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Up until now, when we concerned ourselves with solving different types of equations there was only one equation to
solve at a time. Given an equation \(f(x) = g(x)\), we could check our solutions geometrically by finding where the graphs
of \(y=f(x)\) and \(y=g(x)\) intersect. The \(x\)-coordinates of these intersection points correspond to the solutions to the equation
\(f(x) = g(x)\), and the \(y\)-coordinates were largely ignored. If we modify the problem and ask for the intersection points of
the graphs of \(y=f(x)\) and \(y=g(x)\), where both the solution to \(x\) and \(y\) are of interest, we have what is known as a system of
equations, written as
\[ \left \{ \begin{array}{rcl} y & = & f(x) \\ y & = & g(x) \\ \end{array} \right .\]
The ‘curly bracket’ notation means we are to find all pairs of points \((x,y)\) which satisfy both
equations.
We assume the reader has some experience with systems of equations from high school algebra - specifically systems of
linear equations comprised of two equations and two unknowns. We encourage the reader to read through Section AppLinearSystems before
proceeding if for no other reason than to refresh themselves on the basic mechanics and vocabulary involved. In order to
move this section beyond a review of high school algebra, we define what is meant by a linear equation in \(n\)
variables.
A linear equation in n variables, \(x_1\), \(x_2\), …, \(x_{n}\), is an equation of the form:
\[a_1 x_1 + a_2 x_2 + \ldots + a_{n} x_{n} = c,\]
where \(a_1\), \(a_2\), …\(a_{n}\) and \(c\) are real numbers and at least one of \(a_1\), \(a_2\), …, \(a_{n}\) is nonzero.
Instead of using more familiar variables like \(x\), \(y\), and even \(z\) and/or \(w\) in Definition lineareqnnvariables, we use subscripts to distinguish the different
variables. We have no idea how many variables may be involved, so we use numbers to distinguish them instead of letters.
(There is an endless supply of distinct numbers.)
As an example, the linear equation \(3x_1 - x_2 = 4\) represents the same relationship between the variables \(x_1\) and \(x_2\) as the equation \(3x-y=4\) does
between the variables \(x\) and \(y\). And, just as we cannot combine the terms in the expression \(3x-y\), we cannot combine the terms in
the expression \(3x_1 - x_2\).
Coupling more than one linear equation in \(n\) variables results in a system of linear equations in n variables. When solving
these systems, it becomes increasingly important to keep track of what operations are performed to which equations and to
develop a strategy based on the kind of manipulations (substitution and elimination) taught in high school. To this end, we first
remind ourselves of the maneuvers which can be applied to a system of linear equations that result in an equivalent system.
(That is, a system with the same solution set.)
Given a system of equations, the following moves will result in an equivalent system:
Interchange the position of any two equations.
Replace an equation with a nonzero multiple of itself. (That is, an equation which results from multiplying
both sides of the equation by the same nonzero number.)
Replace an equation with itself plus a nonzero multiple of another equation.
The first move, while it obviously admits an equivalent system, seems silly to state, but our perception will change as we
consider more equations and more variables in this, and later sections.
We have \(z = -1\), so we substitute this into the second equation \(y - \frac {1}{2} (-1) = 4\) to obtain \(y = \frac {7}{2}\). Substituting \(y = \frac {7}{2}\) and \(z=-1\) into the first equation we get \(x - \frac {1}{3}\left (\frac {7}{2}\right ) + \frac {1}{2}(-1) = 1\). This
gives \(x = \frac {8}{3}\). The reader can verify that these values of \(x\), \(y\) and \(z\) satisfy all three original equations.
It is tempting for us to write the solution to this system by extending the usual \((x,y)\) notation to \((x,y,z)\) and list our solution as \(\left (\frac {8}{3},\frac {7}{2},-1\right )\). The
question quickly becomes what does an ‘ordered triple’ like \(\left (\frac {8}{3},\frac {7}{2},-1\right )\) represent? Just as ordered pairs are used to locate points on the
two-dimensional plane, ordered triples can be used to locate points in space. (You were asked to think about this in
Exercise orderedtripleexercise in Section AppCartesianPlane.)
