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This section of the Appendix combines ideas from Section AppLinearEqIneq and AppLines so that we can start to solve systems of linear equations.
Before we get ahead of ourselves, let’s review a few definitions.
A linear equation in two variables is an equation of the form \(a_1 x + a_2 y = c\) where \(a_1\), \(a_2\) and \(c\) are real numbers and at least one of \(a_1\) and \(a_2\) is
nonzero.
For reasons which will become clear when you study Chapter SystemsofEquationsandMatrices, we are using subscripts in Definition lineareqntwovariables to indicate different, but
fixed, real numbers and those subscripts have no mathematical meaning beyond that. For example, \(3x - \frac {y}{2} = 0.1\) is a linear equation in two
variables with \(a_1 = 3\), \(a_2 = -\frac {1}{2}\) and \(c = 0.1\). We can also consider \(x = 5\) to be a linear equation in two variables (Critics may argue that \(x=5\) is clearly an
equation in one variable. It can also be considered an equation in \(117\) variables with the coefficients of \(116\) variables
set to \(0\). As with many conventions in Mathematics, the context will clarify the situation.) by identifying \(a_1 = 1\), \(a_2 = 0\), and
\(c = 5\).
If \(a_1\) and \(a_2\) are both \(0\), then depending on \(c\), we get either an equation which is always true, called an identity, or an
equation which is never true, called a contradiction. (If \(c = 0\), then we get \(0 = 0\), which is always true. If \(c \neq 0\), then we’d
have \(0 \neq 0\), which is never true.) Even though identities and contradictions have a large role to play throughout
Chapter SystemsofEquationsandMatrices, we do not consider them linear equations. The key to identifying linear equations is to note that the
variables involved are to the first power and that the coefficients of the variables are numbers. Some examples of
equations which are non-linear are \(x^2 + y = 1\), \(xy = 5\) and \(e^{2x} + \ln (y) = 1\). The reader should consider why these do not satisfy Definition
lineareqntwovariables.
We know from our work is Sections AppLines that the graphs of linear equations are lines. If we couple two or more linear equations
together, in effect to find the points of intersection of two or more lines, we obtain a system of linear equations in two
variables. Our first example explores the basic techniques for solving these systems. Remember - if we are looking for points
in the plane, then both the \(x\) and \(y\) values are important. This is a key distinction between solving one equation and solving a
system of equations.
Solve the following systems of equations. Check your answer algebraically and graphically. (Said another way, make sure
both \(x\) and \(y\) are correct!)
\(\left \{ \begin{array}{rcr} 2x - y & = & 1 \\ y & = & 3 \\ \end{array} \right .\)
\(\left \{ \begin{array}{rcr} x - y & = & 0 \\ x + y & = & 2 \\ -2x + y & = & -2 \end{array} \right .\)
Our first system is nearly solved for us. The second equation tells us that \(y=3\). To find the corresponding value of
\(x\), we substitute this value for \(y\) into the the first equation to obtain \(2x - 3 = 1\), so that \(x = 2\). Our solution to the system is \((2,3)\). To
check this algebraically, we substitute \(x=2\) and \(y=3\) into each equation and see that they are satisfied. We see \(2(2) - 3 = 1\), and \(3=3\),
as required. To check our answer graphically, we graph the lines \(2x-y = 1\) and \(y=3\) and verify that they intersect at \((2,3)\).
To solve the second system, we use the addition method to eliminate the variable \(x\). We take the two equations as
given and ‘add equals to equals’ to obtain
This gives us \(y = 1\). We now substitute \(y=1\) into either of the two equations, say \(-3x-y = 5\), to
get \(-3x-1 = 5\) so that \(x = -2\). Our solution is \((-2,1)\). Substituting \(x=-2\) and \(y=1\) into the first equation gives \(3(-2) + 4(1) = -2\), which is true, and, likewise, when we
check \((-2, 1)\) in the second equation, we get \(-3(-2) - 1 = 5\), which is also true. Geometrically, the lines \(3x+4y = -2\) and \(-3x-y=5\) intersect at
\((-2,1)\).
The equations in the third system are more approachable if we clear denominators. We multiply both sides of the first
equation by \(15\) and both sides of the second equation by \(18\) to obtain the kinder, gentler system
Adding these
two equations directly fails to eliminate either of the variables, but we note that if we multiply the first
equation by \(4\) and the second by \(-5\), we will be in a position to eliminate the \(x\) term
From this we get \(y = -\frac {1}{2}\). We can
temporarily avoid too much unpleasantness by choosing to substitute \(y = -\frac {1}{2}\) into one of the equivalent equations we
found by clearing denominators, say into \(5x - 12y = 21\). We get \(5x + 6 = 21\) which gives \(x=3\). Our answer is \(\left (3, -\frac {1}{2}\right )\). At this point, we have
no choice \(-\) in order to check an answer algebraically, we must see if the answer satisfies both of the
original equations, so we substitute \(x = 3\) and \(y = -\frac {1}{2}\) into both \(\frac {x}{3} -\frac {4y}{5} = \frac {7}{5}\) and \(\frac {2x}{9} + \frac {y}{3} = \frac {1}{2}\). We leave it to the reader to verify that the
solution is correct. Graphing both of the lines involved with considerable care yields an intersection point of
\(\left (3, -\frac {1}{2}\right )\).
