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Mathematics 12 Online
OpenStudy (anonymous):

Algebra plz help suppose y varies directly with x. Write a direct variation equation that relates x and y. Then find the value of y when x=12. y=-10 when x=2

OpenStudy (jhannybean):

a direct relationship means that as y increases, x increases as well when multiplied by a constant k. On the contrary, an inverse relationship is when y increases and x decreases, vise versa.

OpenStudy (anonymous):

ok I'm with you so far

OpenStudy (jhannybean):

direct variation: \( y = kx\) indirect variation: \(y = \dfrac{k}{x}\)

OpenStudy (jhannybean):

We take our equation for direct variation, \(y=kx\) and solve for \(k\) first.\[k = \frac{y}{x}\] We can solve for the second part of the problem first when x = 2, y = -10.

OpenStudy (jhannybean):

Plug those values in for the equation for k, and tell me what you get :)

OpenStudy (anonymous):

ok give me a second :)

OpenStudy (anonymous):

-10=k2

OpenStudy (jhannybean):

"2k"?

OpenStudy (jhannybean):

Oh no no. solve for \[k= \frac{-10}{2}\]

OpenStudy (anonymous):

oh ok sorry

OpenStudy (jhannybean):

No problem :)

OpenStudy (anonymous):

k=-5

OpenStudy (jhannybean):

Good. Now that we've got our value for k, we can revert back to our original euation, \(y=kx\) and solve for \(y\) when \(x=12\) and \(k=-5\) \[y=kx\]\[y = (-5)(12) = ~?\]

OpenStudy (anonymous):

-60

OpenStudy (jhannybean):

Good job :)

OpenStudy (anonymous):

thank you so much i understand how to do these now

OpenStudy (jhannybean):

Woo!

OpenStudy (anonymous):

:D

OpenStudy (anonymous):

hey can you help me with one more thing?

OpenStudy (jhannybean):

Hmm. I can try :)

OpenStudy (jhannybean):

You can close this question and post up a new question :)

OpenStudy (jhannybean):

That way you wont be spamming one question with multiple questions, haha

OpenStudy (anonymous):

ok

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