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Systems of equationsAlgebraHigh schoolabout 1:12

Solve the system -5x - 5y = 25, -9x + 7y = -3: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -2, y = -3.

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−5y=25−9x+7y=−3\begin{cases} -5x-5y = 25 \\ -9x+7y = -3 \end{cases}
  2. 2. Steps, part 1
    1. {−5x−5y=25−9x+7y=−3\begin{cases} -5x - 5y = 25 \\ -9x + 7y = -3 \end{cases}
      Write every equation in standard form
    2. [−5−525−97−3]\begin{bmatrix}-5 & -5 & 25 \\ -9 & 7 & -3\end{bmatrix}
      Build the augmented matrix [A | b]
    3. Gauss–Jordan elimination
    4. R1→1−5R1: [11−5−97−3]R_{1} \to \tfrac{1}{-5} R_{1}:\ \begin{bmatrix}1 & 1 & -5 \\ -9 & 7 & -3\end{bmatrix}
      Divide R1 by -5
  3. 3. Steps, part 2
    1. R2→R2+9R1: [11−5016−48]R_{2} \to R_{2} + 9R_{1}:\ \begin{bmatrix}1 & 1 & -5 \\ 0 & 16 & -48\end{bmatrix}
      Clear column 1 using the pivot in R1
    2. R2→116R2: [11−501−3]R_{2} \to \tfrac{1}{16} R_{2}:\ \begin{bmatrix}1 & 1 & -5 \\ 0 & 1 & -3\end{bmatrix}
      Divide R2 by 16
    3. R1→R1−1R2: [10−201−3]R_{1} \to R_{1} - 1R_{2}:\ \begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & -3\end{bmatrix}
      Clear column 2 using the pivot in R2
    4. [10−201−3]\begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & -3\end{bmatrix}
      Reduced row-echelon form
  4. 4. Steps, part 3
    1. x=−2, y=−3x = -2,\ y = -3
      Read off the unique solution
  5. 5. The answer
    x=−2, y=−3x = -2,\ y = -3

    -5x - 5y = 25; -9x + 7y = -3 ⇒ x = -2, y = -3

  6. 6. Check
    • Equation 1: -5x - 5y = 25 ✓
    • Equation 2: -9x + 7y = -3 ✓

Transcript

The problem. Here is the problem: negative 5x minus 5y equals 25; negative 9x plus 7y equals negative 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 5.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 16. 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 negative 3. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 5x minus 5y equals 25, which checks out. Equation 2: negative 9x plus 7y equals negative 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 -5x - 5y = 25, -9x + 7y = -3 (Lesson)