An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly integer coefficients
belie something more sinister? We note that if we multiply both sides of the first equation by \(3\) and both
sides of the second equation by \(-2\), we are ready to eliminate the \(x\)
We eliminated not only the \(x\), but the \(y\) as
well and we are left with the identity \(0=0\). This means that these two different linear equations are, in fact,
equivalent. In other words, if an ordered pair \((x,y)\) satisfies the equation \(2x-4y = 6\), it automatically satisfies the equation
\(3x-6y = 9\).
This system has infinitely many solutions and one way to describe the solution set to this system is to use the roster
method (See Section AppSetTheory for a review of this.) and write \(\{(x,y) \, | \, 2x-4y = 6\}\). While this is correct (and corresponds exactly to what’s
happening graphically, as we shall see shortly), we take this opportunity to introduce the notion of a parametric
solution to a system.
Our first step is to solve \(2x-4y = 6\) for one of the variables, say \(y = \frac {1}{2} x - \frac {3}{2}\). For each value of \(x\), the formula \(y = \frac {1}{2} x - \frac {3}{2}\) determines the corresponding
\(y\)-value of a solution. Since we have no restriction on \(x\), it is called a free variable. We let \(x=t\), a so-called ‘parameter’,
and get \(y = \frac {1}{2} t - \frac {3}{2}\). Our set of solutions can then be described as \(\left \{ \left (t, \frac {1}{2} t - \frac {3}{2}\right ) \, | \, -\infty < t < \infty \right \}\). (Note that we could have just as easily
chosen to solve \(2x-4y = 6\) for \(x\) to obtain \(x = 2y + 3\). Letting \(y\) be the parameter \(t\), we have that for any value of \(t\), \(x = 2t+3\), which gives \(\{(2t+3, t) \, | \, - \infty < t < \infty \}\).
There is no one correct way to parameterize the solution set, which is why it is always best to check your
answer.)
For specific values of \(t\), we can generate solutions. For example, \(t=0\) gives us the solution \(\left (0,-\frac {3}{2}\right )\); \(t = 117\) gives us \((117,57)\), and while we can
check that each of these particular solutions satisfy both equations, the question is how do we check our general
answer algebraically? Same as always.
We claim that for any real number \(t\), the pair \(\left (t, \frac {1}{2} t - \frac {3}{2}\right )\) satisfies both equations. Substituting \(x = t\) and \(y = \frac {1}{2} t - \frac {3}{2}\) into \(2x - 4y = 6\) gives \(2t - 4\left (\frac {1}{2} t - \frac {3}{2}\right ) = 6\). Simplifying, we get \(2t - 2t + 6 = 6\),
which is always true. Similarly, when we make these substitutions in the equation \(3x-6y = 9\), we get \(3t - 6\left (\frac {1}{2} t - \frac {3}{2}\right ) = 9\) which reduces to \(3t - 3t + 9 = 9\), so it
checks out, too.
Geometrically, \(2x-4y = 6\) and \(3x-6y=9\) are the same line, which means that they intersect at every point on their graphs. The reader is
encouraged to think about how our parametric solution says exactly that.
Multiplying both sides of the first equation by \(2\) and the both sides of the second equation by \(-3\), we set the stage to
eliminate \(x\)
As in the previous example, both \(x\) and \(y\) dropped out of the equation, but we are left with an irrevocable
contradiction, \(0 = -18\). This tells us that it is impossible to find a pair \((x,y)\) which satisfies both equations; in other
words, the system has no solution. Graphically, the lines \(6x + 3y =9\) and \(4x + 2y = 12\) are distinct and parallel, so they do not
intersect.
We can begin to solve our last system by adding the first two equations
\[ \begin{array}{lrcr} & x - y & = & 0 \\ + & (x + y & = & 2 ) \\ \hline & 2x & = & 2 \end{array}\]
which gives \(x = 1\). Substituting this into the first equation gives \(1 - y = 0\) so that \(y = 1\). We seem to have determined a solution to our
system, \((1,1)\). While this checks in the first two equations, when we substitute \(x=1\) and \(y=1\) into the third equation, we get \(-2(1) + (1) = -2\) which
simplifies to the contradiction \(-1 = -2\). Graphing the lines \(x-y=0\), \(x+y = 2\), and \(-2x+y=-2\), we see that the first two lines do, in fact, intersect at \((1,1)\),
however, all three lines never intersect at the same point simultaneously, which is what is required if a solution to the
system is to be found.
