Question

1. Determine whether the following problems (a-c) show direct variation. If NOT, how could you change it to make it a direct variation? Remember, direct variation takes the form of y=kx.

a.
x 1 3 5 7 9
y 4 10 16 22 28




b. y=10x

c. y=3x-4

Answer 1

@dan815

Answer 2

@pooja195

Answer 3

@iambatman

Answer 4

@Michele_Laino

Answer 5

hint:
direct variation between two variables \(x,\;y\), is modeled by the subsequent equation:
\(y=kx\) for example

Answer 6

i don't get it :c

Answer 7

namely, we can say y varies directly with x, if the subsequent equation holds:
\(y=kx\), where \(k\) is any number, more precisely any real number

Answer 8

" k " is known as the " proportionality constant."
What are your learning resources? Do you have an algebra textbook? Do you have access to online learning materials for your algebra course? Please at least glance through some of this material to find "direct proportion" or "direct variation."

Note that y = 7x represents direct proportion, whereas y = 7 x + 1 does not.

Answer 9

@Michele_Laino is there an easy way or can you give an example

Answer 10

for example, equation c) is similar to this equation \(y=kx\) ?

Answer 11

for example a direct variation can b modeled by this formula:
\(y=2x\)

Answer 12

can be*

Answer 13

whereas an equation like this:
\(y=2x+1\), doesn't represent a direct variation, since it is present the constant value +1, at the right side

Answer 14

okay

Answer 15

so, is option c) a direct variation?

Answer 16

no

Answer 17

that's right!
so, if we cancel the constant -4, at the right side, we get this formula:
\(y=3x\)
and \(y=3x\) is a direct variation?

Answer 18

yes

Answer 19

that's right!
next, is option b) a direct variation?

Answer 20

yes

Answer 21

that's right, since such equation has this form: \(y=kx\), with \(k=10\)

Answer 22

now, for part a), we have to write the equation between x and y, using the data into the table above

Answer 23

okay :D

Answer 24

in order to do that, I consider the first ordered pair \(x=1,\;y=4\), and I replace such values ito this formula:
\(y=ax+b\), where \(a,\;b\) are two constant which have to be determined

Answer 25

so, after a substitution, I get this:
\(4=a \cdot 1+b\) or: \(4=a+b\)

Answer 26

so 4=a*1+b

Answer 27

yes! \(4=a+b\)

Answer 28

then I consider the second ordered pair: \(x=3,\;y=10\), and a fter a substitution, I get:
\(10=a \cdot 3+b\), or: \(10=3a+b\)

Answer 29

10=a*3+b

Answer 30

yes!
so, we have to solve thys algebraic system:
\[\left\{ \begin{gathered}
4 = a + b \hfill \\
10 = 3a + b \hfill \\
\end{gathered} \right.\]

Answer 31

oops.. this*

Answer 32

hehe lol its okay

Answer 33

do you know how to solve such system?

Answer 34

um i don't :c sorry

Answer 35

please wait a moment, I have to answer to my phone

Answer 36

Okay :D

Answer 37

here I am

Answer 38

Yay !! :)

Answer 39

we can solve such system, using this procedure:
I subtract, side by side, the first equation from the second one, so I get:
\[10 - 4 = 30 + b - a - b\]

Answer 40

oops.. I have made a typo:
\[10 - 4 = 3a + b - a - b\]

Answer 41

then I simplify both sides, so I get:
\[6 = 2a\]

Answer 42

next I divide both sides by 2, so I get:
\[\frac{6}{2} = \frac{{2a}}{2}\]
what is \(a\) ?