Solve a problem

Systems of equationsAlgebraHigh schoolabout 1:07

Solve the system y = 2x + 1, y = -x + 4: animated lesson

A narrated, step-by-step systems of equations video lesson with captions. Final answer: x = 1, 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
$\begin{cases} y = 2x+1 \\ y = - x+4 \end{cases}$
2. Steps, part 1
1. $\begin{cases} -2x + y = 1 \\ x + y = 4 \end{cases}$
  Write every equation in standard form
2. $\begin{bmatrix}-2 & 1 & 1 \\ 1 & 1 & 4\end{bmatrix}$
  Build the augmented matrix [A | b]
3. Gauss–Jordan elimination
4. $R_{1} \to \tfrac{1}{-2} R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & 1 & 4\end{bmatrix}$
  Divide R1 by -2
3. Steps, part 2
1. $R_{2} \to R_{2} - 1R_{1}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & \frac{3}{2} & \frac{9}{2}\end{bmatrix}$
  Clear column 1 using the pivot in R1
2. $R_{2} \to \tfrac{1}{\frac{3}{2}} R_{2}:\ \begin{bmatrix}1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 1 & 3\end{bmatrix}$
  Divide R2 by 3/2
3. $R_{1} \to R_{1} + \frac{1}{2} R_{2}:\ \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 3\end{bmatrix}$
  Clear column 2 using the pivot in R2
4. $\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 3\end{bmatrix}$
  Reduced row-echelon form
4. Steps, part 3
1. $x = 1,\ y = 3$
  Read off the unique solution
5. The answer
$x = 1,\ y = 3$
-2x + y = 1; x + y = 4 ⇒ x = 1, y = 3
6. Check
- Equation 1: -2x + y = 1 ✓
- Equation 2: x + y = 4 ✓

Transcript

The problem. Here is the problem: y equals 2x plus 1; y equals negative x plus 4. 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 2.

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

Check. Equation 1: negative 2x plus y equals 1, which checks out. Equation 2: x plus y equals 4, which checks out.

About systems of equations

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

Solve the system y = 2x + 1, y = -x + 4: Step-by-Step Lesson