GlossaryT.Systems2 History

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January 24, 2009, at 04:30 AM by LFS -
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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}
to:
\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}
November 27, 2008, at 02:00 AM by LFS -
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October 05, 2008, at 10:38 PM by LFS -
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[row  bgcolor=#FFDDDD]

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[row  bgcolor=#EEFFEE]

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[row  bgcolor=#EEFFEE]

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[row  bgcolor=#EEEEFF]

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[row  bgcolor=#EEEEEE]

October 05, 2008, at 10:27 PM by LFS -
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  • Exactly one solution. The lines intersect in exactly one point. The solution is that point - see below "Solution Methods".
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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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[table cellspacing=1 cellpadding=3 width=100%]
[row  bgcolor=#FFDDDD]
[]
•  Exactly one solution. The lines intersect in exactly one point. Examples below in "Solution Methods".

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

[row  bgcolor=#EEFFEE]
[]
•  No solution. The lines are parallel and never touch. The system is inconsistent.

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When solving this system you get stuck with 0=0. Answer is: Many solutions.
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When solving this system you get something stupid like 3=5. Answer is: No solution.

[row  bgcolor=#EEEEFF]
[]
•  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.

[tableend]

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October 05, 2008, at 09:48 PM by LFS -
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The solution to this system is   (\color{purple}{x},\color{teal}{y})=(\color{purple}{1},\color{teal}{1}) .
to:
The solution to this system is   (\color{purple}{x},\color{teal}{y})=(\color{purple}{1},\color{teal}{2}) .
October 05, 2008, at 12:07 PM by LFS -
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[table border=1 cellpadding=3 width=825]

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[tableend]

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October 05, 2008, at 12:06 PM by LFS -
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  • Exactly one solution. The lines intersect in exactly one point. The solution is that point.
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  • Exactly one solution. The lines intersect in exactly one point. The solution is that point - see below "Solution Methods".
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When solving this system you get numbers for x and y. Answer is: (x,y)
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When solving this system you get numbers for x and y. Answer is: (number for x, number for y)
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October 05, 2008, at 12:01 PM by LFS -
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  • The substitution method.
  • The addition or elimination method
  • Cramer's rule (determinants).
to:
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[table border=1 cellpadding=3 cellspacing=0 width=100%]

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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.

to:

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

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

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[]

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October 05, 2008, at 08:35 AM by LFS -
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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.

to:

[]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.

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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{#00BB80}{0} \\ \end{array} \right.

to:

[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] 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

to:

[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

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[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

to:

[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] \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
[row]
[ colspan=6]-------------------------------------------------------------------------------
[row]

October 05, 2008, at 08:32 AM by LFS -
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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

to:

[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

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[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

to:

[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

October 05, 2008, at 08:30 AM by LFS -
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[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

to:

[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

October 05, 2008, at 08:27 AM by LFS -
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[] \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

to:

[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

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[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
[row]
[ colspan=6]-------------------------------------------------------------------------------
[row]

October 05, 2008, at 08:25 AM by LFS -
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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.

to:

[ 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.

October 05, 2008, at 08:23 AM by LFS -
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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{#008080}{0} \\ \end{array} \right.

to:

[ 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.

October 05, 2008, at 08:17 AM by LFS -
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[ 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.

to:

[ 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.

October 05, 2008, at 08:12 AM by LFS -
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[ 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.

to:

[ 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.

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[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

to:

[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

October 05, 2008, at 07:50 AM by LFS -
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[ 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.

to:

[ 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.

October 05, 2008, at 07:49 AM by LFS -
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[ 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.

to:

[ 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.

October 05, 2008, at 07:47 AM by LFS -
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A linear system  

to:

A linear system can have:

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[c]Exactly one solution:

to:

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

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[c]No solution:

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[c]No solution

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[c]Infinitely many solutions:

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[c]Infinitely many solutions

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

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

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

to:

[] \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.

to:

[] \left\{ \begin{array}{l} \color{blue}{y=2} \\ \color{navy}{x=1} \\ \end{array} \right.
[] 
[ colspan=2]Solution is: (1,2)

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[  width=110px] Global
[]2x2 System of Equations

to:

[]Cramer's rule. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.
[table border=0 cellpadding=3]

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

to:

[ colspan=6]-------------------------------------------------------------------------------

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

to:

[ 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.

