Systems of linear equations: Two equations with two unknowns
General solution system of linear equations
When solving systems, we have only looked at the case where the two lines have one intersection point. We will now also look at the other options.
A system of two linear equations with two unknowns can be seen as two lines in the flat plane. We distinguish three cases: |
Examples |
1) There is exactly one solution, which is the intersection of the two lines. We call the system consistent. |
#\lineqs{2 \cdot x- y+4=0 \cr x + y -6=0\cr} # |
2) There is no solution, which means that the lines run parallel. We call the system inconsistent. |
#\lineqs{x+ y+2 =0\cr x + y +5=0\cr} # |
3) There are infinite solutions; the lines are the same. We call the system dependent. |
#\lineqs{x+ y-2=0 \cr 2x + 2y -4=0\cr}# |
Recognizing the system
Substitution method
Using the substitution method, we write the first equation in the form #x=\ldots#. When we substitute this in the second equation, we find for
- a consistent system an equation with unknown #y#;
- an inconsistent system an equation without unknown, which is not true.
- a dependent system an equation without unknown, which is true.
It works the same way when we write the second equation as #y=\ldots# and substitute in the first equation, only we find for a consistent system a linear equation with unknown #x#.
Elimination method
The elimination method works in a similar way. After adding both equations (possibly after multiplication of one or both equations by a number), we find for
- a consistent system an equation with one unknown #x# or #y#;
- an inconsistent system an equation without an unknown, which is not true.
- a dependent system an equation without unknown, which is true.
Examples of elimination method
\[\begin{array}{rcl} 2 \cdot x -y+4&=&0 \\ x + y -6&=&0\\ \hline 3 \cdot x-2&=&0\end{array}+\]
This equation has one unknown #x#, so the system is consistent.
\[\begin{array}{rcl} x +y+2&=&0 \\ x + y +5&=&0\\ \hline -3&=&0\end{array}-\]
This equation is not true, so the system is inconsistent.
\[\begin{array}{rcl} x +y-2&=&0 \\ x + y -2&=&0\\ \hline 0&=&0\end{array}-\]
This equation is true, so the system is dependent.
\[\lineqs{y&=&{{7}\over{6}}\cdot x-7\cr
y&=&-4\cdot x+2\cr }\]
- If you think the system is consistent and there is one solution, write #\lineqs{x&=&a\cr y&=&b\cr }# or #x=a \land y=b# with the correct values for #a# and #b#.
- If you think the system is inconsistent and there is no solution, write #none#.
- If you think the system is dependent and there are infinitely many solutions, write #all#.
#\lineqs{x&=&{{54}\over{31}}\cr y&=&-{{154}\over{31}}\cr }#
We can solve the given system using either the substitution method or the elimination method.
Substitution method
The given system of equations is:
\[\lineqs{y&=&{{7}\over{6}}\cdot x-7\cr
y&=&-4\cdot x+2\cr }\]
In this case, both equations are in the form #y=\ldots#. We can therefore start at step 2 and swap the roles of #x# and #y#.
Step 2: We substitute the second equation in the form #y=\ldots# in the first equation. This gives:
\[\lineqs{-4\cdot x+2&=&{{7}\over{6}}\cdot x-7\cr
y&=&-4\cdot x+2\cr }\]
Step 3: We can now solve the first equation from the system in step 2 for #x# by reduction. This goes as follows:
\[\begin{array}{rcllll}
&\lineqs{-4\cdot x+2&=&{{7}\over{6}}\cdot x-7\cr y&=&-4\cdot x+2\cr }&\\ &&\phantom{x}\blue{\text{the system we need to solve}} \\
&\lineqs{-{{31}\over{6}}\cdot x+2&=&-7\cr y&=&-4\cdot x+2\cr }&\\ &&\phantom{x}\blue{\text{both sides of the first equation minus }{{7}\over{6}}\cdot x} \\
&\lineqs{-{{31}\over{6}}\cdot x&=&-9\cr y&=&-4\cdot x+2\cr }&\\ &&\phantom{x}\blue{\text{both sides of the first equation minus }2} \\
&\lineqs{x&=&{{54}\over{31}} \cr y&=&-4\cdot x+2\cr }&\\ &&\phantom{x}\blue{\text{both sides of the first equation divided by }-{{31}\over{6}}}
\end{array}\]
Step 4: We now substitute the obtained value of #x# in the second equation to determine #y#. This goes as follows:
\[\begin{array}{rcllll}
&\lineqs{x&=&{{54}\over{31}} \cr y&=&-4\cdot {{{54}\over{31}}} +2\cr }&\\ &&\phantom{x}\blue{x={{54}\over{31}} \text{ substituted in the second equation}} \\
&\lineqs{x&=&{{54}\over{31}} \cr y&=&-{{154}\over{31}}\cr }&\\ &&\phantom{x}\blue{\text{calculated}}
\end{array}\]
This system is therefore consistent and has one solution: \[\lineqs{x&=&{{54}\over{31}}\cr y&=&-{{154}\over{31}}\cr }\]
Elimination method
The coefficients of the terms in #y# are equal. So we can subtract the two equations from each other. We therefore start at step 2.
Step 2: \[\begin{array}{rcl}y&=&{{7}\over{6}}\cdot x -7\\y&=&-4\cdot x+2 \\ \hline 0&=&{{31}\over{6}}\cdot x-9 \end{array} -\]
Step 3: We solve this equation for #x#. This goes as follows:
\[\begin{array}{rcl}
0&=&{{31}\over{6}}\cdot x-9\\ &&\phantom{x}\blue{\text{the equation we need to solve}} \\
9&=&{{31}\over{6}}\cdot x\\ &&\phantom{x}\blue{\text{both sides minus }-9} \\
{{54}\over{31}} &=&x \\&&\phantom{x}\blue{\text{both sides divided by }{{31}\over{6}}} \\
\end{array}\]
Step 4: We now substitute #x={{54}\over{31}}# in the original second equation to find #y#. This goes as follows:
\[\begin{array}{rcllll}
y&=&-4\cdot {{{54}\over{31}}} +2&\\ &&\phantom{x}\blue{x={{54}\over{31}} \text{ substituted in the second equation}} \\
y&=&-{{154}\over{31}}&\\ &&\phantom{x}\blue{\text{calculated}}
\end{array}\]
This system is therefore consistent and has one solution: \[\lineqs{x&=&{{54}\over{31}}\cr y&=&-{{154}\over{31}}\cr }\]
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