11.06 Linear systems in three unknowns

Solutions to systems of equations

A linear equation is one in which the variables are to the power one. Equations like the following are typical examples.

$2x+\frac{1}{2}y=14$2x+12​y=14

$y=-3x+7$y=−3x+7

$x_1+x_2-x_3=0$x1​+x2​−x3​=0

A solution to any one of these equations is a set of values of the variables that satisfies the equation. For example, the equation $2x+\frac{1}{2}y=14$2x+12​y=14 has solutions $(x,y)=(7,0),(5,8),(3,16),(21,-56)$(x,y)=(7,0),(5,8),(3,16),(21,−56) and countless others.

The second equation, $y=-3x+7$y=−3x+7 also has infinitely many solutions. Among them is the solution $(x,y)=(21,-56)$(x,y)=(21,−56). This pair of numbers also satisfies the first equation. It is a solution to both the first and second equation, so we can say it is a solution to the system containing those two equations.

The third equation has solutions with three numbers. For example, $(x_1,x_2,x_3)=(2,1,3),(5,-1,4),(21,-56,-35)$(x1​,x2​,x3​)=(2,1,3),(5,−1,4),(21,−56,−35), and so on. Again, there are infinitely many solutions.

 

If the equation has two variables, $x$x and $y$y, say, then the relation between them can be displayed as a $2$2-dimensional graph. The graphs of linear equations in two variables are straight lines. A graph contains all the points that satisfy the equation.

If there are three variables, the graph is in $3$3-dimensions and is in the form of a plane.

 

Solutions to systems of equations correspond to intersection points on their graphs. Such points are on both graphs and satisfy both equations. Here is some vocabulary that is important when discussing systems of equations.

If two graphs intersect at one point, then there is a unique solution to the pair of equations. This is the case for the pair 

Here, the two graphs are shown. The intersection is at the point $(21,-56)$(21,−56). This is an independent, consistent system of equations.

 

It can happen that the graphs of two equations coincide. In other words, all the points that satisfy one equation also satisfy the other. Thus, there are infinitely many solutions. This would be the case for the equations $y=-3x+7$y=−3x+7 and $2y+6x-14=0$2y+6x−14=0. This is a dependent, consistent system of equations.

These equations look different as written but they express the same relation and each can be converted into the other by algebraic manipulation.

The possibility of having infinitely many solutions takes on more importance in the case of systems of equations with three variables. These describe planes and two different planes can intersect in a line so that infinitely many solutions exist, all of them lying on that line.

 

It can also happen that a pair of lines does not intersect. Hence, a pair of equations describing the lines has no solution. For example, the graphs of $y=-3x+7$y=−3x+7 and $y=-3x+28$y=−3x+28 do not intersect. Their slopes are the same and one is a vertical translation of the other. This system of equations is inconsistent. The graphs are shown below.

If two equations arranged into the same form have the same coefficients for the variables but different constant terms, then they cannot be true simultaneously. No solution exists and the graphs are parallel. For example, the following are inconsistent:

$2x+5y=19$2x+5y=19 and $2x+5y=16$2x+5y=16

$y=-x+5$y=−x+5 and $y=-x+6$y=−x+6

 

Worked examples

Question 1

One quantity is $1.3$1.3 more than another and together they make $3.7$3.7. What are the two quantities?

We must first express the facts algebraically. Call the quantities $x$x and $y$y. Then, we have $y=x+1.3$y=x+1.3 and $x+y=3.7$x+y=3.7.

One way to find the simultaneous solution is by substituting the $y$y-value from the first equation into the second. Thus, $x+(x+1.3)=3.7$x+(x+1.3)=3.7. This simplifies to $2x=2.4$2x=2.4 and then, $x=1.2$x=1.2.

Finally, we find $y$y by substituting for $x$x in either one of the original equations. Thus, $y=1.2+1.3$y=1.2+1.3  and the solution can be written as $(x,y)=(1.2,2.5)$(x,y)=(1.2,2.5).

Having found a solution, we should check that it does indeed satisfy both equations.

 

Question 2

We wish to find the intersection of the lines $x-3y=-1$x−3y=−1 and $3x-y=13$3x−y=13.

