Help with one pre-calc question.

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

@ganeshie8

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.

Cross multiply?

5/2?

(5/2) is what I get, yeah

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"

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"

n=5/8

So y= 5/8 * 4

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

Or 5/2

Thank you both.

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

yw

Join our real-time social learning platform and learn together with your friends!