Moreover, just as equations involving the variables \(x\) and \(y\) describe graphs of one-dimensional lines and curves in the
two-dimensional plane, equations involving variables \(x\), \(y\), and \(z\) describe objects called surfaces in three-dimensional space.
Each of the equations in the above system can be visualized as a plane situated in three-space. Geometrically, the system is
trying to find the intersection, or common point, of all three planes. If you imagine three sheets of notebook paper each
representing a portion of these planes, you will start to see the complexities involved in how three such planes can
intersect.
Below is a sketch of the three planes. It turns out that any two of these planes intersect in a line, (These
lines are described by ‘parametric solutions’ to the systems formed by taking any two of these equations by
themselves. (Again, see Section AppLinearSystems.) We’ll see an example of this sort of solution in this section shortly.) so our
intersection point is where all three of these lines meet. RIGHT-CLICK and DRAG to rotate the image for a better
view.
Since the geometry for equations involving more than two variables is complicated, we will focus our efforts on the algebra.
Returning to the system
we note the reason it was so easy to solve is because of its structure. The third equation is solved for \(z\) and the second
equation involves only \(y\) and \(z\). Since the coefficient of \(y\) is \(1\), it makes it easy to solve for \(y\) using our known value for \(z\). Lastly, the
coefficient of \(x\) in the first equation is \(1\) making it easy to substitute the known values of \(y\) and \(z\) and then solve for
\(x\).
We formalize this pattern below for the most general systems of linear equations. Again, we use subscripted variables to
describe the general case. The variable with the smallest subscript in a given equation is typically called the leading variable
of that equation.
A system of linear equations with variables \(x_1\), \(x_2\), …\(x_{n}\) is said to be in triangular form provided all of the following conditions
hold:
The subscripts of the variables in each equation are always increasing from left to right.
The leading variable in each equation has coefficient \(1\).
The subscript on the leading variable in a given equation is greater than the subscript on the leading variable
in the equation above it.
Any equation without variables (necessarily an identity or contradiction) cannot be placed above an
equation with variables.
In our previous system, if we make the obvious choices \(x = x_1\), \(y = x_2\), and \(z = x_3\), we see that the system is in triangular form. (If letters are
used instead of subscripted variables, Definition systemtriangularform can be suitably modified using alphabetical order of the variables instead of
numerical order on the subscripts of the variables.) An example of a more complicated system in triangular form
is
Our goal henceforth will be to transform a given system of linear equations into triangular form using the moves in Theorem
equationmoves.
Use Theorem equationmoves to put the following systems into triangular form and then solve the system if possible. Classify each system
as consistent independent, consistent dependent, or inconsistent. (See Section AppLinearSystems for a review of these
terms.)
Solution. For definitiveness, we label the topmost equation in each system \(E1\), the equation beneath that \(E2\), and so
forth.
We put the system in triangular form using an algorithm known as Gaussian Elimination. Starting with \(x\), we
transform the system so that conditions 2 and 3 in Definition systemtriangularform are satisfied. Then we move on to the next
variable, in this case \(y\), and repeat.
Since the variables in all of the equations have a consistent ordering from left to right, our first move is to get
an \(x\) in \(E1\)’s spot with a coefficient of \(1\). While there are many ways to do this, the easiest is to apply the first move
listed in Theorem equationmoves and interchange \(E1\) and \(E3\).
To satisfy Definition systemtriangularform, we need to eliminate the \(x\)’s from \(E2\) and \(E3\). We accomplish this by replacing each of them with
a sum of themselves and a multiple of \(E1\). To eliminate the \(x\) from \(E2\), we need to multiply \(E1\) by \(-2\) then add; to eliminate
the \(x\) from \(E3\), we need to multiply \(E1\) by \(-3\) then add. Applying the third move listed in Theorem equationmoves twice, we get
Now we enforce the conditions stated in Definition systemtriangularform for the variable \(y\). To that end we need to get the coefficient
of \(y\) in \(E2\) equal to \(1\). We apply the second move listed in Theorem equationmoves and replace \(E2\) with itself times \(-\frac {1}{2}\).
