Systems of Linear Equations: Two Linear Equations
Solving systems of equations by substitution
The most obvious method of solving two equations with two unknowns is based upon substitution.
Substitution method for linear equations
A system of two linear equations with unknowns #x# and #y# can be solved as follows.
1. Solve #y# with #x# as a parameter from one of the two equations. This gives an expression of #y# in terms of #x#.
2. Substitute the expression anywhere #y# occurs in the other equation. Then a linear equation with only unknown #x# emerges.
3. Solve that one with help of the chapter Linear equations with one unknown. As a result, #x# is defined.
4. Finally define #y# by substituting the found value of #x# in the equation of the form #y=\ldots#
The method is based on the fact that after each step a system of equations arises that is equivalent to the system from the previous step. We have introduced the notion of equivalence for equations and not for systems of equations, but the definition is the same: two systems are called equivalent if they have the same solution.
Since equivalence for equations means that they have the same solution, the answer is also the solution to all the previous systems, and in particular of the original system.
Three types of systems of two linear equations with two unknowns
Usually a system of two linear equations with two unknowns has exactly one solution. In this case the system is called regular.
Sometimes there is more than one solution: the two equations may be dependent. That means that they have the same line as solution. This occurs when the two equations are multiples of each other. This case occurs after step 2: if we eliminate #y# from the equation, it is possible that #x# disappears. If the equation then becomes \(0=0\), she does not impose a limit the solutions we can remove that equation. There remains one equation with two unknowns whose solutions form a line.
Sometimes there is no solution: the equations may conflict. This means that no single solution to one equation is also a the solution to the other. This case occurs after step 2: if we eliminate #y# from the equation, it is possible that #x# too disappears (as in the previous case). But this time, the equation takes the form #0=c# for a number \(c\) unequal to zero. This equation is never fulfilled: This is a contradiction. Hence the answer is that no point \(\rv{x,y}\) satisfies the system.
There are thus two exceptions to regular systems (dependency and contradiction), which correspond with the coinciding or being parallel of the lines represented by the two equations. More on this in the exercises and in the theory of equations and lines.
Hence there are three possible types of solutions to a system of two linear equations:
- a single point #\rv{x,y}# (regular)
- a straight line (dependent)
- none (conflict)
\[ \lineqs{ 8\cdot x+6\cdot y+10 &=& 0 \cr -x-y+6 &=& 0 \cr} \]
Give the answer in the form #x=a\land y=b]# for suitable values of #a# and #b#. You can enter the system below as a starting point for reducing these to a solution.
\[ 8\cdot x+6\cdot y+10=0 \land -x-y+6=0 \]
We use the substitution method in order to find this solution.
\[ \begin{array}{rcl} 8\cdot x+6\cdot y+10=0 & \land &-x-y+6=0 \\&&\phantom{xxx} \color{blue}{\text{the original equations}} \\
x=-{{3\cdot y}\over{4}}-{{5}\over{4}}& \land &-x-y+6=0 \\&&\phantom{xxx}\color{blue}{x \text{ expressed in terms of }y\text{ in the first equation}}\\
x=-{{3\cdot y}\over{4}}-{{5}\over{4}} & \land & {{29}\over{4}}-{{y }\over{4}}=0 \\&&\phantom{xxx}\color{blue}{x=-{{3\cdot y}\over{4}}-{{5}\over{4}}\text{ substituted in the second equation}}\\
&& \phantom{xxx}\color{blue}{\text{to obtain a linear equation in }y}\\
x = -{{3\cdot y}\over{4}}-{{5}\over{4}} &\land & y = 29\\&&\phantom{xxx}\color{blue}{\text{solved the second equation}}\\
x = -{{3\cdot (29)}\over{4}}-{{5}\over{4}} &\land & y = 29\\&&\phantom{xxx}\color{blue}{\text{substituted }y=29\text{ in the first equation}}\\
x = -23 &\land & y = 29\\&&\phantom{xxx}\color{blue}{\text{solution simplified}}\\ \end{array}
\]Hence, the answer is #x= -23\land y = 29#.
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