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Systems of equationsAlgebraHigh schoolabout 1:13

Solve the system -6x + 3y = 27, -7x + 5y = 30: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -5, 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
    {−6x+3y=27−7x+5y=30\begin{cases} -6x+3y = 27 \\ -7x+5y = 30 \end{cases}
  2. 2. Steps, part 1
    1. {−6x+3y=27−7x+5y=30\begin{cases} -6x + 3y = 27 \\ -7x + 5y = 30 \end{cases}
      Write every equation in standard form
    2. [−6327−7530]\begin{bmatrix}-6 & 3 & 27 \\ -7 & 5 & 30\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−6R1: [1−12−92−7530]R_{1} \to \tfrac{1}{-6} R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{9}{2} \\ -7 & 5 & 30\end{bmatrix}
      Divide R1 by -6
  3. 3. Steps, part 2
    1. R2→R2+7R1: [1−12−92032−32]R_{2} \to R_{2} + 7R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{9}{2} \\ 0 & \frac{3}{2} & -\frac{3}{2}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→132R2: [1−12−9201−1]R_{2} \to \tfrac{1}{\frac{3}{2}} R_{2}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{9}{2} \\ 0 & 1 & -1\end{bmatrix}
      Divide R2 by 3/2
    3. R1→R1+12R2: [10−501−1]R_{1} \to R_{1} + \frac{1}{2} R_{2}:\ \begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & -1\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−501−1]\begin{bmatrix}1 & 0 & -5 \\ 0 & 1 & -1\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−5, y=−1x = -5,\ y = -1
      Read off the unique solution
  5. 5. The answer
    x=−5, y=−1x = -5,\ y = -1

    -6x + 3y = 27; -7x + 5y = 30 ⇒ x = -5, y = -1

  6. 6. Check
    • Equation 1: -6x + 3y = 27 ✓
    • Equation 2: -7x + 5y = 30 ✓

Transcript

The problem. Here is the problem: negative 6x plus 3y equals 27; negative 7x plus 5y equals 30. 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 negative 6.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 3 over 2. 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, y equals negative 1. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 6x plus 3y equals 27, which checks out. Equation 2: negative 7x plus 5y equals 30, 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 -6x + 3y = 27, -7x + 5y = 30 (Lesson)