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Systems of equationsAlgebraHigh schoolabout 1:15

Solve the system 7x - 7y = -14/3, -2x + 8y = -14/3: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -5/3, y = -1.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.5 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Solving an equation
    {7x−7y=−143−2x+8y=−143\begin{cases} 7x-7y = \frac{-14}{3} \\ -2x+8y = \frac{-14}{3} \end{cases}
  2. 2. Steps, part 1
    1. {7x−7y=−143−2x+8y=−143\begin{cases} 7x - 7y = -\frac{14}{3} \\ -2x + 8y = -\frac{14}{3} \end{cases}
      Write every equation in standard form
    2. [7−7−143−28−143]\begin{bmatrix}7 & -7 & -\frac{14}{3} \\ -2 & 8 & -\frac{14}{3}\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→17R1: [1−1−23−28−143]R_{1} \to \tfrac{1}{7} R_{1}:\ \begin{bmatrix}1 & -1 & -\frac{2}{3} \\ -2 & 8 & -\frac{14}{3}\end{bmatrix}
      Divide R1 by 7
  3. 3. Steps, part 2
    1. R2→R2+2R1: [1−1−2306−6]R_{2} \to R_{2} + 2R_{1}:\ \begin{bmatrix}1 & -1 & -\frac{2}{3} \\ 0 & 6 & -6\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→16R2: [1−1−2301−1]R_{2} \to \tfrac{1}{6} R_{2}:\ \begin{bmatrix}1 & -1 & -\frac{2}{3} \\ 0 & 1 & -1\end{bmatrix}
      Divide R2 by 6
    3. R1→R1+1R2: [10−5301−1]R_{1} \to R_{1} + 1R_{2}:\ \begin{bmatrix}1 & 0 & -\frac{5}{3} \\ 0 & 1 & -1\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−5301−1]\begin{bmatrix}1 & 0 & -\frac{5}{3} \\ 0 & 1 & -1\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−53, y=−1x = -\frac{5}{3},\ y = -1
      Read off the unique solution
  5. 5. The answer
    x=−53, y=−1x = -\frac{5}{3},\ y = -1

    7x - 7y = -14/3; -2x + 8y = -14/3 ⇒ x = -5/3, y = -1

  6. 6. Check
    • Equation 1: 7x - 7y = -14/3 ✓
    • Equation 2: -2x + 8y = -14/3 ✓

Transcript

The problem. Here is the problem: 7x minus 7y equals negative 14 over 3; negative 2x plus 8y equals negative 14 over 3. 2 linear equations in 2 unknowns (x, y).

Steps, part 1. Write every equation in standard form. Move all variable terms to the left and constants to the right. Build the augmented matrix [A | b]. Columns: x, y | constant. Gauss–Jordan elimination. Use row operations to make each pivot 1 and clear the rest of its column. Divide R1 by 7.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 6. Clear column 2 using the pivot in R2. Reduced row-echelon form.

Steps, part 3. Read off the unique solution.

The answer. So the answer is x equals negative 5 over 3, y equals negative 1. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 7x minus 7y equals negative 14 over 3, which checks out. Equation 2: negative 2x plus 8y equals negative 14 over 3, which checks out.

About systems of equations

Solve systems of two or three linear equations by elimination and substitution.

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Solve the system 7x - 7y = -14/3, -2x + 8y = -14/3 (Lesson)