# Systems of Linear Equations 2х2 - 2 equations in 2 variables

 Definition: Finding a simultaneous solution to 2 linear equations in 2 variables - values for both variables that make both equations true - is called solving a 2х2 system of linear equations.

Example of a linear system of 2 equations in 2 variables

 Sample 2x2 linear system: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. The solution to this system is   (\color{purple}{x},\color{teal}{y})=(\color{purple}{1},\color{teal}{2}) . Why is this a solution? Because - when we substitute \color{purple}{x=1} and \color{teal}{y=2} , both equations are "true". \begin{array}{c} \color{purple}{1}\color{blue}{+}\color{teal}{2}\color{blue}{ \cdot 2 \,\mathop = \limits^?\, 5} \\ \color{blue}{1 + 4 \,\mathop = \limits^?\, 5} \\ \color{blue}{5 \,=\, 5} \\ \end{array}      \begin{array}{c} \color{red}{2\cdot} \color{purple}{1} \color{red}{- } \color{teal}{2}\color{red}{ \,\mathop = \limits^?\, 0} \\ \color{red}{2 - 2 \,\mathop = \limits^?\, 0} \\ \color{red}{0 = 0} \\ \end{array} Graphically, we have: This browser does not have a Java Plug-in. Get the latest Java Plug-in here.

Regulation: A solution to the system is every intersection (touching) point of the 2 lines.

 InterActivity   Directions for InterActivity 1. Look at the two lines a and b. They intersect at the point E. Check that E satisfies both equations. If you know how, solve the system: a and b and check that you get E.. 2. Click and drag the lines or the points A, B, C or D. If they intersect at one point, E is "good". Look at the left (where the formulas are) to check that E is a point! 3. Click and drag the points A, B, C or D so that a and b are parallel. They are parallel when E says "undefined". Notice that a and b are the same except for the constant after the "=" sign. 4. Click and drag the points A, B, C or D so that a and b coincide. Notice that a and b are completely the same and that E is undefined. This browser does not have a Java Plug-in. Get the latest Java Plug-in here.
Solutions to 2x2 linear systems   Graphs!

A linear system can have:

 •  Exactly one solution. The lines intersect in exactly one point. Examples below in "Solution Methods". When solving this system you get numbers for x and y. Answer is: (number for x, number for y) •  No solution. The lines are parallel and never touch. The system is inconsistent. When solving this system you get something stupid like 3=5. Answer is: No solution. •  Infinitely many solutions. The lines coincide. They are the same line. Every point on this line is a solution. More? When solving this system you get stuck with 0=0. Answer is: Many solutions.
 Exactly one solution: (1,2) \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. No solution \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{x + 2y = 2} \\ \end{array} \right. Infinitely many solutions \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x + 4y = 10} \\ \end{array} \right. This browser does not have a Java Plug-in. Get the latest Java Plug-in here. This browser does not have a Java Plug-in. Get the latest Java Plug-in here. This browser does not have a Java Plug-in. Get the latest Java Plug-in here.
Solution methods

A linear system can be solved using any of the following three methods:

• The substitution method.    Example
• The addition or elimination method   Example
• Cramer's rule (determinants) Example
The substitution method. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.
 ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{x = 5-2y} \\ \color{navy}{2x - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{2\cdot\color{blue}{(5-2y)} - y = 0} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{10-4y - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{10-5y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{10=5y} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{y=2} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{5y=10} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{x = 5-2 \cdot} \color{red}{2} \\ \color{red}{y=2} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{x = 5-4} \\ \color{navy}{y=2} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{x = 1} \\ \color{navy}{y=2} \\ \end{array} \right. Solution is: (1,2)
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The addition method. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.
 ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{red}{4x - 2y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{purple}{5x+0y =5} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{purple}{5x=5} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{purple}{x=1} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{purple}{1} \color{blue}{+2y=5} \\ \color{navy}{x=1} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{2y=4} \\ \color{navy}{x=1} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{y=2} \\ \color{navy}{x=1} \\ \end{array} \right. Solution is: (1,2)
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Cramer's rule. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. = \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = \color{orange}{5} \\ \color{red}{2}x \color{#800080}{ - 1} y = \color{#00BB80}{0} \\ \end{array} \right.
 ------------------------------------------------------------------------------- D = \left| {\matrix{ \color{blue}{1} & \color{#FF8000}{2} \cr \color{red}{2} & \color{#800080}{ - 1} \cr } } \right| = \color{blue}{1} \cdot \color{#800080}{ ( - 1)} - \color{red}{2} \cdot \color{#FF8000}{2} = - 1 - 4 = - 5 ------------------------------------------------------------------------------- \color{purple}{D_x} = \left| {\matrix{ \color{orange}{5} & \color{#FF8000}{2} \cr \color{#00BB80}{0} & \color{#800080}{ - 1} \cr } } \right| = \color{orange}{5} \cdot \color{#800080}{ ( - 1)} - \color{#00BB80}{0} \cdot \color{#FF8000}{2} = - 5 - 0 = - 5 ------------------------------------------------------------------------------- \color{teal}{D_y} = \left| {\matrix{ \color{blue}{1} & \color{orange}{5} \cr \color{red}{2} & \color{#00BB80}{0} \cr } } \right| = \color{blue}{1} \cdot \color{#00BB80}{0} - \color{red}{2} \cdot \color{orange}{5} = 0 - 10 = - 10 ------------------------------------------------------------------------------- x=\frac{\,D_x\,}{D}=\frac{-5}{-5}=1 y=\frac{\,D_y\,}{D}=\frac{-10}{-5}=2 Solution is: (1,2)