Linear Systems in Two Unknowns

What this move does

A linear equation in two unknowns (such as $x + y = 5$ ) has infinitely many solutions. Draw them all and you get a line. Two equations together make a system, and what we want is the $(x, y)$ that satisfies both at once: the intersection of the two lines.

How do you work out the intersection? Multiply each equation by a number and add them:

k_{1} \times ① + k_{2} \times ②

The intersection makes both ① and ② true, so it must make this new equation true as well: the new line always passes through the intersection. If we choose $k_{1}$ and $k_{2}$ so that the coefficient of one unknown becomes exactly $0$ , the new equation has only one unknown left, and the new line is vertical or horizontal. One coordinate of the intersection can then be read off directly. This is elimination by adding or subtracting:

Choose an unknown to eliminate, and multiply so that its coefficients in the two equations are equal or opposite.
Subtract or add the two equations to eliminate it, and solve for the other unknown.
Substitute that value back into either original equation to find the eliminated one.
Check in both original equations.

Predict first

How many solutions does the system $2 x + 3 y = 12$ , $2 x + 3 y = 6$ have?

0

1

2

infinitely many

Choose an answer first; then you can see the result.

Work through one

Example: solve 3x + 2y = 12, 5x − 3y = 1

Step 1 / 6

Just adding does not work
① + ② gives $8 x - y = 13$ , and both unknowns are still there. The new line passes through the intersection, but it is slanted, so you cannot read off a coordinate.

$(3 x + 2 y) + (5 x - 3 y) = 12 + 1$
Aim at y
Add, and y vanishes
Substitute back to find y
You could eliminate x first
Check

Substitution is elimination too

There is more than one way to eliminate. Express one unknown in terms of the other, substitute it into the other equation, and again you are left with a single unknown.

One problem, several solutions

Solve the system $x + 2 y = 7$ (①), $3 x - 2 y = 5$ (②).

Elimination by adding or subtracting

The coefficients of $y$ are $2$ and $- 2$ , so just add:

4 x = 12, x = 3

Substitute into ①: $3 + 2 y = 7$ , $y = 2$ .

CostFastest when the coefficients are already opposite; otherwise multiply first.

Substitution

From ①, $x = 7 - 2 y$ . Substitute into ②:

3 (7 - 2 y) - 2 y = 5, 21 - 8 y = 5, y = 2

So $x = 7 - 2 \times 2 = 3$ .

CostHandy when an unknown has coefficient 1; otherwise fractions appear.

Both methods give the same answer: $x = 3$ , $y = 2$ . Which one to pick depends on which coefficients look friendlier.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: the system has exactly one solution

$x + 2 y = 4$ and $2 x - y = 3$ : $1 \times (- 1) - 2 \times 2 = - 5 \neq = 0$ . The two lines point in different directions, so they must meet at a point, and at only one point.

Practice

Exercise 1Basic

Solve the system $x + y = 10$ , $x - y = 4$ . (Write $x$ first, then $y$ , separated by a comma.)

Exercise 2Basic

Solve the system $2 x + 3 y = 13$ , $4 x - y = 5$ . (Write $x$ first, then $y$ .)

Exercise 3Advanced

Solve the system $3 x + 4 y = 10$ , $5 x - 6 y = 4$ . (Write $x$ first, then $y$ .)

Exercise 4Challenge

The system $2 x + m y = 6$ , $x - 2 y = 3$ in $x$ and $y$ has infinitely many solutions. Find $m$ .

What this move does

Work through one

Example: solve 3x + 2y = 12, 5x − 3y = 1

Substitution is elimination too

Elimination by adding or subtracting

Substitution

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Changing only half the signs when subtracting

Forgetting the right side when multiplying