## GlossaryT.Systems2 History

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- \begin{array}{c} \color{purple}{1}\color{blue}{+}\color{teal}{2}\color{blue}{ \cdot 2 \,≟\, 5} \\ \color{blue}{1 + 4 \,≟\, 5} \\ \color{blue}{5 \,=\, 5} \\ \end{array} \begin{array}{c} \color{red}{2\cdot} \color{purple}{1} \color{red}{- } \color{teal}{2}\color{red}{ \,≟\, 0} \\ \color{red}{2 - 2 \,≟\, 0} \\ \color{red}{0 = 0} \\ \end{array}

- \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}

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- Exactly one solution. The lines intersect in exactly one point. The solution is that point - see below "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.

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• Exactly one solution. The lines intersect in exactly one point. Examples below in "Solution Methods".

- When solving this system you get something stupid like 3=5. Answer is: No solution.

- When solving this system you get numbers for x and y. Answer is: (number for x, number for y)

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• No solution. The lines are parallel and never touch. The system is inconsistent.

- When solving this system you get stuck with 0=0. Answer is: Many solutions.

- When solving this system you get something stupid like 3=5. Answer is: No solution.

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• 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.

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Related topics:

- Exactly one solution. The lines intersect in exactly one point. The solution is that point.

- Exactly one solution. The lines intersect in exactly one point. The solution is that point - see below "Solution Methods".

- When solving this system you get numbers for x and y. Answer is: (x,y)

- When solving this system you get numbers for x and y. Answer is: (number for x, number for y)

- The substitution method.
- The addition or elimination method
- Cramer's rule (determinants).

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[] \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.

[] x=\frac{\,D_x\,}{D}=\frac{-5}{-5}=1

[] y=\frac{\,D_y\,}{D}=\frac{-10}{-5}=2

[]

[]

[]Cramer's rule. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

[]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.

[ colspan=6] Solve: \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.

[colspan=6] 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

[colspan=6] 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

[colspan=6] \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

[colspan=6] \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

[colspan=6] \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

[colspan=6] \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

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[colspan=6] D = \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} = - 1 - 4 = - 5

[colspan=6] \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

[colspan=6] D = \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} = - 1 - 4 = - 5

[colspan=6] \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

[colspan=6] 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

[colspan=6] D = \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} = - 1 - 4 = - 5

[] \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

[colspan=6] D = \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} = - 1 - 4 = - 5

[colspan=6] 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

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[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = \color{orange}{5} \\ \color{red}{2}x \color{#800080}{ - 1} y = \color{#00FF80}{0} \\ \end{array} \right.

[ colspan=6] Solve: \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.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = \color{orange}{5} \\ \color{red}{2}x \color{#800080}{ - 1} y = \color{#008080}{0} \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = \color{orange}{5} \\ \color{red}{2}x \color{#800080}{ - 1} y = \color{#00FF80}{0} \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5 \\ \color{red}{2}x \color{#800080}{ - 1} y = 0 \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = \color{orange}{5} \\ \color{red}{2}x \color{#800080}{ - 1} y = \color{#008080}{0} \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5 \\ \color{red}{2}x \color{purple}{ - 1} y = 0 \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5 \\ \color{red}{2}x \color{#800080}{ - 1} y = 0 \\ \end{array} \right.

[colspan=6] D = \left| {\matrix{ \color{blue}{1} & \color{#FF8000}{2} \cr \color{red}{2} & \color{purple}{ - 1} \cr } } \right| = \color{blue}{1} \cdot \color{purple}{ ( - 1)} - \color{red}{2} \cdot \color{#FF8000}{2} = - 1 - 4 = - 5

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5 \\ \color{red}{2}x \color{purple}{ - 1} y = 0} \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5 \\ \color{red}{2}x \color{purple}{ - 1} y = 0 \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5} \\ \color{red}{2}x \color{purple}{ - 1} y = 0} \\ \end{array} \right.

