Question

Help with one pre-calc question.

Answer 1

If y varies directly as x, and y = 5 when x = 8, find y when x = 4

Answer 2

@ganeshie8

Answer 3

I think you can set that up as a proportion, so you'd have 5/8=y/4 and you'd solve that for y.

Answer 4

Cross multiply?

Answer 5

5/2?

Answer 6

(5/2) is what I get, yeah

Answer 7

something varies directly to "something else" means

something = (some number) * "something else"
y = n * x

what's "n"? we dunno, but we know that when x = 8, y = 5 thus

\(\bf y = n\cdot x\\ \quad \\ \quad \\
x =8\qquad y = 5\\ \quad \\
5 = n\cdot 8\)

to find "n", that is, the "constant of variation", solve for "n"

Answer 8

once you find the "constant of variation", you plug it back in the original function and, you want to know what "y" is when x = 4, well, just plug the "n" value and "x" value and get "y"

Answer 9

n=5/8

Answer 10

So y= 5/8 * 4

Answer 11

\(\bf y = n\cdot x\\ \quad \\ \quad \\
x =8\qquad y = 5\\ \quad \\
5 = n\cdot 8 \implies \cfrac{5}{8}=n\\ \quad \\
\textit{so what's "y" when x = 4?}\\ \quad \\
y = n\cdot x\implies y = \cfrac{5}{8} x\implies y = \cfrac{5}{8} \cdot 4 \implies y = \cfrac{20}{8}\)

yeap

Answer 12

Or 5/2

Answer 13

Thank you both.

Answer 14

\(\bf y = \cfrac{\cancel{20}}{\cancel{8}}\implies\cfrac{5}{2}\)

Answer 15

yw