so solve it.
Then go back and substitute to find the other variable. Answer both — a solution is a pair.
Brackets, every time
Replacing y with 2x + 1 inside 3y gives 3(2x + 1), which is 6x + 3. Writing 3 · 2x + 1 loses the 3 on the constant.
Solving by elimination
Line the equations up. If one variable has coefficients that are opposites, add the equations and it vanishes.
If they are identical, subtract instead. If neither, multiply one or both equations first to make them match.
3x + 2y = 16 and 3x - 2y = 4: add them to get 6x = 20. The y terms cancel.
Subtracting flips every sign
When you subtract an equation, every term on that line changes sign, including the one after the equals. Rewrite the line with flipped signs and add instead.
Word problems with two unknowns
Two unknowns need two equations. One usually counts things; the other usually counts money, or time, or weight.
"8 tickets cost $94; adult tickets are $15 and child tickets $8" gives a + c = 8 and 15a + 8c = 94.
Define your letters in writing before you set anything up. Half of all errors here are forgetting which letter meant which.
Worked examples
Substitution
Solve y = 3x - 4 and 2x + y = 11.
- Substitute: 2x + (3x - 4) = 11 Replace y with what it equals.
- Combine: 5x - 4 = 11 2x + 3x = 5x.
- 5x = 15, so x = 3 Add 4, then divide by 5.
- y = 3(3) - 4 = 5 Back-substitute. Check in the other: 2(3) + 5 = 11 ✓
(3, 5)
Elimination
Solve 4x + 3y = 18 and 4x - y = 2.
- The x coefficients match, so subtract. Subtracting kills the 4x.
- (4x + 3y) - (4x - y) = 18 - 2 gives 4y = 16 3y - (-y) = 4y. Both signs on the second line flipped.
- y = 4 Divide by 4.
- 4x - 4 = 2, so 4x = 6 and x = 1.5 Check the first: 4(1.5) + 3(4) = 6 + 12 = 18 ✓
(1.5, 4)
Tickets
A family buys 8 tickets for $99. Adult tickets cost $15, child tickets $8. How many of each?
- Let a be adult tickets and c child tickets. Name both unknowns before anything else.
- Count: a + c = 8. Cost: 15a + 8c = 99. One equation for how many, one for how much.
- From the first, c = 8 - a. Substitute: 15a + 8(8 - a) = 99 Brackets around the whole expression.
- 15a + 64 - 8a = 99, so 7a = 35 and a = 5 Collect the a terms, then undo.
- c = 8 - 5 = 3 Check: 15(5) + 8(3) = 75 + 24 = 99 ✓
5 adult and 3 child tickets
Words to know
- System
- Two or more equations that must hold at the same time.
- Solution of a system
- The (x, y) pair that works in every equation.
- Substitution
- Solve one equation for a variable, then put that expression into the other.
- Elimination
- Add or subtract the equations so one variable disappears.
Mistakes to watch for
Finding x and stopping. A system's solution is an ordered pair, because it names a point. Half an answer scores nothing, and back-substituting takes one line.
Forgetting brackets when substituting. Replacing y with 2x + 1 inside 3y gives 3(2x + 1) = 6x + 3. Writing 3 · 2x + 1 loses the 3 on the constant and quietly changes the equation.
Flipping only the first sign when subtracting an equation. Subtracting (4x - y = 2) changes every term, including the one after the equals. Safer: rewrite that line with all signs flipped, then add.
Checking in one equation only. A wrong pair often satisfies one of the two by luck. It is the second check that catches it, and it costs ten seconds.
The method of elimination appears in the Chinese text Nine Chapters on the Mathematical Art, compiled around 200 BC.
Its chapter Fangcheng lays out the coefficients in a grid and clears them column by column — essentially the matrix method taught today.
Europe arrived at the same technique roughly 1,800 years later, and now calls it Gaussian elimination.
Problem 1
Solve y = 2x and x + y = 9.
Worked answer
(3, 6)
x + 2x = 9
3x = 9, so x = 3
y = 6
Problem 2
Solve y = x + 4 and y = 3x.
Worked answer
(2, 6)
x + 4 = 3x
4 = 2x, so x = 2
y = 6
Problem 3
Solve x + y = 10 and x - y = 4.
Worked answer
(7, 3)
2x = 14, so x = 7
7 + y = 10, so y = 3
Problem 4
Solve 2x + y = 11 and 2x - y = 5.
Worked answer
(4, 3)
Add: 4x = 16, so x = 4
8 + y = 11, so y = 3
Problem 5
Is (2, 5) a solution of y = 3x - 1 and x + y = 7? Answer yes or no.
Worked answer
yes
3(2) - 1 = 5 ✓
2 + 5 = 7 ✓
Yes
Problem 6
Solve y = 4x - 3 and 2x + y = 15.
Worked answer
(3, 9)
2x + 4x - 3 = 15
6x = 18, so x = 3
y = 9
Problem 7
Solve 3x + 2y = 16 and x - 2y = 0.
Worked answer
(4, 2)
4x = 16, so x = 4
4 - 2y = 0, so y = 2
Problem 8
Solve 5x + 3y = 21 and 5x - y = 9.
Worked answer
(2.4, 3)
4y = 12, so y = 3
5x - 3 = 9, so 5x = 12
x = 2.4
Problem 9
How many solutions does y = 3x + 1 and y = 3x - 4 have?
- None
- Exactly one
- Infinitely many
- Exactly two
Worked answer
None
Same slope 3, different intercepts
The lines are parallel
They never meet
Problem 10
Two lines cross at (5, -2). What is the solution of the system?
Worked answer
(5, -2)
The crossing point satisfies both equations
(5, -2)
Problem 11
Two numbers add to 30 and differ by 8. Find them.
Worked answer
19 and 11
Add: 2x = 38, so x = 19
y = 11
Check: 19 + 11 = 30 and 19 - 11 = 8 ✓
Problem 12
Solve x + 3y = 14 and 2x - y = 7.
Worked answer
(5, 3)
2x + 6y = 28
Subtract: 7y = 21, so y = 3
x + 9 = 14, so x = 5
Problem 13
Adult tickets cost $12 and child tickets $7. 40 tickets sold for $355. How many adult tickets?
Worked answer
15
a + c = 40 and 12a + 7c = 355
c = 40 - a, so 12a + 280 - 7a = 355
5a = 75, so a = 15
Problem 14
A boat travels 36 km downstream in 2 hours and 36 km upstream in 3 hours. Find the speed of the boat in still water.
Worked answer
15
b + c = 18 and b - c = 12
Add: 2b = 30
b = 15 km/h
Problem 15
How many solutions does 2x + 4y = 10 and x + 2y = 5 have?
Worked answer
infinitely many
Divide the first by 2: x + 2y = 5
Identical to the second
The same line, so infinitely many solutions
Problem 16
Solve 4x + 3y = 10 and 3x - 2y = 16.
Worked answer
(4, -2)
× 2: 8x + 6y = 20; × 3: 9x - 6y = 48
Add: 17x = 68, so x = 4
16 + 3y = 10, so y = -2