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Solving systems of equations by addition
We have discussed how you can solve by two linear equations with two unknowns by eliminating an unknown. The method we are discussing here is more suitable for large systems.
Here, the goal is to give the first equation the form #x=a# and to give and the second equation the form #y=b#.
The strategy is to edit the equations; for example multiplication of all terms in the same equation by a number and subtracting one equation from the other.
The addition method for linear equations
A system of two linear equations with unknowns #x# and #y# can be solved as follows.
- Make sure #x# occurs in the first comparison, if this is not the case, then switch the two equations, in such a way that #x# occurs in the first equation.
- Replace the second equation by the difference of this equation with a suitable multiple of the first equation, in such a way that #x# no longer occurs in the second equation.
- Replace first equation by the difference of this equation with a suitable multiple of the second equation in such a way that #y# no longer occurs in the first equation.
- The first equation is now a linear equation with #x# as the only unknown, and the second is a linear equation with #y# as the only unknown. These equations can be solved with the theory of linear equations with one unknown.
The reasoning that this method indeed provides the solution to the original system, can be given in the same manner with the notion of equivalence when discussing the elimination method.
For the system it is assumed that #x# and #y# really occur in the system. If only #x# occurs, then we are dealing with a system of equations with one unknown which has been dealt with earlier, in the examples of the cited theory. For each solution #x=a# of that system, and every real number #b#, #xa\land y=b# is a solution to the system (and these are all solutions).
If only #y# occurs, then the same observations hold, with #x# and #y# interchanged.
If both #x# and #y# do not occur, then each pair #\rv{x,y}# is a solution if all the equations are true (think of #0=0#), and not a single pair is a solution if at least one of the equations is a contradiction (remember #0=1#).
Solving by addition
This method is known as the addition method. After all, you always add one equation to the other.
It may be that after the second step, the second equation becomes true (if #0=0#) or a contraction (if #0=1#) because not only #x# as well as #y# disappears. In that case, the solution is a line given by the first equation.
Solve the following system of linear equations with unknowns #x# and #y#.
\[\lineqs{6x-y -14 &=& 0\cr 2x+3y-18 &=& 0\cr}\]
1. The unknown #x# is already in the upper equation. So we do not need to switch the equations.
2. We subtract #\frac{1}{3}# times the first equation fromt he second one. Because of this the term with #x# in it disappears and the equation #\frac{10}{3}y = \frac{40}{3}# emerges . We hold on to the two equations (but in the mean time get rid of the denominator in the second equation by multiplying #3#):
\[\lineqs{6x-y -14 &=& 0\cr 10y-40 &=& 0\cr}\]
3. We add #\frac{1}{10}# times in the second equation to the first (so the term with #y# disappears) and find the system
\[\lineqs{6x-18 &=& 0\cr 10y -40&=& 0\cr}\]
4. We divide the first equation by #6# and the second by #10# to get the system
\[\lineqs{x-3&=&0\cr y-4 &=& 0\cr}\]
With this, the system is solved: #x=3\land y= 4#.
There are many ways to get to this solution. We will describe one.
- To make sure that the unknown #x# is present in the first equation, we switch the two equations if this was not the case in the original system: # -6\cdot x+7\cdot y-1=0\land x-y=0 #.
- Next we get rid of the term with #x# from the second equation by multiplying the first equation with #\frac{1}{-6}# and subtracting from the second: # -6\cdot x+7\cdot y-1=0\land {{y}\over{6}}=0 #.
- By dividing (left and right hand side) in the second equation by #{{1}\over{6}}# we find #y=1#. We now have the system # -6\cdot x+7\cdot y-1=0\land y=1 #.
- If we enter the solution of #y# (the second equation) in the first equation (or stated different: we subtract #7# times the second equation from the first), then we find the system: # 6-6\cdot x=0\land y=1 #.
- The first equation can be solved as discussed in in Solving by reduction of a linear equation with one unknown. The result is #x= 1\land y = 1#.
There are many variations. For example, we can first set the coefficient of #x# in the first equation equal to 1 by multiplying by #\frac{1}{6}#. If you experiment a lot with the variations, you will see that the difference between this addition method and the elimination method is not as big as it seems at first sight.
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