New Zealand
Level 8 - NCEA Level 3

# 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$$131 all together. The second family consists of 22 adults, 22 students and 22 children. They pay \94$$94 all together.

The third family consists of $2$2 adults, $3$3 students and $1$1 child. They pay currency $\$97$$97 all together. What are the ticket prices at the cinema? ##### SOLUTION After reading the question carefully, we can see that our three unknowns are the ticket prices for adults, students and children, which we can label xx, yy and zz respectively. We want three equations with these variables. Let's look at the first family. They have 44 adults, so they paid 4x4x worth of adult tickets. Similarly, they paid yy worth of student tickets and 3z3z worth of child tickets. If they paid \131$$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$$17000 each. Hartman Rent-A-Car will purchase twice as many hatchbacks as sedans and the total number of cars to be bought is 152152. 1. Let xx be the number of hatchbacks, yy be the number of sedans, and zz be the number of SUVs. Complete the following system of equations:  xx ++ yy ++ zz == \editable{} \editable{}$$x$x$+$+$\editable{}$$yy ++ \editable{}$$z$z$=$=$\editable{}x$x$=$=$\editable{}$$yy 2. Below is one approach to solving the system of equations using the elimination method:  xx ++ yy ++ zz == 152152 ----- equation 11 10000x10000x ++ 13000y13000y ++ 17000z17000z == 20080002008000 ----- equation 22 xx == 2y2y ----- equation 33 Equation 33 gives us a value of xx in terms of yy. This means we can make a substitution in place of xx in equation 11. Step 1: Substitute equation 33 into equation 11.  \editable{} ++ yy ++ zz == 152152 ----- equation 11 \editable{}$$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{}$$yy ++ 17000z17000z == 20080002008000 ----- equation 55 Now we have two linear equations in two unknowns (yy and zz). We would like to use the elimination method here, so we need both of the equations to have the same coefficient for one of the unknowns. Let's make the zz coefficient the same by multiplying every term in equation 44 by 1700017000. Step 3: Multiply equation 44 by 1700017000 so we can use the elimination method.  3y3y ++ zz == 152152 ----- equation 44 \editable{}\times3y×3y ++ \editable{}\times z×z == \editable{}\times152×152 \editable{}yy ++ \editable{}zz == \editable{} ----- equation 66 Now that equation 55 and equation 66 have the same zz coefficient we can subtract one from the other to get an equation with only one unknown. Step 4: Subtract equation 55 from equation 66. Begin by re-writing equation 66 and equation 55.  \editable{}yy ++ \editable{}zz == \editable{} ----- equation 66 \editable{}yy ++ \editable{}zz == \editable{} ----- equation 55 \editable{}yy == \editable{} yy == \editable{} We now have a value for yy. Let's substitute this value in place of yy in equation 44. Step 5: Substitute the value of yy into equation 44.  3\times\editable{}3× ++ zz == 152152 ----- equation 44 zz == \editable{} We now have a value for zz and yy. Let's substitute these values in equation 44. Step 6: Substitute the value of yy and the value of zz into equation 11.  xx ++ \editable{} ++ \editable{} == 152152 ----- equation 11 xx == \editable{} 3. Deduce the number of: Hatchbacks = \editable{}. Sedans = \editable{}. SUVs = \editable{}. ##### Question 3 Paul has made three different investments. In the last financial year he earned 7%7% interest on his savings account, 1%1% interest on his mutual funds investment, and 2%2% interest on his money market investment. Paul earned a total of \45600$$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