Solve a problem

Systems of equationsAlgebraHigh schoolabout 1:31

Solve the system -2x + y + z = -1, 3x - y + 2z = 9, 3y + 3z = 9: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 2, y = 1, z = 2.

SpeedLanguage

Chapters

Transcript

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 (0.6 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Solving an equation
$\begin{cases} -2x+ y+ z = -1 \\ 3x- y+2z = 9 \\ 3y+3z = 9 \end{cases}$
2. Steps, part 1
1. $\begin{cases} -2x + y + z = -1 \\ 3x - y + 2z = 9 \\ 3y + 3z = 9 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}-2 & 1 & 1 & -1 \\ 3 & -1 & 2 & 9 \\ 0 & 3 & 3 & 9\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{1} \to \tfrac{1}{-2} R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 3 & -1 & 2 & 9 \\ 0 & 3 & 3 & 9\end{bmatrix}$
  Divide R1 by -2
3. Steps, part 2
1. $R_{2} \to R_{2} - 3R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & \frac{1}{2} & \frac{7}{2} & \frac{15}{2} \\ 0 & 3 & 3 & 9\end{bmatrix}$
  Clear column 1 using the pivot in R1
2. $R_{2} \to \tfrac{1}{\frac{1}{2}} R_{2}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 7 & 15 \\ 0 & 3 & 3 & 9\end{bmatrix}$
  Divide R2 by 1/2
3. $R_{1} \to R_{1} + \frac{1}{2} R_{2},\ R_{3} \to R_{3} - 3R_{2}:\ \begin{bmatrix}1 & 0 & 3 & 8 \\ 0 & 1 & 7 & 15 \\ 0 & 0 & -18 & -36\end{bmatrix}$
  Clear column 2 using the pivot in R2
4. $R_{3} \to \tfrac{1}{-18} R_{3}:\ \begin{bmatrix}1 & 0 & 3 & 8 \\ 0 & 1 & 7 & 15 \\ 0 & 0 & 1 & 2\end{bmatrix}$
  Divide R3 by -18
4. Steps, part 3
1. $R_{1} \to R_{1} - 3R_{3},\ R_{2} \to R_{2} - 7R_{3}:\ \begin{bmatrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2\end{bmatrix}$
  Clear column 3 using the pivot in R3
2. $\begin{bmatrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2\end{bmatrix}$
  Reduced row-echelon form
3. $x = 2,\ y = 1,\ z = 2$
  Read off the unique solution
5. The answer
$x = 2,\ y = 1,\ z = 2$
-2x + y + z = -1; 3x - y + 2z = 9; 3y + 3z = 9 ⇒ x = 2, y = 1, z = 2
6. Check
- Equation 1: -2x + y + z = -1 ✓
- Equation 2: 3x - y + 2z = 9 ✓
- Equation 3: 3y + 3z = 9 ✓

Transcript

The problem. Here is the problem: negative 2x plus y plus z equals negative 1; 3x minus y plus 2z equals 9; 3y plus 3z equals 9. 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 1 over 2. Clear column 2 using the pivot in R2. Divide R3 by negative 18.

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 2, y equals 1, z equals 2. The system has exactly one solution; substituting it back satisfies every equation.

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

About systems of equations

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

Solve the system -2x + y + z = -1, 3x - y + 2z = 9, 3y +…