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Systems of Equations

Two equations, two unknowns — the solution is the point that makes both true. Learn the three ways to solve a system (graphing, substitution, elimination), how to pick one, and how to spot no-solution and infinite-solution cases.

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The Main Ideas

What a System Is, and Three Ways to Solve One

A system is two (or more) equations at once. The solution is the pair \((x, y)\) that makes every equation true — on a graph, the point where the lines cross. You have three tools; the trick is picking the fast one.

The solution is the intersection

solution

Each equation is a line. The point they share solves both at once.

Method 1: Graphing

Graph both lines and read off where they cross. Great for a quick picture or an estimate, but hard to read exactly when the answer is not a whole number.

Method 2: Substitution

Solve one equation for a variable, then substitute that into the other. Best when a variable is already alone (like \(y = 2x + 1\)) or has a coefficient of 1.

Method 3: Elimination

Add or subtract the equations so one variable cancels. Best when both are in \(ax + by = c\) form. Multiply an equation first if needed to line up opposite coefficients.

Three possible outcomes

  • One solution — lines cross once (different slopes)
  • No solution — parallel lines (same slope, different intercept); you get a false statement like \(0 = 5\)
  • Infinite solutions — the same line twice; you get a true statement like \(6 = 6\)

Always check your answer

Plug your \((x, y)\) back into both original equations. If either one fails, the solution is wrong — this catches almost every arithmetic slip.

See It Worked Out

Worked Examples

Example 1 · Substitution

Solve: \(y = 2x + 1\) and \(3x + y = 11\)

  1. The first equation already gives \(y\). Substitute it into the second.

    \[ 3x + (2x + 1) = 11 \]

  2. Combine and solve for \(x\).

    \[ 5x + 1 = 11 \Rightarrow 5x = 10 \Rightarrow x = 2 \]

  3. Put \(x = 2\) back into \(y = 2x + 1\).

    \[ y = 2(2) + 1 = 5 \]
\( (x, y) = (2, 5) \)

Example 2 · Elimination

Solve: \(2x + 3y = 12\) and \(2x - y = 4\)

  1. Both have \(2x\). Subtract the second equation from the first so \(x\) cancels.

    \[ (2x + 3y) - (2x - y) = 12 - 4 \]

  2. Simplify and solve for \(y\).

    \[ 4y = 8 \Rightarrow y = 2 \]

  3. Substitute \(y = 2\) into \(2x - y = 4\).

    \[ 2x - 2 = 4 \Rightarrow x = 3 \]
\( (x, y) = (3, 2) \)

Example 3 · Elimination with multiplication

Solve: \(2x + 3y = 7\) and \(3x + 2y = 8\)

  1. Nothing cancels yet. Multiply the first by 3 and the second by 2 so both have \(6x\).

    \[ 6x + 9y = 21 \qquad 6x + 4y = 16 \]

  2. Subtract to cancel \(x\), then solve for \(y\).

    \[ 5y = 5 \Rightarrow y = 1 \]

  3. Substitute \(y = 1\) into \(2x + 3y = 7\).

    \[ 2x + 3 = 7 \Rightarrow x = 2 \]
\( (x, y) = (2, 1) \)
Your Turn

Practice Problems

Solve each system, then reveal the solution to check yourself.

\(y = x - 1\) and \(2x + y = 8\)

Show solution

Substitute: \(2x + (x - 1) = 8 \Rightarrow 3x = 9 \Rightarrow x = 3\), so \(y = 2\).

\[ (3, 2) \]

\(x + y = 10\) and \(x - y = 4\)

Show solution

Add the equations: \(2x = 14 \Rightarrow x = 7\), so \(y = 3\).

\[ (7, 3) \]

\(2x + y = 5\) and \(3x - y = 10\)

Show solution

Add to cancel \(y\): \(5x = 15 \Rightarrow x = 3\), so \(y = -1\).

\[ (3, -1) \]

\(y = 2x + 3\) and \(y = 2x - 1\)

Show solution

Same slope, different intercepts — the lines are parallel. Setting them equal gives \(3 = -1\), which is false.

\[ \text{No solution} \]

\(4x + 2y = 6\) and \(y = 3 - 2x\)

Show solution

Substitute: \(4x + 2(3 - 2x) = 6 \Rightarrow 6 = 6\), true for every \(x\) — it is the same line.

\[ \text{Infinitely many solutions} \]
Keep Going

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