Algebra I

Systems of Linear Equations

Lesson

A system of linear equations is a set of equations that share the same variables. A solution is a set of values that makes every equation in the system true at once. With two variables, the solution is an ordered pair $(x,\,y)$ ; with three, it’s an ordered triple $(x,\,y,\,z)$ .

Two standard methods:

Substitution: solve one equation for one variable, then plug that expression into the other equation. Best when one variable is already isolated (or close to it).
Elimination: add or subtract equations (multiplied by suitable constants if needed) to cancel a variable. Best when coefficients line up cleanly.

Worked example 1 — elimination

\begin{cases} x + y = 10 \\ x - y = 4 \end{cases}

Add the equations:

2x = 14 \implies x = 7

Plug back: $7 + y = 10$ , so $y = 3$ .

(x,\,y) = (7,\,3)

Worked example 2 — substitution

\begin{cases} 2x + 3y = 12 \\ x + y = 5 \end{cases}

From the second equation, $x = 5 - y$ . Substitute:

2(5 - y) + 3y = 12 \implies 10 + y = 12 \implies y = 2

Then $x = 5 - 2 = 3$ .

(x,\,y) = (3,\,2)

For three-variable systems, use the same tools repeatedly: eliminate one variable to reduce to a two-variable system, solve that, then back-substitute.

How to type your answer

Type the values in order, separated by commas — no parentheses. For two-variable systems: $x$ first, then $y$ . For three-variable: $x$ , $y$ , $z$ . Examples: 7,3, -1,3, 1,2,3.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

\begin{cases} x + y = 5 \\ x - y = 1 \end{cases}

Problem 2

\begin{cases} 2x + y = 7 \\ x = 2 \end{cases}

Problem 3

\begin{cases} x + y = 10 \\ x = y \end{cases}

Problem 4

\begin{cases} y = 2x \\ x + y = 9 \end{cases}

Practice

Standard problems matching the lesson.

Problem 5

\begin{cases} x + y = 10 \\ x - y = 4 \end{cases}

Problem 6

\begin{cases} 2x + 3y = 12 \\ x + y = 5 \end{cases}

Problem 7

\begin{cases} 2x + y = 11 \\ x + y = 8 \end{cases}

Problem 8

\begin{cases} x + y = 8 \\ 2x - y = 7 \end{cases}

Problem 9

\begin{cases} 3x - y = 10 \\ x + y = 6 \end{cases}

Problem 10

\begin{cases} x + y = 7 \\ y - x = 3 \end{cases}

Problem 11

\begin{cases} 3x + y = 7 \\ x - y = -3 \end{cases}

Problem 12

\begin{cases} x - y = 4 \\ 2x + y = 14 \end{cases}

Problem 13

\begin{cases} x + y = 1 \\ 2x - y = -7 \end{cases}

Problem 14

\begin{cases} x + y = 3 \\ 3x - y = 13 \end{cases}

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 15

\begin{cases} 2x + 3y = 18 \\ 3x - 2y = 1 \end{cases}

Problem 16

\begin{cases} 4x + y = 1 \\ 3x + 2y = 7 \end{cases}

Problem 17

\begin{cases} 5x + 2y = 4 \\ x - y = 5 \end{cases}

Problem 18

\begin{cases} 2x - 3y = 8 \\ 3x + y = 23 \end{cases}

Problem 19

\begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ 2x + y - z = 1 \end{cases}

Problem 20

\begin{cases} x + y + z = 2 \\ x - y - z = 2 \\ 2x + y - z = 6 \end{cases}

Problem 21

\begin{cases} x + y = -1 \\ y + z = 1 \\ x + z = 4 \end{cases}

Problem 22

\begin{cases} x + y + z = 1 \\ x - y + z = -3 \\ x + y - z = 3 \end{cases}

Practice

Standard problems matching the lesson.

Problem 23

Two numbers sum to 12 and differ by 4. Find them as x,y (x larger).

Problem 24

Adult $10, child $6, 50 tickets total $360. Find x = adults, y = kids.

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 25

Solve x + y + z = 6, x − y + 2z = 5, 2x + y − z = 1. Give x,y,z.

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Quiz

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