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Systems of equationsAlgebraHigh schoolabout 1:11

Solve the system 6x + 4y = 10, 4x - 7y = -32: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -1, 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 (0.5 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Solving an equation
    {6x+4y=104x−7y=−32\begin{cases} 6x+4y = 10 \\ 4x-7y = -32 \end{cases}
  2. 2. Steps, part 1
    1. {6x+4y=104x−7y=−32\begin{cases} 6x + 4y = 10 \\ 4x - 7y = -32 \end{cases}
      Write every equation in standard form
    2. [64104−7−32]\begin{bmatrix}6 & 4 & 10 \\ 4 & -7 & -32\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→16R1: [123534−7−32]R_{1} \to \tfrac{1}{6} R_{1}:\ \begin{bmatrix}1 & \frac{2}{3} & \frac{5}{3} \\ 4 & -7 & -32\end{bmatrix}
      Divide R1 by 6
  3. 3. Steps, part 2
    1. R2→R2−4R1: [123530−293−1163]R_{2} \to R_{2} - 4R_{1}:\ \begin{bmatrix}1 & \frac{2}{3} & \frac{5}{3} \\ 0 & -\frac{29}{3} & -\frac{116}{3}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−293R2: [12353014]R_{2} \to \tfrac{1}{-\frac{29}{3}} R_{2}:\ \begin{bmatrix}1 & \frac{2}{3} & \frac{5}{3} \\ 0 & 1 & 4\end{bmatrix}
      Divide R2 by -29/3
    3. R1→R1−23R2: [10−1014]R_{1} \to R_{1} - \frac{2}{3} R_{2}:\ \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 4\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−1014]\begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 4\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−1, y=4x = -1,\ y = 4
      Read off the unique solution
  5. 5. The answer
    x=−1, y=4x = -1,\ y = 4

    6x + 4y = 10; 4x - 7y = -32 ⇒ x = -1, y = 4

  6. 6. Check
    • Equation 1: 6x + 4y = 10 ✓
    • Equation 2: 4x - 7y = -32 ✓

Transcript

The problem. Here is the problem: 6x plus 4y equals 10; 4x minus 7y equals negative 32. 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 6.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 29 over 3. 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 1, y equals 4. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: 6x plus 4y equals 10, which checks out. Equation 2: 4x minus 7y equals negative 32, 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 + 4y = 10, 4x - 7y = -32 (Lesson)