Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Elementary algebra
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
=== System of linear equations === {{Main|System of linear equations}} There are different methods to solve a system of linear equations with two variables. ==== Elimination method ==== [[File:Intersecting Lines.svg|thumb|right|The solution set for the equations <math>x - y = -1</math> and <math>3x + y = 9</math> is the single point (2, 3).]] An example of solving a system of linear equations is by using the elimination method: : <math>\begin{cases}4x + 2y&= 14 \\ 2x - y&= 1.\end{cases} </math> Multiplying the terms in the second equation by 2: : <math>4x + 2y = 14 </math> : <math>4x - 2y = 2. </math> Adding the two equations together to get: : <math>8x = 16 </math> which simplifies to : <math>x = 2. </math> Since the fact that <math>x = 2</math> is known, it is then possible to deduce that <math>y = 3</math> by either of the original two equations (by using ''2'' instead of {{mvar|x}} ) The full solution to this problem is then : <math>\begin{cases} x = 2 \\ y = 3. \end{cases}</math> This is not the only way to solve this specific system; {{mvar|y}} could have been resolved before {{mvar|x}}. ==== Substitution method ==== Another way of solving the same system of linear equations is by substitution. : <math>\begin{cases}4x + 2y &= 14 \\ 2x - y &= 1.\end{cases} </math> An equivalent for {{mvar|y}} can be deduced by using one of the two equations. Using the second equation: : <math>2x - y = 1 </math> Subtracting <math>2x</math> from each side of the equation: : <math>\begin{align}2x - 2x - y & = 1 - 2x \\ - y & = 1 - 2x \end{align}</math> and multiplying by β1: : <math> y = 2x - 1. </math> Using this {{mvar|y}} value in the first equation in the original system: : <math>\begin{align}4x + 2(2x - 1) &= 14\\ 4x + 4x - 2 &= 14 \\ 8x - 2 &= 14 \end{align}</math> Adding ''2'' on each side of the equation: : <math>\begin{align}8x - 2 + 2 &= 14 + 2 \\ 8x &= 16 \end{align}</math> which simplifies to : <math>x = 2 </math> Using this value in one of the equations, the same solution as in the previous method is obtained. : <math>\begin{cases} x = 2 \\ y = 3. \end{cases}</math> This is not the only way to solve this specific system; in this case as well, {{mvar|y}} could have been solved before {{mvar|x}}.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)