NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Applications of Linear Systems in 3 (or more) unknowns
Lesson

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

Hartman Rent-A-Car are about to spend $\$2008000$$2008000 buying a fleet of new vehicles. The fleet is made up of hatchbacks, sedans and SUVs. Hatchbacks cost $\$10000$$10000 each, sedans cost $\$13000$$13000 each and SUVs cost $\$17000$$17000 each.

Hartman Rent-A-Car will purchase twice as many hatchbacks as sedans and the total number of cars to be bought is $152$152.

  1. Let $x$x be the number of hatchbacks, $y$y be the number of sedans, and $z$z be the number of SUVs.

    Complete the following system of equations:

    $x$x $+$+ $y$y $+$+ $z$z $=$= $\editable{}$
    $\editable{}$$x$x $+$+ $\editable{}$$y$y $+$+ $\editable{}$$z$z $=$= $\editable{}$
            $x$x $=$= $\editable{}$$y$y
  2. Below is one approach to solving the system of equations using the elimination method:

    $x$x $+$+ $y$y $+$+ $z$z $=$= $152$152 ----- equation $1$1
    $10000x$10000x $+$+ $13000y$13000y $+$+ $17000z$17000z $=$= $2008000$2008000 ----- equation $2$2
    $x$x $=$= $2y$2y ----- equation $3$3

    Equation $3$3 gives us a value of $x$x in terms of $y$y. This means we can make a substitution in place of $x$x in equation $1$1.

    Step 1: Substitute equation $3$3 into equation $1$1.

    $\editable{}$ $+$+ $y$y $+$+ $z$z $=$= $152$152 ----- equation $1$1
    $\editable{}$$y$y $+$+ $z$z $=$= $152$152 ----- equation $4$4

    Using the same tactic as we did in step 1, we can make a substitution in place of $x$x in equation $2$2.

    Step 2: Substitute equation $3$3 into equation $2$2.

    $10000$10000$\times$×$\editable{}$ $+$+ $13000y$13000y $+$+ $17000z$17000z $=$= $2008000$2008000 ----- equation $2$2
    $\editable{}$$y$y $+$+ $17000z$17000z $=$= $2008000$2008000 ----- equation $5$5

    Now we have two linear equations in two unknowns ($y$y and $z$z).

    We would like to use the elimination method here, so we need both of the equations to have the same coefficient for one of the unknowns. Let's make the $z$z coefficient the same by multiplying every term in equation $4$4 by $17000$17000.

    Step 3: Multiply equation $4$4 by $17000$17000 so we can use the elimination method.

    $3y$3y $+$+ $z$z $=$= $152$152 ----- equation $4$4
    $\editable{}\times3y$×3y $+$+ $\editable{}\times z$×z $=$= $\editable{}\times152$×152
    $\editable{}y$y $+$+ $\editable{}z$z $=$= $\editable{}$ ----- equation $6$6

    Now that equation $5$5 and equation $6$6 have the same $z$z coefficient we can subtract one from the other to get an equation with only one unknown.

    Step 4: Subtract equation $5$5 from equation $6$6. Begin by re-writing equation $6$6 and equation $5$5.

    $\editable{}y$y $+$+ $\editable{}z$z $=$= $\editable{}$ ----- equation $6$6
    $\editable{}y$y $+$+ $\editable{}z$z $=$= $\editable{}$ ----- equation $5$5
    $\editable{}y$y $=$= $\editable{}$
    $y$y $=$= $\editable{}$

    We now have a value for $y$y. Let's substitute this value in place of $y$y in equation $4$4.

    Step 5: Substitute the value of $y$y into equation $4$4.

    $3\times\editable{}$3× $+$+ $z$z $=$= $152$152

    ----- equation $4$4

    $z$z $=$= $\editable{}$

    We now have a value for $z$z and $y$y. Let's substitute these values in equation $4$4.

    Step 6: Substitute the value of $y$y and the value of $z$z into equation $1$1.

    $x$x $+$+ $\editable{}$ $+$+ $\editable{}$ $=$= $152$152

    ----- equation $1$1

    $x$x $=$= $\editable{}$
  3. Deduce the number of:

    Hatchbacks = $\editable{}$.

    Sedans = $\editable{}$.

    SUVs = $\editable{}$.

Question 3

Paul has made three different investments. In the last financial year he earned $7%$7% interest on his savings account, $1%$1% interest on his mutual funds investment, and $2%$2% interest on his money market investment. Paul earned a total of $\$45600$$45600 in interest from the investments.

Paul has twice as much money invested in the money market than he has in his savings account. Further, he earned half as much from his money market investment than he did from his mutual funds investment.

  1. Let $x$x be the amount of money Paul has in his saving accounts, $y$y the amount he has invested in mutual funds, and $z$z the amount he has invested in the money market.

    Complete the following system of equations:

    $\editable{}x$x $+$+ $\editable{}y$y $+$+ $\editable{}z$z $=$= $\editable{}$
    $z$z $=$= $\editable{}x$x
    $y$y $=$= $\editable{}z$z
  2. Work through the following approach to solving the system of equations:

    $0.07x$0.07x $+$+ $0.01y$0.01y $+$+ $0.02z$0.02z $=$= $45600$45600 ----- equation $1$1
            $z$z $=$= $2x$2x ----- equation $2$2
            $y$y $=$= $4z$4z ----- equation $3$3

    Step 1: Substitute equation $2$2 into equation $3$3.

            $y$y $=$= $4z$4z

    ----- equation $3$3

            $y$y $=$= $4\times\editable{}$4×  
            $y$y $=$= $\editable{}$ ----- equation $4$4

    Step 2: Substitute equation $2$2 and equation $4$4 into equation $1$1.

    $0.07x$0.07x $+$+ $0.01y$0.01y $+$+ $0.02z$0.02z $=$= $45600$45600 ----- equation $1$1
    $0.07x$0.07x $+$+ $0.01\times\editable{}$0.01× $+$+ $0.02\times\editable{}$0.02× $=$= $45600$45600  
    $0.07x$0.07x $+$+ $\editable{}$ $+$+ $\editable{}$ $=$= $45600$45600  
            $\editable{}x$x $=$= $45600$45600  
            $x$x $=$= $\editable{}$  

    Step 3: Substitute the value of $x$x into equation $2$2.

            $z$z $=$= $2x$2x

    ----- equation $2$2

            $z$z $=$= $2\times\editable{}$2×  
            $z$z $=$= $\editable{}$  

    Step 4: Substitute the value of $z$z into equation $3$3.

            $y$y $=$= $4z$4z

    ----- equation $3$3

            $y$y $=$= $4\times\editable{}$4×  
            $y$y $=$= $\editable{}$  
  3. As a result of the above, enter the amounts Paul has invested in the following places:

    Saving account: $\editable{}$ dollars.

    Mutual funds: $\editable{}$ dollars.

    Money market: $\editable{}$ dollars.

Outcomes

M8-8

Form and use systems of simultaneous equations, including three linear equations and three variables, and interpret the solutions in context

91587

Apply systems of simultaneous equations in solving problems

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