A few remarks about Example reviewsubelim are in order. Notice that some of the systems of linear equations had solutions while others did
not. Those which have solutions are called consistent, those with no solution are called inconsistent. We also distinguish
between the two different types of behavior among consistent systems. Those which admit free variables are called
dependent and those with no free variables are called independent. (In the case of systems of linear equations,
regardless of the number of equations or variables, consistent independent systems have exactly one solution.
The reader is encouraged to think about why this is the case for linear equations in two variables. Hint: think
geometrically.)
Using this new vocabulary, we classify numbers 1, 2 and 3 in Example reviewsubelim as consistent independent systems, number 4 is
consistent dependent, and numbers 5 and 6 are inconsistent. (The adjectives ‘dependent’ and ‘independent’
apply only to consistent systems – they describe the type of solutions. Is there a free variable (dependent) or
not (independent)?) The system in 6 above is called overdetermined, since we have more equations than
variables. (If we think if each variable being an unknown quantity, then ostensibly, to recover two unknown
quantities, we need two pieces of information - i.e., two equations. Having more than two equations suggests we
have more information than necessary to determine the values of the unknowns. While this is not necessarily
the case, it does explain the choice of terminology ‘overdetermined’.) Not surprisingly, a system with more
variables than equations is called underdetermined. While the system in number 6 above is overdetermined and
inconsistent, there exist overdetermined consistent systems (both dependent and independent) and we leave it to the
reader to think about what is happening algebraically and geometrically in these cases. Likewise, there are both
consistent and inconsistent underdetermined systems, (We need more than two variables to give an example
of the latter.) but a consistent underdetermined system of linear equations is necessarily dependent. (Again, experience with systems with more variables helps to see this here, as does a solid course in Linear
Algebra.)
We end this section with a story problem. It is an example of a classic “mixture” problem and should be familiar to most
readers. The basic goal here is to create two equations: one which represents
\[\text {value of stuff} + \text {value of other stuff} = \text {value of total stuff.}\]
The Dude-Bros want to create a highly caffeinated, yet still drinkable, fruit punch for their annual “Disturb the Neighbors BBQ
and Dance Competition”. They plan to add Sasquatch SweatTM Energy Drink, which has 100 mg. of caffeine per fluid ounce,
to Frooty Giggle DelightTM, which has only 3 mg. of caffeine per fluid ounce. How much of each component is required to
make 5 gallons (Warning: unit conversion ahead!) of a fruit punch that has 80 mg. of caffeine per fluid
ounce.
Solution. Let \(S\) stand for the number of fluid ounces of Sasquatch SweatTM Energy Drink and let \(F\) be the number of fluid
ounces of Frooty Giggle DelightTM that will be added together. The goal is to make 5 gallons and there are 128 fluid ounces
per gallon so the first equation is
\[S + F = 640.\]
That equation describes “stuff + other stuff = total stuff” measured in fluid ounces. Now we
need to consider the value of the stuff - in this case we need to see how much caffeine is being contributed by each
component. Each fluid ounce of Sasquatch SweatTM contains 100 mg. of caffeine so \(S\) fluid ounces would contain \(100S\) mg./ of
caffeine.
Similarly, the \(F\) fluid ounces of Frooty Giggle DelightTM add \(3F\) mg. of caffeine to the total mixture. Thus when we go to express
“value of stuff + value of other stuff = value of total stuff” we need to figure out how much caffeine is supposed
to be in the end product. Well, the goal was 5 gallons of punch that had 80 mg. of caffeine per fluid ounce
so the Dude-Bros need to end up with \(5*128*80 = 51200\) mg. of caffeine when they’re done. Hence the second is equation is
\[100S + 3F = 51200.\]
By turning the first equation into \(F = 640 - S\) and substituting that into the second equation we get
\[100S + 3(640 - S) = 51200\]
which yields \(S = \frac {49280}{97} \approx 508.04\) fluid ounces.
Back-substituting this value of \(S\) into the first equation gives us \(F = \frac {12800}{97} \approx 131.96\) fluid ounces.
The reader should take the time to verify that \(S = \frac {49280}{97}\) and \(F = \frac {12800}{97}\) do indeed satisfy both equations and thus are the solution to
the problem. Those are fairly unattractive numbers so we end this example by discussing a way to verify an
approximate answer which is reasonable without having to fight with fractions. Round \(S\) down to 508 and round \(F\)
up to 132. Clearly \(508 + 132 = 640\) so the first equation is still satisfied. Notice that \(100*508 + 3*132 = 51196\) which is really close to 51200. Thus the
second equation is nearly satisfied which means the values \(S = 508\) and \(F = 132\), while not precise, are reasonable. (Just be careful here - sometimes “close enough for the Dude-Bros” is not good enough for your Professor!)
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PreCalculus