Middle Years

Lesson

Equations are useful because they can help us find values that we don't know yet. This can range from unknown numbers in a number sentence to missing dimensions in a shape. However, in order to use equations to help us with these problems, we will first need to be able to convert our problems into equations and then interpret our results.

Converting our real world problems into mathematical equations that we can solve is very similar to how we convert between word statements and number sentences.

When converting our problems into mathematical language, the most important thing to understand is what the problem is and which operations we can use to represent it.

Let's have a look at a simple example.

The sum of an unknown number and $46$46 is $74$74. What is the unknown number?

**Think**: To solve this problem, we need to identify what operation(s) we should use to build our equation so that our mathematical sentence matches the real world problem.

**Do**: We can use the pronumeral $x$`x` to represent the unknown number. The word 'sum' indicates addition so there should be a $+$+ symbol in our equation. The word "is" is short for "is equal to" so there should also be an $=$= symbol in our equation. Putting this all together in an equation gives us:

$x+46=74$`x`+46=74

Once we have converted the problem into an equation, we can solve it to find that:

$x=28$`x`=28

Since $x$`x` represents the unknown number, we can interpret our solution to mean that the unknown number was $28$28.

**Reflect**: When converting from the problem into an equation we used the key words "sum" and "is" to help us determine what mathematical symbols we would need.

Sometimes the equations that we make cannot be solved immediately because we do not have enough information. This happens when we end up with more than one unknown value. In equations, this is represented by there being more than one pronumeral.

However, these equations are still useful because they can show us the relationship between the two unknown values.

Consider the following scenario.

Hana and Curt eat a box of $12$12 biscuits together.

We do not know how many biscuits each person gets but we can still represent this scenario using an equation.

We can represent the number of biscuits Hana gets with the pronumeral $h$`h`, and we can do the same for Curt with the pronumeral $c$`c`. Since their total will be equal to $12$12, we can express this scenario with the equation:

$h+c=12$`h`+`c`=12

We can now use this equation to answer some questions we might have.

If Curt gets $5$5 biscuits, how many does Hana get?

In other words, if $c=5$`c`=5, what is the value for $h$`h`?

We can find the answer by substituting our new value into the equation. Replacing $c$`c` with $5$5 in our equation gives us:

$h+5=12$`h`+5=12

Solving this equation tells us that, when $c=5$`c`=5, $h=7$`h`=7. In other words, if Curt gets $5$5 biscuits then Hana gets $7$7.

What if $c$`c` is equal to some other values? We can represent this using a table of values:

$c$c |
$6$6 | $7$7 | $8$8 | $9$9 | $10$10 |
---|---|---|---|---|---|

$h$h |
$6$6 | $5$5 | $4$4 | $3$3 | $2$2 |

As we can see from the table, as the value for $c$`c` increases, the value for $h$`h` decreases. In other words, the more biscuits Curt gets, the less Hana gets.

We can see from this example that equations with more than one unknown value can be useful, either by substituting new information into the equation or by testing a range of values to find a relationship between the unknown values.

Consider the word statement "$y$`y` is equal to the product of $3$3 and the sum of $x$`x` and $4$4".

Write an equation in the form $y=\editable{}$

`y`= that describes the word statement.Using the equation found in part (a), complete the table.

$x$ `x`$1$1 $2$2 $3$3 $5$5 $10$10 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$

An important application of equations and substitution is for using formulas.

Formula

A formula is an equation that defines the relationship between two or more variables, usually representing a real world or mathematical relationship.

In mathematics, we often use formulas to find unknown values based on relationships that we know are true for certain situations.

We know that the area of a rectangle will always be equal to the product of its length and width.

We can represent this relationship using the equation:

$A=l\times w$`A`=`l`×`w`

where $A=\text{Area }$`A`=Area , $l=\text{length }$`l`=length and $w=\text{width }$`w`=width .

Using this equation, we can find any of the three variables as long as we know the other two.

Suppose the length is $4$4 units and the width is $7$7 units. Substituting this into the equation gives us:

$A=4\times7$`A`=4×7

Solving this tells us that a rectangle with these dimensions will have an area of $28$28 square units.

What about a rectangle that has an area of $20$20 square units and a length of $4$4 units?

We can find the height of such a rectangle by substituting our known values into the equation giving us:

$20=4w$20=4`w`

Solving this tells us that a rectangle with this area and length must have a width of $5$5 units.

The perimeter of a square is given by the formula $P=4s$`P`=4`s`, where $P$`P` represents the perimeter and $s$`s` represents the side length of the square.

What is the side length of the square when the perimeter is $32$32 units?

**Think**: We know that the formula $P=4s$`P`=4`s` defines a relationship between perimeter and side length. If we want to find the side length when the perimeter is $32$32 units, then we can use the formula to find the value for $s$`s` when $P=32$`P`=32.

**Do**: We should start by writing out an equation and substituting in any values that we know. Since we know that the perimeter is $32$32 units, we can replace $P$`P` with $32$32 in the equation. This gives us:

$32=4s$32=4`s`

Solving this gives us:

$s=8$`s`=8

We can interpret this result to mean that the side length of a square will be $8$8 units when the perimeter of that square is $32$32 units.

**Reflect**: To find the unknown value, we used the formula as our equation and substituted in the known variable in order to find the unknown variable.

The area of a rhombus is given by the formula $A=\frac{1}{2}xy$`A`=12`x``y`, where $x$`x` and $y$`y` are the lengths of the diagonals.

A particular rhombus has a short diagonal length $x=6$`x`=6 m and area $A=33$`A`=33 m^{2}.

Set up an equation and rearrange to solve for the unknown $y$

`y`.

The speed of a plane can be calculated using the formula $S=\frac{D}{T}$`S`=`D``T`, where $D$`D` is distance travelled, $T$`T` is time taken and $S$`S` is speed.

If a plane travels $3600$3600 kilometres in $6$6 hours, what is its speed?