openfreebooks-bookgenerator/content/math/linear-equations-two-unknowns/core.html

<div class="book-prose">
  <p>
    A <strong>system of linear equations in two unknowns</strong> is a pair of equations in
    variables \(x\) and \(y\), where each equation is linear (the highest power of each variable is
    1). A <strong>solution</strong> is an ordered pair \((x, y)\) that satisfies
    <em>both</em> equations at the same time.
  </p>

  <div class="book-callout" role="note">
    <p class="book-callout__label">Assumed knowledge</p>
    <ul>
      <li>Expand and collect like terms, e.g. \(2(x+3) = 2x + 6\).</li>
      <li>Substitute a value into an expression, e.g. replace \(x\) by \(4\) in \(3x - 1\).</li>
      <li>Add or subtract equations term by term when both sides are equal.</li>
    </ul>
  </div>

  <h2 id="standard-form" class="book-prose__heading">Standard form</h2>
  <p>
    A convenient form is
    \[ ax + by = c, \qquad dx + ey = f, \]
    where \(a, b, c, d, e, f\) are constants. The same system can be written in other equivalent
    forms — for example \(y = 2x + 1\) is the same as \(-2x + y = 1\).
  </p>

  <h2 id="substitution" class="book-prose__heading">Substitution</h2>
  <p>
    <strong>Substitution</strong> means isolate one unknown in one equation, then replace that
    expression in the other equation.
  </p>
  <p><strong>Example.</strong> Solve</p>
  <p class="book-formula">
    \[ \begin{cases} y = 2x + 1 \\ x + y = 7 \end{cases} \]
  </p>
  <p>
    From the first equation, \(y = 2x + 1\). Substitute into the second:
    \[ x + (2x + 1) = 7 \Rightarrow 3x + 1 = 7 \Rightarrow 3x = 6 \Rightarrow x = 2. \]
    Then \(y = 2(2) + 1 = 5\). The solution is \((2, 5)\).
  </p>
  <p>
    Check: \(2(2)+1 = 5\) and \(2 + 5 = 7\).
  </p>

  <h2 id="elimination" class="book-prose__heading">Elimination</h2>
  <p>
    <strong>Elimination</strong> adds or subtracts the equations so that one unknown cancels. You
    may need to multiply one or both equations first so that the coefficients of \(x\) or \(y\) match
    in size (one positive, one negative).
  </p>
  <p><strong>Example.</strong> Solve</p>
  <p class="book-formula">
    \[ \begin{cases} x + y = 5 \\ 2x - y = 1 \end{cases} \]
  </p>
  <p>
    The coefficients of \(y\) are \(1\) and \(-1\), so add the equations:
    \[ (x+y) + (2x-y) = 5 + 1 \Rightarrow 3x = 6 \Rightarrow x = 2. \]
    Substitute into \(x + y = 5\): \(2 + y = 5\), so \(y = 3\). Solution: \((2, 3)\).
  </p>

  <h2 id="graphical" class="book-prose__heading">Graphical interpretation</h2>
  <p>
    Each equation \(ax + by = c\) (with \(b \neq 0\)) can be rearranged to
    \(y = -\frac{a}{b}x + \frac{c}{b}\), which is a straight line. The solution of the system is
    the point where the two lines meet — their <strong>intersection</strong>.
  </p>
  <div class="book-table-wrap">
    <table class="book-table">
      <thead>
        <tr>
          <th scope="col">Lines</th>
          <th scope="col">Algebraic picture</th>
          <th scope="col">Solutions</th>
        </tr>
      </thead>
      <tbody>
        <tr>
          <td>Intersect at one point</td>
          <td>Different slopes</td>
          <td>Exactly one \((x, y)\)</td>
        </tr>
        <tr>
          <td>Parallel, distinct</td>
          <td>Same slope, different intercepts</td>
          <td>No solution</td>
        </tr>
        <tr>
          <td>Coincident (same line)</td>
          <td>One equation is a multiple of the other</td>
          <td>Infinitely many</td>
        </tr>
      </tbody>
    </table>
  </div>

  <h2 id="try-graph" class="book-prose__heading">Try it — graph explorer</h2>
  <p>
    Adjust the two lines or use presets. Read off the intersection and how the classification changes
    when the lines are parallel or coincident.
  </p>
</div>

<div class="math-widget-mount" data-widget="simultaneous-graph-explorer" data-pagefind-ignore></div>

<div class="book-prose">
  <h2 id="try-solver" class="book-prose__heading">Try it — step-by-step solver</h2>
  <p>
    Pick a preset system and watch substitution or elimination steps. Systems with no unique
    solution show an explanation instead of incorrect steps.
  </p>
</div>

<div class="math-widget-mount" data-widget="simultaneous-solver-workbench" data-pagefind-ignore></div>

<div class="book-prose">
  <h2 id="word-problems" class="book-prose__heading">Word problems</h2>
  <p>
    For story problems, define unknowns clearly, write <em>two</em> independent equations, then solve
    by substitution, elimination, or a graph.
  </p>
  <p><strong>Example (tickets).</strong> Adult tickets cost \$12 and child tickets \$7. A group buys
    10 tickets for \$95. How many of each type?</p>
  <p>
    Let \(a\) be the number of adult tickets and \(c\) the number of child tickets. Then
    \(a + c = 10\) and \(12a + 7c = 95\). From the first, \(c = 10 - a\). Substitute:
    \[ 12a + 7(10 - a) = 95 \Rightarrow 12a + 70 - 7a = 95 \Rightarrow 5a = 25 \Rightarrow a = 5, \]
    so \(c = 5\). Five adult and five child tickets.
  </p>
</div>