⚙️ Function Kingdom · Functions

Systems of Equations

Solve two equations at once by graphing, substitution or elimination, and interpret the intersection point.

In short

A solution satisfies every equation at once, and graphically it is the point where the lines cross.
Substitution suits a system where one variable is already isolated; elimination suits matching coefficients.
A system has one solution, none (parallel lines) or infinitely many (identical lines).
Word problems usually give one equation counting items and one counting value.

Two conditions at once

A system of equations is two (or more) equations that must hold at the same time. A solution is a pair $(x, y)$ that satisfies both.

Graphically, each linear equation is a line, and a solution is a point where the lines cross. That picture explains everything that can happen:

the lines cross once — exactly one solution
the lines are parallel — no solution
the lines are identical — infinitely many solutions

Checking an answer is easy and always worth doing: substitute the pair into both original equations. If it satisfies only one of them, it is not a solution.

Answers are written as an ordered pair, $(x, y)$ , so the order matters.

Substitution

Substitution works best when one equation already gives a variable on its own, or can easily be rearranged to do so.

$y = 2x$ - $1 3x + 2y = 12$

The first equation says what y is, so replace y in the second one by that whole expression:

$3x + 2(2x$ - $1) = 12 3x + 4x$ - $2 = 12 7x = 14 x = 2$

Now go back for $y: y = 2(2)$ - $1 = 3$ . The solution is $(2, 3)$ .

Two habits prevent nearly all errors here: put the substituted expression in brackets, and never stop at x — the answer needs both numbers.

Elimination

Elimination adds or subtracts the equations so that one variable disappears.

$3x + 2y = 19 5x$ - $2y = 5$

The y-terms are $+2y$ and -2y, so adding the two equations removes y entirely:

$8x = 24$ , so $x = 3$

Now substitute back into either original equation: $3(3) + 2y = 19$ gives $2y = 10$ , so $y = 5$ . The solution is $(3, 5)$ , and the second equation confirms it: 5(3) - $2(5) = 5$ .

When the coefficients do not already match, multiply one or both equations first:

$2x + 3y = 13 5x + 2y = 16$

Multiply the first by 2 and the second by 3 to make the y-terms 6y and 6y, then subtract. Whatever you multiply an equation by must be applied to every term, including the number on the right.

Word problems: naming the unknowns

Most systems in the wild come from a sentence with two unknowns and two facts.

"Adult tickets cost 12 gold and child tickets 5 gold. 40 tickets were sold for 347 gold."

Name the unknowns first: let a be the number of adult tickets and c the number of child tickets. Now each fact becomes an equation:

counting tickets: $a + c = 40$ counting money: $12a + 5c = 347$

From the first, $c = 40$ - a. Substituting gives $12a + 5(40$ - $a) = 347$ , so $7a + 200 = 347$ , so $7a = 147$ and $a = 21$ . Then $c = 40$ - $21 = 19$ .

The pattern repeats everywhere: one equation counts how many things, the other counts how much they are worth. Coin problems, mixture problems and "two kinds of item" problems are all this shape.

Finish by checking against the story, not just the algebra: $21 + 19 = 40$ tickets, and $12(21) + 5(19) = 252 + 95 = 347$ gold. Both facts hold, so the answer is right.

Worked examples

Example 1

Solve: $y = 3x$ - 4 and $2x + y = 11$ .

The first equation gives y directly, so substitute it into the second.
$2x + (3x$ - $4) = 11$ , which becomes 5x - $4 = 11$ .
$5x = 15$ , so $x = 3$ .
Then $y = 3(3)$ - $4 = 5$ . Solution: $(3, 5)$ . Check: $2(3) + 5 = 11$ . Correct.

Example 2

Solve by elimination: $4x + 3y = 27$ and 4x - $y = 7$ .

The x-coefficients are both 4, so subtracting removes x.
$(4x + 3y)$ - (4x - $y) = 27$ - 7, giving $4y = 20$ .
So $y = 5$ .
Substitute back: 4x - $5 = 7$ , so $4x = 12$ and $x = 3$ . Solution: $(3, 5)$ .

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Solve the system by substitution: $y = 3x$ - $10 4x + 2y = 20$ Give your answer as the pair $(x, y)$ .

Answer: $(4, 2)$

Substitute $y = 3x$ - 10 into $4x + 2y = 20$ :
$4x + 2(3x$ - $10) = 20$
10x - $20 = 20$
$x = 4$
$y = 3(4)$ - $10 = 2$
Solution: $(4, 2)$

Problem 2

Difficulty 3 of 5

Solve the system by elimination: $7x + 3y = 24 6x$ - $3y = 54$ Give your answer as the pair $(x, y)$ .

Answer: $(6, -6)$

$(1) 7x + 3y = 24$
$(2) 6x$ - $3y = 54$
Add the equations to eliminate y.
This gives $x = 6$ .
Substitute into $(1): 7(6) + 3y = 24$ , so $y = -6$ .
Solution: $(6, -6)$

Problem 3

Difficulty 4 of 5

Tickets to the Tournament of Numbers cost 13 gold for an adult and 1 gold for a child. 102 tickets were sold and 642 gold was taken. How many of each kind were sold? Give your answer as the pair (adults, children).

Answer: $(45, 57)$

Let $a =$ adult tickets, $c =$ child tickets.
$a + c = 102$
$13a + 1c = 642$
$c = 102$ - a, so $13a + 1(102$ - $a) = 642$
$12a + 102 = 642$ , so $a = 45$
$c = 102$ - $45 = 57$
Answer: $(45, 57)$

Common mistakes

Solving for x and stopping, without finding y.
Forgetting brackets when substituting, so a sign is lost.
Multiplying only part of an equation when scaling it for elimination.
Writing the pair in the wrong order, or not checking it in both original equations.

What you should be able to do

Solve a system by substitution.
Solve a system by elimination.
Interpret a system with no solution or infinitely many solutions.
Model two-quantity word problems with a system.

Where this fits in the curriculum

Common Core

8.EE.C.8.B
Grade 8 — Solve systems of two linear equations algebraically, and solve simple cases by inspection.
HSA-REI.C.6
High school — Solve systems of linear equations exactly and approximately, focusing on pairs in two variables.
HSA-CED.A.3
High school — Represent constraints by systems of equations and interpret solutions as viable options in a modelling context.

Ontario

MTH1W.C3.3
Grade 9 de-streamed — Compare two linear relations, including identifying and interpreting their point of intersection.
MTH1W reaches the intersection of two lines graphically only — substitution and elimination are Grade 10 (MPM2D) — so this tag covers the meaning of a solution rather than the method this skill teaches.

SAT

Heart of Algebra
Systems of two linear equations and the meaning of their solution.