Applications of linear systems in three 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. Or also solve by substitution where we sub 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 equations you will need.

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

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

Examples

Question 1

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?

SOLUTION

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.

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

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.

Question 2

Question 3