Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (k_lynn):

x+38x=y2+12 24x-64=12

OpenStudy (k_lynn):

@dan815 can you help me?

jagr2713 (jagr2713):

@amistre64

Nnesha (nnesha):

statement ?

OpenStudy (dan815):

Hello I am qualified helper reporting for duty

OpenStudy (k_lynn):

Thanks! :D

OpenStudy (dan815):

x+38x=y2+12 24x-64=12 can you make this more clear, are these supposed to be exponents

OpenStudy (dan815):

\[x+38^x=y^2+12\\ 24x-64=12\] something like this?

OpenStudy (k_lynn):

nope. No exponents. It looks like this. \[x+38x=y2+12\] \[24x-64=12\]

OpenStudy (k_lynn):

Would I subtract 38x from both sides to get x by itself.

OpenStudy (k_lynn):

Because I asked a Qualified Helper.

OpenStudy (anonymous):

What do you need help with? I'm not sure what you need

OpenStudy (dan815):

okay so in that case simple add all the like terms you see in the equation first

OpenStudy (dan815):

simply*

OpenStudy (k_lynn):

then would it just be 38x=y2+12?

OpenStudy (k_lynn):

because you add x+38x?

OpenStudy (dan815):

39*x

OpenStudy (k_lynn):

\[39x\] ?

OpenStudy (dan815):

\[Equation~1\\39x=2y+12\\ \\ \\ Equation~2\\ 24x-64 + 64=12+64\\ 24x=76\\ x=76/24\]

OpenStudy (k_lynn):

ohhh. Thank you!

OpenStudy (k_lynn):

now do I plug 76/24 in for every x I see in equation 2?

Nnesha (nnesha):

\(\color{blue}{\text{Originally Posted by}}\) @k_lynn \[39x\] ? \(\color{blue}{\text{End of Quote}}\) \[\large\rm \color{red}{1}x+38x\] yes there is invisible one so (1+38)x =39x :-)

OpenStudy (dan815):

Yes that's is right, now you can solve for y, as you know what x is: x=76/24 \[39*x=2y+12\\ 39*\color {red}{(76/24)}=2y+12\\ 39*76/24 \color {red} {-12} = 2y+12\color {red}{-12}\\ \frac{(39*76/24-12)}{\color {red}{2}}=\frac{2y}{\color {red}{2}}\\ y=\frac{(39*76/24-12)}{{2}}\]

OpenStudy (k_lynn):

|dw:1431298577831:dw|

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!