## GlossaryT.Systems2 History

January 24, 2009, at 04:30 AM by LFS -
Changed line 27 from:
\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 -
Changed line 3 from:
to:
October 05, 2008, at 10:38 PM by LFS -
Changed line 67 from:

[row  bgcolor=#FFDDDD]

to:

[row  bgcolor=#EEFFEE]

Changed line 72 from:

[row  bgcolor=#EEFFEE]

to:

[row  bgcolor=#FFDDDD]

Changed line 77 from:

[row  bgcolor=#EEEEFF]

to:

[row  bgcolor=#EEEEEE]

October 05, 2008, at 10:27 PM by LFS -
Changed line 62 from:
to:
Deleted lines 64-65:
• Exactly one solution. The lines intersect in exactly one point. The solution is that point - see below "Solution Methods".
Changed lines 66-67 from:
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.
to:

[row  bgcolor=#FFDDDD]
[]
•  Exactly one solution. The lines intersect in exactly one point. Examples below in "Solution Methods".

Changed lines 71-72 from:
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.

Changed lines 76-82 from:
When solving this system you get stuck with 0=0. Answer is: Many solutions.
to:
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]

Changed lines 111-113 from:
to:
Deleted lines 233-234:
October 05, 2008, at 09:48 PM by LFS -
Changed line 23 from:
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 -
Changed lines 226-240 from:
to:

[tableend]

Changed lines 295-296 from:
to:
October 05, 2008, at 12:06 PM by LFS -
Changed line 65 from:
• Exactly one solution. The lines intersect in exactly one point. The solution is that point.
to:
• Exactly one solution. The lines intersect in exactly one point. The solution is that point - see below "Solution Methods".
Changed line 68 from:
When solving this system you get numbers for x and y. Answer is: (x,y)
to:
When solving this system you get numbers for x and y. Answer is: (number for x, number for y)
October 05, 2008, at 12:01 PM by LFS -
Changed line 99 from:
to:
Changed lines 103-105 from:
• The substitution method.
• The addition or elimination method
• Cramer's rule (determinants).
to:
Changed lines 107-109 from:

to:
Changed lines 211-213 from:

[] \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

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

Changed line 217 from:
to:

[]

Deleted line 218:
October 05, 2008, at 08:35 AM by LFS -
Changed line 182 from:

[]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.

Changed line 187 from:

[ 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

Changed line 191 from:

[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

Changed line 195 from:

[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

Deleted lines 198-201:

[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 -
Changed line 195 from:

[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

Changed line 199 from:

[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 -
Changed line 199 from:

[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 -
Changed lines 195-200 from:

[] \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

[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 -
Changed line 187 from:

[ 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 -
Changed line 187 from:

[ 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 -
Changed line 187 from:

[ 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 -
Changed line 187 from:

[ 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.

Changed line 191 from:

[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 -
Changed line 187 from:

[ 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 -
Changed line 187 from:

[ 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 -
Changed line 62 from:
to:
Changed line 64 from:

A linear system

to:

A linear system can have:

Changed line 80 from:

[c]Exactly one solution:

to:

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

Changed line 82 from:

[c]No solution:

to:

[c]No solution

Changed line 84 from:

[c]Infinitely many solutions:

to:

[c]Infinitely many solutions

Changed lines 99-100 from:
to:
Changed line 102 from:

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

to:

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

Changed line 104 from:
• The addition method (a.k.a. method of elimination).
to:
• The addition or elimination method
Changed line 106 from:
to:
Changed line 112 from:

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

to:

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

Changed line 121 from:

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

to:

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

Changed line 130 from:

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

to:

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

Changed line 139 from:

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

to:

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

Changed lines 143-145 from:

[ 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)

Changed line 153 from:

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

to:

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

Changed line 162 from:

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

to:

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

Changed line 171 from:

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

to:

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

Changed lines 175-177 from:

[ 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)

Deleted lines 180-200:
Changed lines 182-183 from:

[  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.

Changed lines 185-186 from:

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

to:

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

Changed lines 187-188 from:

[]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.

Changed lines 189-190 from:

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

to:

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

Changed lines 191-192 from:
to:

[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

Changed lines 193-194 from:

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

to:

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

Changed lines 195-196 from:

[]none

to:

[] \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

Changed lines 202-203 from:

[]

to:

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

Changed lines 204-211 from:

[]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)

Changed lines 210-212 from:
to:
Deleted lines 213-224:
October 03, 2008, at 03:02 PM by LFS -
Changed line 144 from:

[c valign="middle"]

to:

[c valign="middle"]

Changed line 174 from:

[c valign="middle"]

to:

[c valign="middle"]

October 03, 2008, at 01:10 PM by LFS -
Changed line 108 from:

[]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.

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

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

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

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

Changed line 147 from:

[]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.

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

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

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

October 03, 2008, at 11:09 AM by LFS -
Changed lines 24-27 from:
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
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 -
Changed lines 24-25 from:
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 -
Changed lines 78-80 from:

[]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 -
Changed line 23 from:
The solution to this system is   (%up%x%%,%bgr%y%%)=(%up%1%%,%bgr%2%%) .
to:
The solution to this system is   (\color{purple}{x},\color{teal}{y})=(\color{purple}{1},\color{teal}{1}) .
October 02, 2008, at 11:13 PM by LFS -
Changed line 11 from:

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.

to:

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.

Changed line 17 from:
to:
Changed lines 23-24 from:
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 -
Changed line 60 from:
to:
Changed lines 63-70 from:
• 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.

[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.

to:
• Exactly one solution. The lines intersect in exactly one point. The solution is that point.
Changed lines 65-70 from:
to:
Changed lines 69-73 from:

[tableend]

to:
Changed lines 75-76 from:

to:
Changed lines 78-85 from:
to:

[]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.

Changed lines 83-84 from:

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

to:
October 02, 2008, at 11:48 AM by LFS -
Changed line 156 from:
to:
October 02, 2008, at 11:47 AM by LFS -
October 01, 2008, at 12:10 PM by LFS -
October 01, 2008, at 08:47 AM by LFS -
September 29, 2008, at 03:40 PM by LFS -
Changed line 3 from:
to:
September 29, 2008, at 02:57 PM by LFS -
Changed line 120 from:

[c]

to:

[c valign="middle"]

Changed line 143 from:

[c]

to:

[c valign="middle"]

September 28, 2008, at 02:14 PM by LFS -
Changed line 120 from:

[c]

to:

[c]

Changed line 143 from:

[c]

to:

[c]

September 28, 2008, at 02:10 PM by LFS -
Changed line 8 from:

to:

Changed line 14 from:

to:

Changed line 35 from:

to:

Changed line 58 from:

to:

Changed line 81 from:

to:

September 28, 2008, at 02:04 PM by LFS -

[c]
[row]

[c]

September 28, 2008, at 09:43 AM by LFS -
Changed line 161 from:

[]Slope of a line

to:

[]2x2 System of Equations

September 28, 2008, at 09:42 AM by LFS -
Changed line 64 from:
• The system has no solution, i.e. the lines are parallel and never touch.
to:
• The system is inconsistent and has no solution, i.e. the lines are parallel and never touch.
Changed line 87 from:
to:
• The addition method (a.k.a. method of elimination).
September 28, 2008, at 09:40 AM by LFS -
Changed line 173 from:
to:
September 28, 2008, at 09:29 AM by LFS -
Changed lines 150-151 from:
•
to:
Changed line 173 from:
to:
September 28, 2008, at 09:26 AM by LFS -
Changed lines 154-157 from:

[table width=100%]

to: