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Systems of equationsAlgebraHigh schoolabout 1:12

Solve the system -6x + 6y = 6, -5x - 5y = 5: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = -1, y = 0.

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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. The problem
Solving an equation
$\begin{cases} -6x+6y = 6 \\ -5x-5y = 5 \end{cases}$
2. Steps, part 1
1. $\begin{cases} -6x + 6y = 6 \\ -5x - 5y = 5 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}-6 & 6 & 6 \\ -5 & -5 & 5\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{1} \to \tfrac{1}{-6} R_{1}:\ \begin{bmatrix}1 & -1 & -1 \\ -5 & -5 & 5\end{bmatrix}$
  Divide R1 by -6
3. Steps, part 2
1. $R_{2} \to R_{2} + 5R_{1}:\ \begin{bmatrix}1 & -1 & -1 \\ 0 & -10 & 0\end{bmatrix}$
  Clear column 1 using the pivot in R1
2. $R_{2} \to \tfrac{1}{-10} R_{2}:\ \begin{bmatrix}1 & -1 & -1 \\ 0 & 1 & 0\end{bmatrix}$
  Divide R2 by -10
3. $R_{1} \to R_{1} + 1R_{2}:\ \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 0\end{bmatrix}$
  Clear column 2 using the pivot in R2
4. $\begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 0\end{bmatrix}$
  Reduced row-echelon form
4. Steps, part 3
1. $x = -1,\ y = 0$
  Read off the unique solution
5. The answer
$x = -1,\ y = 0$
-6x + 6y = 6; -5x - 5y = 5 ⇒ x = -1, y = 0
6. Check
- Equation 1: -6x + 6y = 6 ✓
- Equation 2: -5x - 5y = 5 ✓

Transcript

The problem. Here is the problem: negative 6x plus 6y equals 6; negative 5x minus 5y equals 5. 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 negative 10. 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 0. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 6x plus 6y equals 6, which checks out. Equation 2: negative 5x minus 5y equals 5, which checks out.

About systems of equations

Solve systems of two or three linear equations by elimination and substitution.

Solve the system -6x + 6y = 6, -5x - 5y = 5: Animated Lesson