This example has a simple whole-number solution and it may be possible to find it by trying out various possibilities. Other cases are not so simple. So, we show a method that always works.

If both sides of an equation are multiplied by the same number, the result must also be a true statement. In this case, we could multiply the first equation by $3$3 to get $3x-9y=-3$3x−9y=−3. This was done to make the $x$x coefficients the same in the two equations.

Now, we can eliminate the x-term by subtracting the left and right sides of this new equation from the second equation. That is

$3x-y-(3x-9y)=13-(-3)$3x−y−(3x−9y)=13−(−3)

This simplifies to $8y=16$8y=16 and hence, $y=2$y=2.

We can now substitute this $y$y-value back into one of the original equations to find the $x$x-value. If $x-3y=-1$x−3y=−1, then $x-3\times2=-1$x−3×2=−1 and so, $x=5$x=5.

 

This method is particularly suited to implementation in computer programs for solving much more complicated systems of linear equations.

Practice questions

Question 3

Question 4

Question 5

Examine solutions to systems with three unknowns

A system of three linear equations in three unknowns can have no solutions, one solution or infinitely many solutions. In this chapter, we explore the conditions for these three possibilities.

One equation

A single linear equation in three unknowns represents a plane in a $3$3-dimensional coordinate system. For example, in the equation $3x-y+2z=0$3x−y+2z=0, we could choose a fixed value for $z$z, say $z=t$z=t. Then, $y=3x-2t$y=3x−2t. But this is the equation of a line in the $x$x-$y$y plane at the level of $t$t. There must be such a line at every possible level of $t$t and we might imagine these infinitely many lines, side by side, forming a plane in $3$3-space.

Worked example

Question 6

There are infinitely many sets of numbers $(x,y,z)$(x,y,z) that satisfy the equation $3x-y+2z=0$3x−y+2z=0 but there are also infinitely many sets of numbers that do not satisfy the equation. We can specify the solution set by choosing a parameter $t$t for $z$z and another parameter $s$s for $y$y. Then, we see that $x$x is constrained by the equation to be $x=\frac{1}{3}s-\frac{2}{3}t$x=13​s−23​t.

The solution set can be written in vector form as

The two vectors on the right define a plane and the solution set is the set of linear combinations of the two vectors as $s$s and $t$t range over the real numbers.

 

Two equations

Two equations represent two planes. The planes might be parallel, in which case they have no points in common and there can be no solution that satisfies both equations. If the planes are not parallel, they intersect along a line and there are infinitely many solutions, all belonging to the line.

Worked example

Question 7

The equations $3x-y+2z=0$3x−y+2z=0 and $3x-y+2z=6$3x−y+2z=6 represent parallel planes. Any attempt at a simultaneous solution leads to an absurdity like $6=0$6=0 and we must conclude that no solution exists.

On the other hand, a pair of equations like $3x-y+2z=0$3x−y+2z=0 and $x+y+z=0$x+y+z=0 can be solved to find a line of solutions. We might observe immediately that $(0,0,0)$(0,0,0) satisfies both equations. That is, both planes pass through the origin.

We could multiply the second equation by $-3$−3 and add it to the first equation. This gives $0x-4y-z=0$0x−4y−z=0.  We now have the equations $x+y+z=0$x+y+z=0 and $4y+z=0$4y+z=0 and these must have the same solution set as the original pair.

As before, we could choose to put $z=t$z=t. Then, from the $4y+z=0$4y+z=0 we find $y=-\frac{t}{4}$y=−t4​. Finally, from $x+y+z=0$x+y+z=0 we have $x=\frac{t}{4}-t=-\frac{3t}{4}$x=t4​−t=−3t4​. All three variables have been expressed in terms of the single parameter $t$t.

This time, the solution set in vector form is

The solution set is the set of scalar multiples of the vector on the right. The solution $(0,0,0)$(0,0,0) corresponds to $t=1$t=1.

 

Three equations

Three linear equations in three unknowns represent three planes in coordinate space. There are four possibilities.

