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Systems of equationsAlgebraHigh schoolabout 1:17

Solve the system 5x - 6y = -3/2, -7x - 9y = 63: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -9/2, y = -7/2.

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
    {5x−6y=−32−7x−9y=63\begin{cases} 5x-6y = \frac{-3}{2} \\ -7x-9y = 63 \end{cases}
  2. 2. Steps, part 1
    1. {5x−6y=−32−7x−9y=63\begin{cases} 5x - 6y = -\frac{3}{2} \\ -7x - 9y = 63 \end{cases}
      Write every equation in standard form
    2. [5−6−32−7−963]\begin{bmatrix}5 & -6 & -\frac{3}{2} \\ -7 & -9 & 63\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→15R1: [1−65−310−7−963]R_{1} \to \tfrac{1}{5} R_{1}:\ \begin{bmatrix}1 & -\frac{6}{5} & -\frac{3}{10} \\ -7 & -9 & 63\end{bmatrix}
      Divide R1 by 5
  3. 3. Steps, part 2
    1. R2→R2+7R1: [1−65−3100−87560910]R_{2} \to R_{2} + 7R_{1}:\ \begin{bmatrix}1 & -\frac{6}{5} & -\frac{3}{10} \\ 0 & -\frac{87}{5} & \frac{609}{10}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−875R2: [1−65−31001−72]R_{2} \to \tfrac{1}{-\frac{87}{5}} R_{2}:\ \begin{bmatrix}1 & -\frac{6}{5} & -\frac{3}{10} \\ 0 & 1 & -\frac{7}{2}\end{bmatrix}
      Divide R2 by -87/5
    3. R1→R1+65R2: [10−9201−72]R_{1} \to R_{1} + \frac{6}{5} R_{2}:\ \begin{bmatrix}1 & 0 & -\frac{9}{2} \\ 0 & 1 & -\frac{7}{2}\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−9201−72]\begin{bmatrix}1 & 0 & -\frac{9}{2} \\ 0 & 1 & -\frac{7}{2}\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−92, y=−72x = -\frac{9}{2},\ y = -\frac{7}{2}
      Read off the unique solution
  5. 5. The answer
    x=−92, y=−72x = -\frac{9}{2},\ y = -\frac{7}{2}

    5x - 6y = -3/2; -7x - 9y = 63 ⇒ x = -9/2, y = -7/2

  6. 6. Check
    • Equation 1: 5x - 6y = -3/2 ✓
    • Equation 2: -7x - 9y = 63 ✓

Transcript

The problem. Here is the problem: 5x minus 6y equals negative 3 over 2; negative 7x minus 9y equals 63. 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 5.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 87 over 5. 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 9 over 2, y equals negative 7 over 2. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 5x minus 6y equals negative 3 over 2, which checks out. Equation 2: negative 7x minus 9y equals 63, 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 5x - 6y = -3/2, -7x - 9y = 63 (Lesson)