Solve a problem

Systems of equationsAlgebraHigh schoolabout 1:38

Solve the system -3x - 3y - 3z = 9, -3x - y - 3z = 1, -x + 3y + 3z = -25: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 4, y = -4, z = -3.

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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 (0.7 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Solving an equation
$\begin{cases} -3x-3y-3z = 9 \\ -3x- y-3z = 1 \\ - x+3y+3z = -25 \end{cases}$
2. Steps, part 1
1. $\begin{cases} -3x - 3y - 3z = 9 \\ -3x - y - 3z = 1 \\ -x + 3y + 3z = -25 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}-3 & -3 & -3 & 9 \\ -3 & -1 & -3 & 1 \\ -1 & 3 & 3 & -25\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{1} \to \tfrac{1}{-3} R_{1}:\ \begin{bmatrix}1 & 1 & 1 & -3 \\ -3 & -1 & -3 & 1 \\ -1 & 3 & 3 & -25\end{bmatrix}$
  Divide R1 by -3
3. Steps, part 2
1. $R_{2} \to R_{2} + 3R_{1},\ R_{3} \to R_{3} + 1R_{1}:\ \begin{bmatrix}1 & 1 & 1 & -3 \\ 0 & 2 & 0 & -8 \\ 0 & 4 & 4 & -28\end{bmatrix}$
  Clear column 1 using the pivot in R1
2. $R_{2} \to \tfrac{1}{2} R_{2}:\ \begin{bmatrix}1 & 1 & 1 & -3 \\ 0 & 1 & 0 & -4 \\ 0 & 4 & 4 & -28\end{bmatrix}$
  Divide R2 by 2
3. $R_{1} \to R_{1} - 1R_{2},\ R_{3} \to R_{3} - 4R_{2}:\ \begin{bmatrix}1 & 0 & 1 & 1 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 4 & -12\end{bmatrix}$
  Clear column 2 using the pivot in R2
4. $R_{3} \to \tfrac{1}{4} R_{3}:\ \begin{bmatrix}1 & 0 & 1 & 1 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & -3\end{bmatrix}$
  Divide R3 by 4
4. Steps, part 3
1. $R_{1} \to R_{1} - 1R_{3}:\ \begin{bmatrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & -3\end{bmatrix}$
  Clear column 3 using the pivot in R3
2. $\begin{bmatrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & -3\end{bmatrix}$
  Reduced row-echelon form
3. $x = 4,\ y = -4,\ z = -3$
  Read off the unique solution
5. The answer
$x = 4,\ y = -4,\ z = -3$
-3x - 3y - 3z = 9; -3x - y - 3z = 1; -x + 3y + 3z = -25 ⇒ x = 4, y = -4, z = -3
6. Check
- Equation 1: -3x - 3y - 3z = 9 ✓
- Equation 2: -3x - y - 3z = 1 ✓
- Equation 3: -x + 3y + 3z = -25 ✓

Transcript

The problem. Here is the problem: negative 3x minus 3y minus 3z equals 9; negative 3x minus y minus 3z equals 1; negative x plus 3y plus 3z equals negative 25. 3 linear equations in 3 unknowns (x, y, z).

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, z | constant. Gauss–Jordan elimination. Use row operations to make each pivot 1 and clear the rest of its column. Divide R1 by negative 3.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 2. Clear column 2 using the pivot in R2. Divide R3 by 4.

Steps, part 3. Clear column 3 using the pivot in R3. Reduced row-echelon form. Read off the unique solution.

The answer. So the answer is x equals 4, y equals negative 4, z equals negative 3. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 3x minus 3y minus 3z equals 9, which checks out. Equation 2: negative 3x minus y minus 3z equals 1, which checks out. Equation 3: negative x plus 3y plus 3z equals negative 25, which checks out.

About systems of equations

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

Solve the system -3x - 3y - 3z = 9, -3x - y - 3z = 1, -x +…