Sorry this kinda a lot but pls help

Question

1. If y varies with x and y is 20 when x is 4, what is the constant variation?

2. if y varies directly with x and y is 7 when x is 35, write a function that represents this relationship. 

3. if y varies directly with x and y is 25 when x is 15, what is x when y is 30?

 

4. if p varies directly q and p=26 when q=2, find p when q=5

Answer 1

Hi, I hope you’re doing well!

1) If y varies directly with x, and y is 20 when x is 4, what is the constant of variation?

(I’m solving this assuming it mentioned direct variance.)

We can start by writing the equation for “y varies directly with x.” Then we’ll solve for the constant of variation, k.

y = kx ← Direct variance. (Indirect is y = k/x.)

20 = k·4 ← Since given y is 20 when x is 4.

20 = 4k

÷4 ÷4 ← Divide by 4 to “undo” the multiplication

5 = k

2) If y varies directly with x, and y is 7 when x is 35, write a function that represents this relationship.

This is basically the same problem, except after we find the constant, k, we will plug it into y = kx.

y = kx

7 = k·35

7 = 35k

÷35 ÷35

7/35 = k

1/5 = k ← From reducing the fraction 7/35 by dividing the top and bottom by 7

Last, we want to write the direction variance equation, y = kx, using the k value we found.

y = kx

y = 1/5·x

3) If y varies directly with x, and y is 25 when x is 15, what is x when y is 30?

Since y varies directly with x:

y = kx

Then we plug in the given values and solve for k:

25 = k·15

÷15 ÷15

25/15 = k

We can reduce this fraction by dividing 5 out of the top and bottom to get this k value:

5/3 = k

Then we write y = kx using the k value we found:

y = 5/3·x

Last, we want to find x when y is 30. So, we plug in 30 for y and solve for x:

30 = 5/3·x

÷5/3 ÷5/3

18 = x

4) If p varies directly with q, and p = 26 when q = 2, find p when q = 5.

This is the same process as the last problem, we just have p and q instead of y and x. We’ll start by writing y = max with p and q in place of y and x:

p = kq

Then we can the k value by plugging in the two values we’re first given.

26 = k·2

÷2 ÷2

13 = k

Then we plug the k value in:

p = kq

p = 13q

Last, we find p when q is 5. So, we plug in 5 for q and solve for p:

p = 13q

p = 13·5

p = 65

I hope that helped! If you have questions, or if you’d like to have a tutoring lesson, feel free to let me know. Take care!

Answer 2

Direct variation:

y = kx