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Difference between revisions of "System of equations"

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A '''system of equations''' is a set of [[equation]]s which share the same [[variable]]s.  An example of a system of equations is
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A '''system of equations''' is a set of [[equation]]s which share the same [[variable]]s.  Below is an example of a system of equations.
  
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{| class="wikitable" style="margin: 1em auto 1em auto"
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<cmath>\left\{\begin{array}{l}a^2 + b^2 = 25\\a+b=7\end{array}\right.</cmath>
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| <math>2a - 3b</math> || <math>= 4</math>
 
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|-
 
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| <math>3a - 2b</math> || <math>= 3</math>
 
−
|}
 
  
 
== Solving Linear Systems ==
 
== Solving Linear Systems ==
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Find the ordered pair <math>(x,y)</math> for which
 
Find the ordered pair <math>(x,y)</math> for which
  
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{| class="wikitable" style="margin: 1em auto 1em auto"
+
<cmath>\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.</cmath>
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| <math>x - 12y</math> || <math>= 2</math>
 
−
|-
 
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| <math>3x + 6y</math> || <math>= 6</math>
 
−
|}
 
  
 
==== Solution ====
 
==== Solution ====
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|}
 
|}
  
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Thus <math>x=2</math>.  We can then plug in for <math>x</math> in either of the equations: <center><math> (2)-12y = 2 \Rightarrow y = 0</math>.</center>
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Thus <math>x=2</math>.  We can then plug in for <math>x</math> in either of the equations:
 +
<cmath>\begin{align*}
 +
(2)-12y &= 2 \\
 +
y &= 0
 +
\end{align*}</cmath>
  
 
Thus, the solution to the system is <math>(2,0)</math>.
 
Thus, the solution to the system is <math>(2,0)</math>.
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Find the ordered pair <math>(x,y)</math> for which
 
Find the ordered pair <math>(x,y)</math> for which
  
−
{| class="wikitable" style="margin: 1em auto 1em auto"
+
<cmath>\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.</cmath>
−
| <math>x - 12y</math> || <math>= 2</math>
 
−
|-
 
−
| <math>3x + 6y</math> || <math>= 6</math>
 
−
|}
 
  
 
==== Solution ====
 
==== Solution ====
Line 62: Line 54:
 
Find the ordered pair <math>(x,y)</math> for which
 
Find the ordered pair <math>(x,y)</math> for which
  
−
{| class="wikitable" style="margin: 1em auto 1em auto"
+
<cmath>\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.</cmath>
−
| <math>x - 12y</math> || <math>= 2</math>
 
−
|-
 
−
| <math>3x + 6y</math> || <math>= 6</math>
 
−
|}
 
  
 
==== Solution ====
 
==== Solution ====
Line 79: Line 67:
  
 
== Convenient Systems ==
 
== Convenient Systems ==
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Convenient systems usually seem very tough to solve at first.  Often times a clever insight will make things much easier.
 
  
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=== Problem ===
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Some systems can be solved by taking advantage of specific forms.  Such systems can often seem tough to solve at first, however.
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Solve the following system:
+
 
 +
===Symmetry===
 +
 
 +
Consider the below system.
  
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{| class="wikitable" style="margin: 1em auto 1em auto"
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<cmath>\left\{\begin{array}{l}a+b+c+d=4\\a+b+c+e=8\\a+b+d+e=12\\a+c+d+e=16\\b+c+d+e=20\end{array}\right.</cmath>
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| <math>a+</math> || <math>b+</math> || <math>c+</math> || <math>d</math> || || <math>=4</math>
 
−
|-
 
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| <math>a+</math> || <math>b+</math> || <math>c+</math> || || <math>e</math> || <math>=8</math>
 
−
|-
 
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| <math>a+</math> || <math>b+</math> || || <math>d+</math> || <math>e</math> || <math>=12</math>
 
−
|-
 
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| <math>a+</math> || || <math>c+</math> || <math>d+</math> || <math>e</math> || <math>=16</math>
 
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|-
 
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| || <math>b+</math> || <math>c+</math> || <math>d+</math> || <math>e</math> || <math>=20</math>
 
−
|}
 
  
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=== Solution ===
 
 
The key here is to take advantage of the [[symmetry]].  If we add up all 5 equations we will have a total of 4 of each variable on the [[LHS]].  On the [[RHS]] we will have <math>4+8+12+16+20 = 60</math>.  Thus <center><math> 4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15. </math></center>
 
The key here is to take advantage of the [[symmetry]].  If we add up all 5 equations we will have a total of 4 of each variable on the [[LHS]].  On the [[RHS]] we will have <math>4+8+12+16+20 = 60</math>.  Thus <center><math> 4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15. </math></center>
  
 
So then subtracting the first equation from this leaves <math>e</math> on the LHS and <math>15-4=11</math> on the RHS.  Subtracting this equation from the second equation leaves <math>d</math> on the LHS and <math>15-8=7</math> on the RHS.  And thus we continue on in this way to find that <math>(a,b,c,d,e)=(-5,-1,3,7,11).</math>
 
