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Systems of equationsAlgebraHigh schoolabout 1:36

Solve the system -2x - 2y - 3z = -22, -x + 2y + z = 2, 2x - 3y - 3z = -7: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 4, y = 1, z = 4.

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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
$\begin{cases} -2x-2y-3z = -22 \\ - x+2y+ z = 2 \\ 2x-3y-3z = -7 \end{cases}$
2. Steps, part 1
1. $\begin{cases} -2x - 2y - 3z = -22 \\ -x + 2y + z = 2 \\ 2x - 3y - 3z = -7 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}-2 & -2 & -3 & -22 \\ -1 & 2 & 1 & 2 \\ 2 & -3 & -3 & -7\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{1} \to \tfrac{1}{-2} R_{1}:\ \begin{bmatrix}1 & 1 & \frac{3}{2} & 11 \\ -1 & 2 & 1 & 2 \\ 2 & -3 & -3 & -7\end{bmatrix}$
  Divide R1 by -2
3. Steps, part 2
1. $R_{2} \to R_{2} + 1R_{1},\ R_{3} \to R_{3} - 2R_{1}:\ \begin{bmatrix}1 & 1 & \frac{3}{2} & 11 \\ 0 & 3 & \frac{5}{2} & 13 \\ 0 & -5 & -6 & -29\end{bmatrix}$
  Clear column 1 using the pivot in R1
2. $R_{2} \to \tfrac{1}{3} R_{2}:\ \begin{bmatrix}1 & 1 & \frac{3}{2} & 11 \\ 0 & 1 & \frac{5}{6} & \frac{13}{3} \\ 0 & -5 & -6 & -29\end{bmatrix}$
  Divide R2 by 3
3. $R_{1} \to R_{1} - 1R_{2},\ R_{3} \to R_{3} + 5R_{2}:\ \begin{bmatrix}1 & 0 & \frac{2}{3} & \frac{20}{3} \\ 0 & 1 & \frac{5}{6} & \frac{13}{3} \\ 0 & 0 & -\frac{11}{6} & -\frac{22}{3}\end{bmatrix}$
  Clear column 2 using the pivot in R2
4. $R_{3} \to \tfrac{1}{-\frac{11}{6}} R_{3}:\ \begin{bmatrix}1 & 0 & \frac{2}{3} & \frac{20}{3} \\ 0 & 1 & \frac{5}{6} & \frac{13}{3} \\ 0 & 0 & 1 & 4\end{bmatrix}$
  Divide R3 by -11/6
4. Steps, part 3
1. $R_{1} \to R_{1} - \frac{2}{3} R_{3},\ R_{2} \to R_{2} - \frac{5}{6} R_{3}:\ \begin{bmatrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4\end{bmatrix}$
  Clear column 3 using the pivot in R3
2. $\begin{bmatrix}1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4\end{bmatrix}$
  Reduced row-echelon form
3. $x = 4,\ y = 1,\ z = 4$
  Read off the unique solution
5. The answer
$x = 4,\ y = 1,\ z = 4$
-2x - 2y - 3z = -22; -x + 2y + z = 2; 2x - 3y - 3z = -7 ⇒ x = 4, y = 1, z = 4
6. Check
- Equation 1: -2x - 2y - 3z = -22 ✓
- Equation 2: -x + 2y + z = 2 ✓
- Equation 3: 2x - 3y - 3z = -7 ✓

Transcript

The problem. Here is the problem: negative 2x minus 2y minus 3z equals negative 22; negative x plus 2y plus z equals 2; 2x minus 3y minus 3z equals negative 7. 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 2.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by 3. Clear column 2 using the pivot in R2. Divide R3 by negative 11 over 6.

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 1, z equals 4. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: negative 2x minus 2y minus 3z equals negative 22, which checks out. Equation 2: negative x plus 2y plus z equals 2, which checks out. Equation 3: 2x minus 3y minus 3z equals negative 7, which checks out.

About systems of equations

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

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