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Systems of equationsAlgebraHigh schoolabout 1:11

Solve the system x/2 + y/3 = 4, x - y = 3: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 6, y = 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 (0.5 MB) · Captions (WebVTT)

Steps on screen

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

Transcript

The problem. Here is the problem: x/2 plus y/3 equals 4; x minus y equals 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 1 over 2.

Steps, part 2. Clear column 1 using the pivot in R1. Divide R2 by negative 5 over 3. 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 6, y equals 3. The system has exactly one solution; substituting it back satisfies every equation.

Check. Equation 1: (1 over 2)x plus (1 over 3)y equals 4, which checks out. Equation 2: x minus y equals 3, which checks out.

About systems of equations

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

Solve the system x/2 + y/3 = 4, x - y = 3: Animated Lesson