MathCompass

Systems of Linear Equations

Algebra Intermediate

πŸ“‹ Prerequisites

When you have two variables, you need two equations to find both values. Systems of equations show up everywhere β€” coordinate geometry, word problems, even number theory. Master substitution and elimination, and you'll solve 90% of competition system problems.

πŸ“š Key Concepts

What Is a System?

A system of equations is a set of equations that share the same variables. A solution is a pair (or tuple) that makes all equations true at the same time.

For two linear equations with two variables, there are three possibilities:

Method 1: Substitution

Best when one equation already has a variable isolated, or can be easily isolated.

  1. Solve one equation for one variable
  2. Substitute that expression into the other equation
  3. Solve the resulting single-variable equation
  4. Plug back to find the other variable

Example: $ y = 2x + 1 $ and $ 3x + y = 11 $ \(\to\) substitute: $ 3x + (2x+1) = 11 $ \(\to\) $ x = 2 $, $ y = 5 $

Method 2: Elimination (Addition)

Best when coefficients line up nicely, or when substitution would create fractions.

  1. Line up equations with like terms stacked
  2. Multiply one or both equations so one variable has opposite coefficients
  3. Add the equations to eliminate that variable
  4. Solve, then plug back to find the other variable

Key idea: if $ax + by = c$ and $dx + ey = f$, you can scale to make the $x$ (or $y$) coefficients cancel.

\(Three Variables (3 \\times 3 Systems)\)

Same elimination idea, just more steps:

  1. \(Take two equations, eliminate one variable \\to get a 2-variable equation\)
  2. \(Take a different pair, eliminate the same variable \\to get another 2-variable equation\)
  3. \(Now you have a 2 \\times 2 system\) β€” solve with substitution or elimination
  4. Plug back to find the third variable

Always check your answer in all three original equations!

Word Problems with Systems

Most system word problems fall into these categories:

⚠️ Common Mistake: When multiplying an equation for elimination, multiply every term on both sides. Forgetting a term is the #1 elimination error. Also watch your signs β€” the goal is to get opposite coefficients that add to zero.
πŸ’‘ Key Insight: When you see "find the value of $x + y$" or "what is $2x + 3y$?" in a competition problem, you often DON'T need to find $x$ and $y$ individually. Look for a way to add or subtract the equations directly to get the expression they're asking for. It's usually faster.

✏️ Example Problems

πŸ“ Example 1 (Substitution)

Solve the system:
$ y = 3x - 2 $
$ 2x + 3y = 16 $
What is the value of $x + y$?

Method: Substitution (first equation already has y isolated)

Step 1: Substitute $3x - 2$ for $y$ in the second equation.

\[ 2x + 3(3x - 2) = 16 \]

Step 2: Simplify and solve for $x$.

\[ 2x + 9x - 6 = 16 \\ 11x - 6 = 16 \\ 11x = 22 \\ x = 2 \]

Step 3: Find $y$ using $y = 3x - 2$.

$ y = 3(2) - 2 = 6 - 2 = 4 $

Step 4: $x + y = 2 + 4 = 6$

Check: $2(2) + 3(4) = 4 + 12 = 16$. βœ“

Answer: 6

πŸ“ Example 2 (Elimination)

Solve the system:
$ 2x + 3y = 13 $
$ 3x - 2y = 0 $
What is the value of $x$?

Method: Elimination. Eliminate $y$ first.

Step 1: Multiply first equation by 2, second by 3 to get opposite $y$ coefficients.

\(Eq1 \\times 2:\) $ 4x + 6y = 26 $

\(Eq2 \\times 3:\) $ 9x - 6y = 0 $

Step 2: Add the two new equations.

\[ (4x + 6y) + (9x - 6y) = 26 + 0 \\ 13x = 26 \\ x = 2 \]

Step 3 (check $y$): Plug $x=2$ into $3x - 2y = 0$.

$ 6 - 2y = 0 $ \(\to\) $ y = 3 $

Check in Eq1: $ 2(2) + 3(3) = 4 + 9 = 13 $. βœ“

Answer: 2

πŸ“ Example 3 (Word Problem β€” Ticket Sales)

A theater sells adult tickets for $8 and child tickets for $5. A total of 100 tickets were sold, bringing in $686. How many adult tickets were sold?

Step 1: Define variables.

Let $a$ = adult tickets, $c$ = child tickets.

Step 2: Set up the system.

Total tickets: $ a + c = 100 $

Total money: $ 8a + 5c = 686 $

Step 3: Solve. Use substitution from first equation: $ c = 100 - a $

\[ 8a + 5(100 - a) = 686 \\ 8a + 500 - 5a = 686 \\ 3a + 500 = 686 \\ 3a = 186 \\ a = 62 \]

Check: \(62 adults + 38 children = 100 tickets. Revenue: 62 \\times\) $8 + 38 \\times $5 = $496 + $190 = $686. βœ“

Answer: 62 adult tickets

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