Multiplicative Relationship

Find the constants of the multiplicative relationships below.

1. 4 and 16
2. 2 and 6

Solution

a. 4 and 16

4 is the input and 16 is the output. You can write this in the form of the multiplicative relationship equation.

$$\begin{gathered} y=ax \\ 16=4a \end{gathered}$$

Dividing both sides of the equation by 4 to isolate $a$ we get, $$\begin{gathered} \frac{16}{4}=\frac{4a}{4} \\ a=\frac{16}{4}=4 \end{gathered}$$

The constant is 4.

b. 2 and 6

2 is the input and 6 is the output. You can write this in form of the multiplicative relationship equation to get the constant.

$$\begin{gathered} y=ax \\ 6=2a \end{gathered}$$

Dividing both sides of the equation by 2 to isolate $a$ we get,

$$\begin{gathered} \frac{6}{2}=\frac{2a}{2} \\ a=\frac{6}{2}=3 \end{gathered}$$

The constant is 3.

Determine which set in the table below has a multiplicative relationship.

Let's see what we will have if the numbers in set A are substituted in our equation.

The first pair of numbers are 6 and 12. Our equation suggests that 6 multiplied by a number will give 12. Substituting in the equation gives

$12=6a$

Let's simplify further to get the constant $a$,

$$\begin{gathered} 12=6a \\ \frac{12}{6}=\frac{6a}{6} \\ a=2 \end{gathered}$$

We can't conclude now that set A has a multiplicative relationship because there are still other pairs of numbers to consider and the constant $a$ has to be the same for all of them.

The next pair of numbers are 8 and 13. 13 is the output $y$ and 8 is the input $x$. Let's substitute in the formula to see what constant $a$ would be.

$$\begin{gathered} y=ax \\ 13=8a \end{gathered}$$

Divide both sides by 8 to isolate $a$, we get

$$\begin{gathered} \frac{13}{8}=\frac{8a}{8} \\ a=1.63 \end{gathered}$$

The constant we got here is different from the one we got before which means that the numbers in Set A do not have a multiplicative relationship.

Divide both sides by 3 to isolate $a$ to get,

$$\begin{gathered} \frac{12}{3}=\frac{3a}{3} \\ a=4 \end{gathered}$$

The constant $a$ is 4. Let's see if it will be the same for the other numbers in the set.

The next pair of numbers are 5 and 20. Substituting in the equation will give,

$$\begin{gathered} y=ax \\ 20=5a \end{gathered}$$

The constant here is 4.

Show the multiplicative relationship between the pairs below.

1. 15 and 17
2. 35 and 27

Solution

a. 15 and 17

Here, 15 is the input, and 17 is the output. To show the multiplicative relationship between 15 and 17, we have to multiply 15 by a fraction. The numerator of the fraction must be 17 and the denominator must be 15 so we can cancel out to get 17 as the answer.

Let's recall the multiplicative relationship equation.

$y=ax$

where $y$ is 17

$x$ is 15

$a$ is the constant,

$$\begin{gathered} y=15a \\ y=15\times \frac{17}{15} \\ y=\cancel{15}\times \frac{17}{\cancel{15}} \\ y=17 \end{gathered}$$

$y$ is 17 as desired and $a$ is $\frac{17}{15}$.

b. 35 and 27

Here, 35 is the input, and 27 is the output. To show the multiplicative relationship between 25 and 27, we have to multiply 35 by a fraction. The numerator of the fraction must be 27 and the denominator must be 35 so we can cancel out to get 27 as the answer.

Recall, the multiplicative relationship equation.

$y=ax$

where $y$ is 27

$x$ is 35

$a$ is the constant

$$\begin{gathered} y=35a \\ y=35\times \frac{27}{35} \\ y=\cancel{35}\times \frac{27}{\cancel{35}} \\ y=27 \end{gathered}$$

$yis27$ as desired and $ais\frac{27}{35}$.

Determine which set in the table has a multiplicative relationship.

Solution

We want to know which set of numbers have a multiplicative relationship. Let's start with Set A.

The first pair of numbers in set A are 3 and 7 where 3 is the input and 7 is the output.

Recall the multiplicative relationship equation,

$y=ax$

We will substitute in the equation to get the constant,

$$\begin{gathered} 7=3a \\ \frac{7}{3}=\frac{3a}{3} \\ a=\frac{7}{3} \end{gathered}$$

The constant is $\frac{7}{3}$.

Let's find out if the next pair will give the same constant.

The next pair is 4 and 5 where 4 is the input and 5 is the output.