Finally, we apply the second move from Theorem equationmoves one last time and multiply \(E3\) by \(-1\) to satisfy the conditions of
Definition systemtriangularform for the variable \(z\).
Substituting \(z=6\) into \(E2\) gives \(y - 3 = -3\) so that \(y = 0\). With \(y=0\) and \(z=6\), \(E1\) becomes \(x - 0 + 6 = 5\), or \(x = -1\). Hence, our solution is \((-1,0,6)\). We leave it to the reader
to check that substituting the respective values for \(x\), \(y\), and \(z\) into the original system results in three identities.
Since there is a solution to the system, the system is classified as consistent. Since there are no free variables, (Again, see Section AppLinearSystems for a review of this concept, if needed.) the system is classified as independent.
Proceeding as above, our first step is to get an equation with \(x\) in the \(E1\) position with \(1\) as its coefficient. Since there
is no easy fix, we multiply \(E1\) by \(\frac {1}{2}\).
The last equation, \(0=6\), is a contradiction so the system has no solution. According to Theorem equationmoves, since this system
has no solutions, neither does the original, thus we have an inconsistent system.
For our last system, we begin by multiplying \(E1\) by \(\frac {1}{3}\) to get a coefficient of \(1\) on \(x_1\).
Equation \(E3\) reduces to \(0=0\),which is always true. Since we have no equations with \(x_3\) or \(x_4\) as leading variables, they are
both ‘free’ variables so we have a consistent dependent system.
We ‘parametrize’ the solution set by letting \(x_3 = s\) and \(x_4 = t\). From \(E2\), we get \(x_2 = 3s + 2t\). Substituting this and \(x_4 = t\) into \(E1\), we have \(x_1 + \frac {1}{3}\left ( 3s+2t \right ) + \frac {1}{3}t = 2\) which
gives \(x_1 = 2 - s - t\). Our solution is the set \(\{ (2-s-t,2s+3t,s,t) \, | \, -\infty < s, t < \infty \}\). (Here, any choice of \(s\) and \(t\) determines a point in \(4\)-dimensional space. Yeah,
we have trouble visualizing that, too.) We leave it to the reader to verify that the substitutions \(x_{1} = 2-s-t\), \(x_{2} = 3s+2t\), \(x_{3} = s\) and \(x_{4} = t\) satisfy
the equations in the original system regardless of the choices made for the parameters \(s\) and \(t\). □
Like all algorithms, Gaussian Elimination has the advantage of always producing what we need, but it can also be inefficient at
times. For example, when solving the second system in Example GaussEqnEx, it is clear after we eliminated the \(x\)’s in the second step to
get the system
that equations \(E2\) and \(E3\), taken together, produce a contradiction. (We have identical left hand sides and different right hand
sides.) However, the algorithm takes an additional two steps to reach this conclusion.
We also note that substitution in Gaussian Elimination is delayed until all the elimination is done, whence the name
back-substitution. This may also be inefficient in many cases.
Lastly, we note that the last system in Example GaussEqnEx is underdetermined, (Recall this means we have fewer equations than
unknowns.) and as it is consistent, we necessarily have free variables in our answer. We close this section with a standard
‘mixture’ type application of systems of linear equations which features an application of a consistent dependent
system.
Lucas needs to create a \(500\) milliliters (mL) of a \(40 \%\) acid solution. He has stock solutions of \(30 \%\) and \(90 \%\) acid as well as all of the distilled
water he wants. Set-up and solve a system of linear equations which determines all of the possible combinations of the stock
solutions and water which would produce the required solution.