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[]Strand
[]Algebra; Expressions, Equations and Inequalities

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[ colspan=6]-------------------------------------------------------------------------------

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[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

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

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[ colspan=6]-------------------------------------------------------------------------------

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[]Comments
[]none

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[] \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

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[]Download
[] 

to:

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[]Author
[]LFS - contact - website
[row]
[]Type
[]Freeware - Available for Offline and Online Use - Translatable (html)
[row]
[]Use
[]Requires sunJava player

to:

[] \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)

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October 03, 2008, at 03:02 PM by LFS -
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[c valign="middle"]

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[c valign="middle"]

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[c valign="middle"]

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[c valign="middle"]

October 03, 2008, at 01:10 PM by LFS -
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[]The substitution method. \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.

to:

[]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.

to:

[]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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October 03, 2008, at 11:09 AM by LFS -
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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 \,≟\, 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}

[c width=300]Graphically, we have:

to:
Why is this a solution? Because - when we
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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 \,≟\, 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}

[c width=300]Graphically, we have:

October 03, 2008, at 11:05 AM by LFS -
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Why is this a solution? Because, when we substitute \color{purple}{x=1} and \color{teal}{y=2} , both equations are "true":
to:
Why is this a solution? Because:
When we substitute \color{purple}{x=1} and \color{teal}{y=2} , both equations are "true":
October 03, 2008, at 11:01 AM by LFS -
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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.

to:

[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.

October 03, 2008, at 10:59 AM by LFS -
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The solution to this system is   (%up%x%%,%bgr%y%%)=(%up%1%%,%bgr%2%%) .
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The solution to this system is   (\color{purple}{x},\color{teal}{y})=(\color{purple}{1},\color{teal}{1}) .
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October 02, 2008, at 11:13 PM by LFS -
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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.

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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.

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The solution to this system is   (x,y)=(1,2) .
To see this we substitute \color{purple}{x=1} and \color{teal}{y=2} and check whether both equations are "true":
to:
The solution to this system is   (%up%x%%,%bgr%y%%)=(%up%1%%,%bgr%2%%) .
Why is this a solution? Because, when we substitute \color{purple}{x=1} and \color{teal}{y=2} , both equations are "true":
October 02, 2008, at 11:07 PM by LFS -
October 02, 2008, at 11:07 PM by LFS -
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  • 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.

[table border=1 cellpadding=3 cellspacing=0 width=100%]
[row]
[]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.

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  • Exactly one solution. The lines intersect in exactly one point. The solution is that point.
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[tableend]

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[table border=1 cellpadding=3 width=825]

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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.

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[]The substitution method. \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.
[table border=0 cellpadding=3]

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October 02, 2008, at 11:48 AM by LFS -
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October 02, 2008, at 11:47 AM by LFS -
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October 01, 2008, at 12:10 PM by LFS -
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September 29, 2008, at 03:40 PM by LFS -
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September 29, 2008, at 02:57 PM by LFS -
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[c]

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[c valign="middle"]

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[c]

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[c valign="middle"]

September 28, 2008, at 02:14 PM by LFS -
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[c]

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[c]

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[c]

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[c]

September 28, 2008, at 02:10 PM by LFS -
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[table border=1 cellpadding=3 width=825]

September 28, 2008, at 02:04 PM by LFS -
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[c]
[row]

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[c]

September 28, 2008, at 09:43 AM by LFS -
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[]Slope of a line

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[]2x2 System of Equations

September 28, 2008, at 09:42 AM by LFS -
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  • The system has no solution, i.e. the lines are parallel and never touch.
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  • The system is inconsistent and has no solution, i.e. the lines are parallel and never touch.
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  • The addition method.
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  • The addition method (a.k.a. method of elimination).
September 28, 2008, at 09:40 AM by LFS -
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  •  
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September 28, 2008, at 09:26 AM by LFS -
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[table width=100%]

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September 27, 2008, at 04:42 AM by LFS -
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Page last modified on January 24, 2009, at 04:30 AM