• All three may be parallel, in which case there is no intersection and no solution. The system of equations would be found to be inconsistent.
• The three planes could intersect pairwise in three different parallel lines, in which case there is no simultaneous solution. Again, the system would be inconsistent.
• All three could intersect in a single line so that there are infinitely many solutions. Solutions would be found as in Question $7$7.
• The three planes could intersect at a single point.

 

Worked example

Question 8

Consider the system of three equations

If we subtract the first equation from the second and also subtract twice the first equation from the third, we obtain

Next, subtract $\frac{1}{2}$12​ the second equation from the third.

From the third of these equations, we find $z=5$z=5. Then, when this is substituted into the second equation, we get $y=3$y=3. And when these values of $z$z and $y$y are substituted into the first equation, we have $x=2$x=2.

So, the solution set is the single point

 

Practice questions

Question 9

Question 10

Question 11

Linear systems in three unknowns

A system of equations is a group of $2$2 or more equations, we call it a system when maybe the equations are related by context or purpose, kind of like that the system is working together.  The individual equations can be linear, quadratic or indeed take any form but for now we will concentrate on just the linear ones.  There are two useful features about systems of equations, the first is the solution to the system and the second relates to regions within the system.

The solution to a system is the point where all the equations are satisfied.  When we had $2$2 linear equations in $2$2 unknowns the solution was the intersection of the two lines because at this point both lines are satisfied.

An equation in $3$3 unknowns takes us up a dimension.  No longer are we working in $2$2 dimensions of the $x$x and $y$y axis, but in $3$3 dimensions with the $x,y$x,y and $z$z axis. This is the $3$3 dimensional coordinate plane.  You can see the $x$x and $y$y axis in red and green, and the $3$3rd dimension is $z$z, which is in blue.

The line $y=x$y=x appears in the $3$3 dimensions in what we call a plane.  It is a surface.  It is still taking on the familiar $y=x$y=x shape, but the third dimension gives us this shape.  It is similar to imaging holding up a piece of paper.  

The equation $x+y+z=1$x+y+z=1, produces another plane surface. 

 

The following applet will allow you to create different planes using the sliders for different values of $x,y,z$x,y,z.  

So now we know what an equation in $3$3 dimensions produces.  

Let's have a brief look at the behavior of planes.

 

Parallel planes

For example here we have pictured $x+y+z=1,x+y+z=-3$x+y+z=1,x+y+z=−3 and $x+y+z=5$x+y+z=5

Intersecting planes

Here we have pictured $x+y+z=1$x+y+z=1 and $-2x+y+5z=1$−2x+y+5z=1.

We can see here that we don't have a single point of intersection, but we have a line. 

 

With three planes intersecting we end up with a point. It's at this point that the solution to the system of $3$3 equations and $3$3 unknowns provides us.

This image shows the line that was created by the intersection of $x+y+z=1$x+y+z=1 and $-2x+y+5z=1$−2x+y+5z=1 and also the plane created by $3x+0.5y-0.5z=1$3x+0.5y−0.5z=1.  We can see that these intersect at just the one point. The place where the line passes through the plane.

 

Here is the original image of the $3$3 planes.

The point of intersection for this set is $(0.29,0.5,0.21)$(0.29,0.5,0.21).

 

Solving systems of equations in three unknowns.

Now we know what we are finding, (the intersection of the planes), then we need to set about doing it.

The good news is, we use all the tools we have already gained solving systems in $2$2 unknowns.  We can use substitution, elimination, or take a graphical approach using technology as we saw above.

Let's do one together.

Solve the system of equations

$2x+y-z$2x+y−z	$=$=	$9$9	$(1)$(1)
$x-3y+z$x−3y+z	$=$=	$-2$−2	$(2)$(2)
$-x+y+3z$−x+y+3z	$=$=	$-8$−8	$(3)$(3)

To solve this let's use elimination, first on equations $1$1 and $2$2

$2x+y-z$2x+y−z	$=$=	$9$9	$(1)$(1)
$x-3y+z$x−3y+z	$=$=	$-2$−2	$(2)$(2)
$3x-2y$3x−2y	$=$=	$7$7	$(1)+(2)->(4)$(1)+(2)−>(4)

and a similar operation with equations $1$1 and $3$3.

