Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (calculusxy):

MEDAL WILL BE AWARDED. Please help. See attachment

OpenStudy (calculusxy):

OpenStudy (calculusxy):

@amistre64 @hartnn @ranga @AkashdeepDeb

OpenStudy (tkhunny):

Step #1 - Name Stuff. Q1 - Name What? A1 - What does it want? Name that. G = Price of a stick of gum Start traslating the problem statement.

OpenStudy (anonymous):

Let \(x\) be the quantity of gum purchased, and \(y\) be the price of one piece. You have the system, \[\begin{cases}xy=2.16\\ (x+3)(y-1)=2.16\end{cases}\] Solve for \(x\).

OpenStudy (calculusxy):

How do I solve for x?

OpenStudy (anonymous):

Substitution should work. From the first equation, you have \(y=\dfrac{2.16}{x}\). Expanding the second equation, you have \[xy+3y-x-3=2.16\] Plug in the expression for \(y\): \[x\left(\frac{2.16}{x}\right)+3\left(\frac{2.16}{x}\right)-x-3=2.16\\ 2.16+\frac{6.48}{x}-x-3=2.16\\ 0=x+3-\frac{6.48}{x}\]

OpenStudy (calculusxy):

Is there any easier way than that please? I am not up to that level yet :)

OpenStudy (anonymous):

It's a quadratic in disguise. Have you worked with quadratic equations? Multiply both sides by \(x\): \[0=x^2+3x-6.48\]

OpenStudy (calculusxy):

No

OpenStudy (anonymous):

So you've never seen this formula before? \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

OpenStudy (calculusxy):

no

OpenStudy (ranga):

Are you allowed to plot graphs and see where they intersect and find the solution that way?

OpenStudy (calculusxy):

well i dont know. i mean is there anything easier than a quadratic equation or something?

OpenStudy (ranga):

The $2.16 may have to be converted to pennies in the equations.

OpenStudy (calculusxy):

so that would be 216 pennies right?

OpenStudy (ranga):

Yes. You can write various factors of 216 and create a table and from there you may be able to find the answer.

OpenStudy (calculusxy):

any easier way than that?

OpenStudy (ranga):

Try out each answer choice. And you can eliminate the ones that don't work.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

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