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

Math help

OpenStudy (anonymous):

OpenStudy (anonymous):

A & B

OpenStudy (amistre64):

are you able to solve an equation when there is just one variable?

OpenStudy (anonymous):

yeah

OpenStudy (amistre64):

then substitution is a method that should work for the first one: we see from the setup that x+y = 65, and by annoyance they say y is a vegetable instead of using a variable named v ... let x = 65 - y for all x parts in the setup

OpenStudy (amistre64):

16x + 12y = 972 substitute in for x 16(65-y) + 12y = 972 now you have a one variable equation ... solve for y

OpenStudy (anonymous):

how i solve for x?

OpenStudy (amistre64):

x = 65 - y so once you know what y is ...

OpenStudy (anonymous):

wait im confused

OpenStudy (amistre64):

tell me what your confusion is ... im terrible at reading minds

OpenStudy (anonymous):

what u just told me x = 65 - y am i suppose find y or x?

OpenStudy (amistre64):

you are spose to find both the idea of substitution is that you can solve one equation for one variable after knowing what its value is you can then determine the other value by default

OpenStudy (amistre64):

these are your equations that must be true for some xy values right? 16x + 12y = 972 x+y = 65

OpenStudy (amistre64):

we can find an equation for x (or y) ... i chose x since x+y = 65; we know that for some value of y: x = 65 - y this has to be true for every single x part that we have in the setup, so lets use that in the first equation: 16x + 12y = 972 ^^ replace x with 65-y 16(65-y) + 12y = 972 now we can solve for y ... and once we know y; x = 65-y

OpenStudy (amistre64):

once x and y are known, then we can start assessing the options: does: 12y = 48 ? does: x = 48 ? does: 3x = y ? does: y = 48 ?

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!