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5 \\ \color{red}{2}x \color{purple}{ - 1} y = 0} \\ \end{array} \right.

A linear system

A linear system can have:

[c]Exactly one solution:

[c]Exactly one solution: (1,2)

[c]No solution:

[c]No solution

[c]Infinitely many solutions:

[c]Infinitely many solutions

A linear system with exactly one solution can be solved using any of the following three methods:

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

- The addition method (a.k.a. method of elimination).

- The addition or elimination method

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[ colspan=3] \left\{ \begin{array}{l} \color{blue}{x = 1} \\ \color{navy}{y=2} \\ \end{array} \right.

[] \left\{ \begin{array}{l} \color{blue}{x = 1} \\ \color{navy}{y=2} \\ \end{array} \right.

[]

[ colspan=2]Solution is: (1,2)

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[ colspan=3] \left\{ \begin{array}{l} \color{blue}{y=2} \\ \color{navy}{x=1} \\ \end{array} \right.

[] \left\{ \begin{array}{l} \color{blue}{y=2} \\ \color{navy}{x=1} \\ \end{array} \right.

[]

[ colspan=2]Solution is: (1,2)

[ width=110px] **Global**

[]2x2 System of Equations

[]Cramer's rule. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

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[]Brief

[]Mathcasts and Interactivities to understand and practice solving systems of 2 equations in 2 variables (unknowns).

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[]Grade

[]7-10 Interactivities start at 8th grade level on up

[ colspan=6] Solve: \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = 5} \\ \color{red}{2}x \color{purple}{ - 1} y = 0} \\ \end{array} \right.

[]Strand

[]Algebra; Expressions, Equations and Inequalities

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[]Standard

[]Algebra 1 3.1, Algebra 1 3.2, Algebra 1 3.3, ACT EE 28-32

[colspan=6] D = \left| {\matrix{ \color{blue}{1} & \color{#FF8000}{2} \cr \color{red}{2} & \color{purple}{ - 1} \cr } } \right| = \color{blue}{1} \cdot \color{purple}{ ( - 1)} - \color{red}{2} \cdot \color{#FF8000}{2} = - 1 - 4 = - 5

[]Keywords

[]system, linear system, variable, unknown, linear equation, elimination, addition, substitution

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[]Comments

[]none

[] \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

[]Download

[]

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[]Author

[]LFS - contact - website

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[]Type

[]Freeware - Available for Offline and Online Use - Translatable (html)

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[]Use

[]Requires sunJava player

[] \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.

[]

[ colspan=2]Solution is: (1,2)

[c valign="middle"]

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[]The substitution method. \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

[]The substitution method. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

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[]The addition method. \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

[]The addition method. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

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[c width=300]Graphically, we have:

[c width=300]Graphically, we have:

[]Exactly one solution: \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.

[c]Exactly one solution:

\left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

[c]No solution:

\left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{x + 2y = 2} \\ \end{array} \right.

[c]Infinitely many solutions:

\left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x + 4y = 10} \\ \end{array} \right.

Definition: Given 2 linear equations in 2 unknowns (variables). Finding a solution - that is, values for both variables that make both equations true - is called solving a 2х2 system of linear equations.

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.

- The system has exactly one solution, i.e. the lines intersect in exactly one point and the solution is that point.
- The system is inconsistent and has no solution, i.e. the lines are parallel and never touch.
- The system has infinitely many solutions, i.e. the lines coincide (are one and the same) and every point on this line is a solution.

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[]Exactly one solution: \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.

- Exactly one solution. The lines intersect in exactly one point. The solution is that point.

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[]Exactly one solution: \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.

[]The substitution method. \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

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[c]

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[c]

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[c]

[c]

[c]

[c]

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[c]

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[c]

[]Slope of a line

[]2x2 System of Equations

- The system has no solution, i.e. the lines are parallel and never touch.

- The system is inconsistent and has no solution, i.e. the lines are parallel and never touch.

- The addition method.

- The addition method (a.k.a. method of elimination).

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