$2x+y-z$2x+y−z	$=$=	$9$9	$(1)$(1)
$-x+y+3z$−x+y+3z	$=$=	$-8$−8	$(3)$(3)
$6x+3y-3z$6x+3y−3z	$=$=	$27$27	$3\times(1)->(5)$3×(1)−>(5)
$5x+4y$5x+4y	$=$=	$19$19	$(5)+(3)->(6)$(5)+(3)−>(6)

Now we have this smaller system in $2$2 unknowns that we can solve for $x$x and $y$y.

$3x-2y$3x−2y	$=$=	$7$7	$(4)$(4)
$5x+4y$5x+4y	$=$=	$19$19	$(6)$(6)
$11x$11x	$=$=	$33$33	$(6)+2\times(4)$(6)+2×(4)
$x$x	$=$=	$3$3

So $x=3$x=3, which means that 

$3x-2y$3x−2y	$=$=	$7$7
$3\cdot3-2y$3·3−2y	$=$=	$7$7
$9-2y$9−2y	$=$=	$7$7
$-2y$−2y	$=$=	$-2$−2
$y$y	$=$=	$1$1

Substituting back into (1), $2x+y-z=9$2x+y−z=9 we discover that $2\times(3)+1-z=9$2×(3)+1−z=9, which means that $z=-2$z=−2

So our final solution is that 

$x=3$x=3, $y=1$y=1 and $z=-2$z=−2.  We can write this as an ordered triple of $(3,1,-2)$(3,1,−2).

The great thing about systems of equations is that you can always substitute back in and check your answer. 

 

We can check that a point (in $3$3 dimensions we call this point an ordered triple) is a solution by substituting into each of the equations.  If the values for $x,y$x,y and $z$z satisfy all three equations, then the ordered triple is a solution. 

Practice question

Question 12

 

Systems with no solution

I already eluded to a special case of systems of equations in 3 unknowns, when they are all parallel.  If the solution to a system is the point (or line) that occurs on their intersection, then what happens when there are no intersections?  

In the case of parallel lines, the system is what we call inconsistent.  When you try to solve a system that is parallel, you end up with what we call contradictions.  Take our parallel line example from earlier.  

For example here we have pictured $x+y+z=1,x+y+z=-3$x+y+z=1,x+y+z=−3 and $x+y+z=5$x+y+z=5

See how $x+y+z$x+y+z is $1$1, but it is also $-3$−3 and $5$5.  Clearly, there is not a set of numbers that when we add them together we can have three different answers.  So this is the contradiction, and it is what indicates parallel lines. 

 

Systems with infinitely many solutions

Infinite solutions result when one of the equations ends up being (through some kind of algebraic manipulation) the same as one of the other equations.  This means we really only have two unique equations in three unknowns so we are not actually able to solve it. . 

For example, this system has infinitely many solutions as equation $(1)$(1) and equation $(3)$(3) are actually a multiple of each other. 

$3x-2y+z=5$3x−2y+z=5 $1$1

$5x+y+z=-2$5x+y+z=−2 $2$2

$-3x+2y-z=-5$−3x+2y−z=−5 $3$3

If a system has infinitely many solutions we say that the system is dependent.  

Systems with one solution

If a system has one solution, than we say that the system is independent.  

 

 

 

Practice questions

Question 13

Question 14

Applications of linear systems in three or more unknowns

We've now covered how to solve linear systems of equations in three unknowns. We can solve by elimination where we can multiply/divide any equation and add/subtract equations. We can also solve by substitution where we substitute one equation into another. Alternatively, we can look at systems of equations graphically as regions or curves in space (for three unknowns) and curves on the plane (for two unknowns).

Now it's time to apply these concepts to certain situations. For any problem involving systems of equations, just remember to follow this three-step process.

1) Identify - Read the question carefully and identify the necessary equations.

2) Construct - Construct the system of equations. Try to line up the variables to make solving them easier.

3) Solve - Choose a method and solve the system.

Worked example

Question 15

At a local cinema, there are different ticket prices for adults, students and children. Three families come in to see a movie.