So then subtracting the first equation from this leaves <math>e</math> on the LHS and <math>15-4=11</math> on the RHS.  Subtracting this equation from the second equation leaves <math>d</math> on the LHS and <math>15-8=7</math> on the RHS.  And thus we continue on in this way to find that <math>(a,b,c,d,e)=(-5,-1,3,7,11).</math>
  
−
== See also ==
+
===Clever Substitution===
 +
 
 +
Consider the below system.
 +
 
 +
<cmath>\left\{\begin{array}{l}x+y+3xy = 5\\ 2x+2y+4xy=8 \end{array}\right.</cmath>
 +
 
 +
We can let <math>x+y=a</math> and <math>b = xy</math> to get the two-variable linear system below.
 +
 
 +
<cmath>\left\{\begin{array}{l}a+3b = 5\\ 2a+4b=8 \end{array}\right.</cmath>
 +
 
 +
Solving the system results in <math>b = 1</math> and <math>a = 2</math>.  Substituting that back results in <math>xy = 1</math> and <math>x + y = 2</math>.  We can do another substitution by letting <math>y = 2 - x</math> and substituting to get <math>x(2-x) = 1</math>.  Rearranging results in <math>0 = x^2 - 2x + 1</math>, so <math>x = 1</math>.  Finally, by substituting <math>x</math> back in, we get <math>y = 1</math>.  Plugging <math>(1,1)</math> back satisfies the system.
 +
 
 +
== Problems ==
 +
 
 +
=== Introductory ===
 +
 
 +
* [[2002 AMC 8 Problems/Problem 17]]
 +
* [[2007 iTest Problems/Problem 2]]
 +
 
 +
=== Intermediate ===
 +
 
 +
* [[1989 AIME Problems/Problem 8]]
 +
* [[1993 AIME Problems/Problem 3]]
 +
 
 +
== See Also ==
 
* [[Algebra]]
 
* [[Algebra]]
 +
* [[Substitution]]
 +
 +
[[Category:Algebra]]

Revision as of 17:39, 10 April 2020

A system of equations is a set of equations which share the same variables. Below is an example of a system of equations.

\[\left\{\begin{array}{l}a^2 + b^2 = 25\\a+b=7\end{array}\right.\]

Solving Linear Systems

A system of linear equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations.

Gaussian Elimination

Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example:

Problem

Find the ordered pair $(x,y)$ for which

\[\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.\]

Solution

We can eliminate $y$ by adding twice the second equation to the first:

$x - 12y= 2$
$+2($ $3x + 6y = 6)$
$\overline{7x + 0=14}$

Thus $x=2$. We can then plug in for $x$ in either of the equations: \begin{align*} (2)-12y &= 2 \\ y &= 0 \end{align*}

Thus, the solution to the system is $(2,0)$.

Substitution

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution.

Problem

Find the ordered pair $(x,y)$ for which

\[\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.\]

Solution

The first equation can be solved for $x$:

$x = 12y + 2.$

Plugging this into the second equation yields

$3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0.$

Thus $y=0$. Plugging this into either of the equations and solving for $x$ yields $x=2$.

Graphing

The third method for solving a system of linear equations is to graph them in the plane and observe where they intersect. We'll go back to our same example to illustrate this.

Problem

Find the ordered pair $(x,y)$ for which

\[\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.\]

Solution

We graph the two lines as follows:

Twolineintersect.PNG

From the graph, we can see that the solution to the system is $(2,0)$.

Advanced Methods

Matrices can also be used to solve systems of linear equations. In fact, they provide a way to make much broader statements about systems of linear equations.

There is a whole field of mathematics devoted to the study of linear equations called linear algebra.

Convenient Systems

Some systems can be solved by taking advantage of specific forms. Such systems can often seem tough to solve at first, however.

Symmetry

Consider the below system.

\[\left\{\begin{array}{l}a+b+c+d=4\\a+b+c+e=8\\a+b+d+e=12\\a+c+d+e=16\\b+c+d+e=20\end{array}\right.\]

The key here is to take advantage of the symmetry. If we add up all 5 equations we will have a total of 4 of each variable on the LHS. On the RHS we will have $4+8+12+16+20 = 60$. Thus

$4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15.$

So then subtracting the first equation from this leaves $e$ on the LHS and $15-4=11$ on the RHS. Subtracting this equation from the second equation leaves $d$ on the LHS and $15-8=7$ on the RHS. And thus we continue on in this way to find that $(a,b,c,d,e)=(-5,-1,3,7,11).$

Clever Substitution

Consider the below system.

\[\left\{\begin{array}{l}x+y+3xy = 5\\ 2x+2y+4xy=8 \end{array}\right.\]

We can let $x+y=a$ and $b = xy$ to get the two-variable linear system below.

\[\left\{\begin{array}{l}a+3b = 5\\ 2a+4b=8 \end{array}\right.\]

Solving the system results in $b = 1$ and $a = 2$. Substituting that back results in $xy = 1$ and $x + y = 2$. We can do another substitution by letting $y = 2 - x$ and substituting to get $x(2-x) = 1$. Rearranging results in $0 = x^2 - 2x + 1$, so $x = 1$. Finally, by substituting $x$ back in, we get $y = 1$. Plugging $(1,1)$ back satisfies the system.

Problems

Introductory

Intermediate

See Also

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