Let's substitute in the multiplicative relationship formula.

$$\begin{gathered} y=ax \\ 5=4a \\ \frac{5}{4}=\frac{4a}{4} \\ a=\frac{5}{4} \end{gathered}$$

The constant here is different from the first. This means that the numbers in set A do not have a multiplicative relationship.

Let's take a look at set B.

The first pair of numbers are 5 and 2 where 5 is the input and 2 is the output.

Let's substitute in the multiplicative relationship formula.

$$\begin{gathered} y=ax \\ 2=5a \\ \frac{2}{5}=\frac{5a}{5} \\ a=\frac{2}{5} \end{gathered}$$

The constant is $\frac{2}{5}$. If the rest of the pair have $\frac{2}{5}$ as their constant, then there is a multiplicative relationship.

The next pair is 10 and 4 where 10 is the input and 4 is the output.

Let's substitute in the multiplicative relationship formula.

$$\begin{gathered} y=ax \\ 4=10a \\ \frac{4}{10}=\frac{10a}{10} \\ a=\frac{4}{10}=\frac{2}{5} \end{gathered}$$

We have a constant of $\frac{2}{5}$ which is the same as the first. Let's try the last two pair.

The third pair is 20 and 8 where 20 is the input and 8 is the output.

Substituting in the formula will give,

$$\begin{gathered} y=ax \\ 8=20a \\ \frac{8}{20}=\frac{20a}{20} \\ a=\frac{8}{20}=\frac{2}{5} \end{gathered}$$

The constant is $\frac{2}{5}$ again.

The last pair is 15 and 6 where 15 is the input and 6 is the output.

Let's substitute in the multiplicative relationship formula.

$$\begin{gathered} y=ax \\ 6=15a \\ \frac{6}{15}=\frac{15a}{15} \\ a=\frac{6}{15}=\frac{2}{5} \end{gathered}$$

Again, we have the constant to be $\frac{2}{5}$. This means that the numbers in set B have a multiplicative relationship.

If $y=4x$ complete the table below and plot the graph.

Solution

The equation given is $y=4x$ and you will notice that it is in the form of the multiplicative relationship equation

$y=ax$

What we need to do is substitute the values of $x$ in the equation given to get $y$.

When $x=0$,

$$\begin{gathered} y=4\times 0 \\ y=0 \end{gathered}$$

When $x=1$,

$$\begin{gathered} y=4\times 1 \\ y=4 \end{gathered}$$

When $x=2$,

$$\begin{gathered} y=4\times 2 \\ y=8 \end{gathered}$$

When $x=3$,

$$\begin{gathered} y=4\times 3 \\ y=12 \end{gathered}$$

Now let's put the results we've gotten in the table.

Now that we've completed the table, let's plot the graph.

We can see that the graph has the characteristics of a multiplicative relationship graph which is that it is linear and it starts from the origin (0, 0).

A company pays its workers £13 per hour. Show the multiplicative relationship here, create a table and plot a graph from the information on the table.

Solution

The question says a company pays £13 per hour. Believe it or not, there are two quantities to consider here. You have to consider the money earned and the hours worked.

Now, we have to figure out what x and y are.

Recall that x is the input or independent quantity and y is the output or dependent quantity.

The independent quantity here is the hours worked. This is because the worker has control over it. The worker decides how many hours they can work. The dependent quantity is the money earned because it depends on how many hours are worked.

Recall the multiplicative relationship equation

$y=ax$

A worker will earn £13 per hour meaning that: $y=13$ and $x=1$. We can substitute this in the equation to get the constant,

$$\begin{gathered} y=ax \\ 13=1\times a \\ a=13 \end{gathered}$$

For a question like this, you may not need to go through this step to find the constant because it is quite clear that you will have to multiply 13 by the number of hours worked to get the total money earned.

Now let's move on to creating a table. The table will contain the number of hours worked and the money earned. The number of hours will be multiplied by the constant 13 to get the money earned.

We have

$y=ax$

when $x=1$,

$$\begin{gathered} y=13\times 1 \\ y=13 \end{gathered}$$

when $x=2$,

$$\begin{gathered} y=13\times 2 \\ y=26 \end{gathered}$$

when $x=5$,

$$\begin{gathered} y=13\times 5 \\ y=65 \end{gathered}$$

when $x=6,$

$$\begin{gathered} y=13\times 6 \\ y=78 \end{gathered}$$

So, we will use the values we've got to fill the table.

With this information, we can now plot a graph.

We know that the graph is correct because it is linear.