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Systems of equationsAlgebraHigh schoolabout 1:09

Solve the system x - 2y = -14, x - 5y = -26: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -6, y = 4.

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 (0.4 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Solving an equation
    {x−2y=−14x−5y=−26\begin{cases} x-2y = -14 \\ x-5y = -26 \end{cases}
  2. 2. Steps, part 1
    1. {x−2y=−14x−5y=−26\begin{cases} x - 2y = -14 \\ x - 5y = -26 \end{cases}
      Write every equation in standard form
    2. [1−2−141−5−26]\begin{bmatrix}1 & -2 & -14 \\ 1 & -5 & -26\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R2→R2−1R1: [1−2−140−3−12]R_{2} \to R_{2} - 1R_{1}:\ \begin{bmatrix}1 & -2 & -14 \\ 0 & -3 & -12\end{bmatrix}
      Clear column 1 using the pivot in R1
  3. 3. Steps, part 2
    1. R2→1−3R2: [1−2−14014]R_{2} \to \tfrac{1}{-3} R_{2}:\ \begin{bmatrix}1 & -2 & -14 \\ 0 & 1 & 4\end{bmatrix}
      Divide R2 by -3
    2. R1→R1+2R2: [10−6014]R_{1} \to R_{1} + 2R_{2}:\ \begin{bmatrix}1 & 0 & -6 \\ 0 & 1 & 4\end{bmatrix}
      Clear column 2 using the pivot in R2
    3. [10−6014]\begin{bmatrix}1 & 0 & -6 \\ 0 & 1 & 4\end{bmatrix}
      Reduced row-echelon form
    4. x=−6, y=4x = -6,\ y = 4
      Read off the unique solution
  4. 4. The answer
    x=−6, y=4x = -6,\ y = 4

    x - 2y = -14; x - 5y = -26 ⇒ x = -6, y = 4

  5. 5. Check
    • Equation 1: x - 2y = -14 ✓
    • Equation 2: x - 5y = -26 ✓

Transcript

The problem. Here is the problem: x minus 2y equals negative 14; x minus 5y equals negative 26. 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. Clear column 1 using the pivot in R1.

Steps, part 2. Divide R2 by negative 3. Clear column 2 using the pivot in R2. Reduced row-echelon form. Read off the unique solution.

The answer. So the answer is x equals negative 6, y equals 4. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: x minus 2y equals negative 14, which checks out. Equation 2: x minus 5y equals negative 26, 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 x - 2y = -14, x - 5y = -26: Animated Lesson