The first family consists of $4$4 adults, $1$1 student and $3$3 children. They pay $\$131$$131 all together.

The second family consists of $2$2 adults, $2$2 students and $2$2 children. They pay $\$94$$94 all together.

The third family consists of $2$2 adults, $3$3 students and $1$1 child. They pay currency $\$97$$97 all together.

What are the ticket prices at the cinema?

 

Think: After reading the question carefully, we can see that our three unknowns are the ticket prices for adults, students and children, which we can label $x$x, $y$y and $z$z respectively. We want three equations with these variables.

Do: Let's look at the first family. They have $4$4 adults, so they paid $4x$4x worth of adult tickets. Similarly, they paid $y$y worth of student tickets and $3z$3z worth of child tickets. If they paid $\$131$$131 all together, we can come up with this equation.

$4x+y+3z=131$4x+y+3z=131 ... (1)

In the same way, we get the following equations for the other two families.

$2x+2y+2z=94$2x+2y+2z=94 ... (2)

$2x+3y+z=97$2x+3y+z=97 ... (3)

Now that we have our equations, we need to line them up in a system and choose a method for solving them. For this particular system, let's use elimination.

$4x$4x	$+$+	$y$y	$+$+	$3z$3z	$=$=	$131$131	...	(1)
$2x$2x	$+$+	$2y$2y	$+$+	$2z$2z	$=$=	$94$94	...	(2)
$2x$2x	$+$+	$3y$3y	$+$+	$z$z	$=$=	$97$97	...	(3)

Noticing that equations (2) and (3) both have $2x$2x at the start, let's subtract them to eliminate $x$x altogether.

$2x$2x	$+$+	$3y$3y	$+$+	$z$z	$=$=	$97$97	...	(3)
$2x$2x	$+$+	$2y$2y	$+$+	$2z$2z	$=$=	$94$94	...	(2)
		$y$y	$-$−	$z$z	$=$=	$3$3	...	(3)$-$−(2)$=$=(4)

Now we have an equation in $y$y and $z$z only. To solve fully, we're going to need another equation in $y$y and $z$z. We will have eliminate $x$x somewhere else too.

To get this, we could multiply equation (2) by $2$2 to get $4x$4x at the start, then subtract equation (1) from this.

$4x$4x	$+$+	$4y$4y	$+$+	$4z$4z	$=$=	$188$188	...	(2) $\times$×$2$2
$4x$4x	$+$+	$y$y	$+$+	$3z$3z	$=$=	$131$131	...	(1)
		$3y$3y	$+$+	$z$z	$=$=	$57$57	...	(2)$\times$×$2$2$-$−(1)$=$=(5)

Now, in these new equations (4) and (5) that we've derived, notice that the signs are opposite, so if we add these equations together, $z$z should vanish, leaving $y$y ready to solve.

$y$y	$-$−	$z$z	$=$=	$3$3	...	(4)
$3y$3y	$+$+	$z$z	$=$=	$57$57	...	(5)
$4y$4y			$=$=	$60$60	...	(4)$+$+(5)$=$=6

And after dividing both sides by $4$4, we finally get:

$y=15$y=15

Now it's time to use this to retrieve $x$x and $z$z. We can substitute backwards like dominoes to get our remaining solutions.

Substituting $y=15$y=15 into equation (4) gives:

$y=15$y=15     →     $\left(15\right)-z=3$(15)−z=3     →     $z=12$z=12

And substituting $y=15$y=15 and $z=12$z=12 into equation (2) gives:

$y=15$y=15, $z=12$z=12     →     $2x+2\times\left(15\right)+2\times\left(12\right)=94$2x+2×(15)+2×(12)=94     →     $2x+54=94$2x+54=94     →     $x=20$x=20

And so we have our solutions:

$x=20$x=20
$y=15$y=15
$z=12$z=12

Reflect: Which means the cinema charges the following prices:

Adults	$\$20$$20
Students	$\$15$$15
Children	$\$12$$12

Elimination is not the only method we could have used, so it will be up to you to identify a method that you think will be efficient.

Practice questions

Question 16

Question 17