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Systems of equationsAlgebraHigh schoolabout 1:12

Solve the system -7x + 2y = 16, -2x - y = 3: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -2, 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+2y=16−2x−y=3\begin{cases} -7x+2y = 16 \\ -2x- y = 3 \end{cases}
  2. 2. Steps, part 1
    1. {−7x+2y=16−2x−y=3\begin{cases} -7x + 2y = 16 \\ -2x - y = 3 \end{cases}
      Write every equation in standard form
    2. [−7216−2−13]\begin{bmatrix}-7 & 2 & 16 \\ -2 & -1 & 3\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−7R1: [1−27−167−2−13]R_{1} \to \tfrac{1}{-7} R_{1}:\ \begin{bmatrix}1 & -\frac{2}{7} & -\frac{16}{7} \\ -2 & -1 & 3\end{bmatrix}
      Divide R1 by -7
  3. 3. Steps, part 2
    1. R2→R2+2R1: [1−27−1670−117−117]R_{2} \to R_{2} + 2R_{1}:\ \begin{bmatrix}1 & -\frac{2}{7} & -\frac{16}{7} \\ 0 & -\frac{11}{7} & -\frac{11}{7}\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→1−117R2: [1−27−167011]R_{2} \to \tfrac{1}{-\frac{11}{7}} R_{2}:\ \begin{bmatrix}1 & -\frac{2}{7} & -\frac{16}{7} \\ 0 & 1 & 1\end{bmatrix}
      Divide R2 by -11/7
    3. R1→R1+27R2: [10−2011]R_{1} \to R_{1} + \frac{2}{7} R_{2}:\ \begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & 1\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−2011]\begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & 1\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−2, y=1x = -2,\ y = 1
      Read off the unique solution
  5. 5. The answer
    x=−2, y=1x = -2,\ y = 1

    -7x + 2y = 16; -2x - y = 3 ⇒ x = -2, y = 1

  6. 6. Check
    • Equation 1: -7x + 2y = 16 ✓
    • Equation 2: -2x - y = 3 ✓

Transcript

The problem. Here is the problem: negative 7x plus 2y equals 16; negative 2x minus y equals 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 negative 7.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 11 over 7. 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 2, y equals 1. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 7x plus 2y equals 16, which checks out. Equation 2: negative 2x minus y equals 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 + 2y = 16, -2x - y = 3: Animated Lesson