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Section 2.4 Gaussian Reduction

Subsection 1. Write an equation in standard form

Before we can use Gaussian reduction, we must write each equation in standard form.

Subsubsection Examples

Example 2.25.

Write the equation in standard form.

  1. \(\displaystyle 3y-7=4z+x\)
  2. \(\displaystyle 6=-5z+2x\)
Solution

The standard form is \(ax+by+cz=d\text{.}\) We add or subtract appropriate terms on both sides of the equation.

  1. \(-x+3y-4z=7,~\) or \(~x-3y+4z=-7\)
  2. \(2x+0y-5z=6,~\) or \(~-2x+0y+5z=-6\)

Subsubsection Exercise

Write the equation in standard form.

  1. \(\displaystyle 5-3x+4y=2z\)
  2. \(\displaystyle y=8-2z\)
Answer
  1. \(\displaystyle -3x+4y-2z=-5\)
  2. \(\displaystyle 0x+y+2z=8\)

Subsection 2. Clear fractions from an equation

It is easier to use Gaussian reduction if the equations have integer coefficients.

Subsubsection Examples

Example 2.27.

Write the equation with integer coefficients.

  1. \(\displaystyle \dfrac{1}{4}x+z=\dfrac{3}{4}\)
  2. \(\displaystyle \dfrac{2}{3}x-2y+\dfrac{1}{2}z=3\)
Solution

  1. We multiply both sides of the equation by \(\alert{4}\text{.}\)

    \begin{align*} \alert{4} \cdot \left(\dfrac{1}{4}x+z\right) \amp = \left(\dfrac{3}{4}\right) \cdot \alert{4}\\ x+4z \amp = 3 \end{align*}

  2. We multiply both sides of the equation by the LCD of the fractions, \(\alert{6}\text{.}\)

    \begin{align*} \alert{6} \cdot \left(\dfrac{2}{3}x-2y+\dfrac{1}{2}z\right) \amp = (3) \cdot \alert{6}\\ 4x-12y+3z \amp = 18 \end{align*}

Subsubsection Exercise

Write the equation with integer coefficients.

  1. \(\displaystyle \dfrac{1}{5}x-\dfrac{2}{5}y+z=-1\)
  2. \(\displaystyle \dfrac{3}{4}x-y+\dfrac{5}{6}z=6\)
Answer
  1. \(\displaystyle x-2y+5z=-5\)
  2. \(\displaystyle 9x-12y+10z=72\)