Solution. We are after three unknowns, the amount (in mL) of the \(30 \%\) stock solution (which we’ll call \(x\)), the amount (in mL) of the \(90 \%\)
stock solution (which we’ll call \(y\)) and the amount (in mL) of water (which we’ll call \(w\)). We now need to determine some
relationships between these variables.
Our goal is to produce \(500\) milliliters of a \(40 \%\) acid solution. This product has two defining characteristics. First, it must be \(500\) mL;
second, it must be \(40 \%\) acid. We take each of these qualities in turn.
First, the total volume of \(500\) mL must be the sum of the volumes of the two stock solutions and the water:
\[ \mbox {amount of $30 \%$ stock solution} + \mbox {amount of $90 \%$ stock solution} + \mbox {amount of water} = 500 \, \mbox {mL}\]
Using our defined
variables, this reduces to \(x+y+w = 500\).
Next, we need to make sure the final solution is \(40 \%\) acid. Since water contains no acid, the acid will come from the stock
solutions only. We find \(40 \%\) of \(500\) mL to be \(200\) mL which means the final solution must contain \(200\) mL of acid. We have
\[ \mbox {amount of acid in $30 \%$ stock solution} + \mbox {amount of acid $90 \%$ stock solution} = 200 \, \mbox {mL}\]
The amount of
acid in \(x\) mL of \(30 \%\) stock is \(0.30x\) and the amount of acid in \(y\) mL of \(90 \%\) solution is \(0.90y\). We have \(0.30x + 0.90y = 200\). Converting to fractions, (We do this only
because we believe students can use all of the practice with fractions they can get!) our system of equations
becomes
Notice that we have no equation to determine \(w\), and as such, \(w\) is free. Setting \(w = t\) in \(E2\), we get \(y = \frac {1}{2} t + \frac {250}{3}\). Substituting for \(w\) and \(y\) in \(E1\) gives \(x + \left (\frac {1}{2} t + \frac {250}{3}\right ) + t = 500\) so
that \(x = -\frac {3}{2} t + \frac {1250}{3}\).
This system is consistent, dependent and its solution set is \(\{ \left (-\frac {3}{2} t + \frac {1250}{3}, \frac {1}{2} t + \frac {250}{3}, t\right ) \, | \, - \infty < t < \infty \}\).
While this answer checks algebraically, we have neglected to take into account that \(x\), \(y\) and \(w\), being amounts of acid and water,
need to be nonnegative. That is, \(x \geq 0\), \(y \geq 0\) and \(w \geq 0\).
The constraint \(x \geq 0\) gives us \(-\frac {3}{2} t + \frac {1250}{3} \geq 0\), or \(t \leq \frac {2500}{9}\). From \(y \geq 0\), we get \(\frac {1}{2} t + \frac {250}{3} \geq 0\) or \(t \geq -\frac {500}{3}\). The condition \(z \geq 0\) yields \(t \geq 0\), and we see that when we take the set theoretic
intersection of these intervals, we get \(0 \leq t \leq \frac {2500}{9}\). This gives our final answer is \(\{ \left (-\frac {3}{2} t + \frac {1250}{3}, \frac {1}{2} t + \frac {250}{3}, t\right ) \, | \,0 \leq t \leq \frac {2500}{9} \}\).
Of what practical use is our answer? Suppose there is only \(100\) mL of the \(90 \%\) solution remaining and it is due to expire. Can we use
all of it to make our required solution? We would have \(y = 100\) so that \(\frac {1}{2} t + \frac {250}{3} = 100\), and we get \(t = \frac {100}{3}\). This means the amount of \(30 \%\) solution required is \(x = -\frac {3}{2} t + \frac {1250}{3} = -\frac {3}{2} \left (\frac {100}{3}\right ) + \frac {1250}{3} = \frac {1100}{3}\)
mL, and for the water, \(w = t = \frac {100}{3}\) mL. The reader is invited to check that mixing these three amounts of our constituent solutions
produces the required \(40 \%\) acid mix. □
PreCalculus