Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

x + 2y = -9 4x - 2y = 14 A. (-5,1) B. (-4,1) C. (1,-5) D. (1,-4)

OpenStudy (anonymous):

@Brill @Decarr432 @foolofatook @mthompson440

OpenStudy (calculusxy):

this seems like a system of equations problem... is there a certain way (substitution or elimination) that you are asked to solve this problem?

OpenStudy (anonymous):

idk @calculusxy

OpenStudy (calculusxy):

okay. so i think that using elimination will be much better. do you know how to solve systems using the elimination method?

OpenStudy (anonymous):

not realy

OpenStudy (calculusxy):

the elimination method means that we get rid of one variable (x or y) and then solve for the other variable that is present through zero pairs. for instance, if we have the equations \(x+3y=-14\) and \(-4x-3y=2\), we can get rid of the y variable because \(3y - 3y = 0y \text{ } \text{or simply 0}\).

OpenStudy (anonymous):

o got -4,1

OpenStudy (calculusxy):

then we would have the x variable to solve for \(x + (-4x) = -4x\) and \(-14 + 2 = -12\) \(-3x=-12\) \(x=4\)

OpenStudy (calculusxy):

sorry \(x + (-4x) = -3x\)

OpenStudy (anonymous):

ohh

OpenStudy (calculusxy):

similarly, in this situation \(x + 2y = -9\) \(4x - 2y = 14\) we can easily get rid of y because \(2y + (-2y) = 0y \text{ } \text{or 0}\) to solve for x we can have \(x + 4x = 5x\) and \(-9+14=5\) \(5x=5\) \(x=\text{?}\)

OpenStudy (calculusxy):

do you understand ?

OpenStudy (anonymous):

wount x be be 0

OpenStudy (calculusxy):

can you show me your work?

OpenStudy (anonymous):

well i dont really know how that stuffs works u know so used to saying it out load

OpenStudy (calculusxy):

do you understand the work that i showed?

OpenStudy (anonymous):

yeah man thank u

OpenStudy (calculusxy):

okay and remember to plug back in that x = 1 to x + 2y = -9 or the other equation and solve for y!

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!
Latest Questions
Arriyanalol: i need help
1 day ago 7 Replies 1 Medal
Arriyanalol: CONGRATS to me
1 day ago 8 Replies 2 Medals
Legomaster346: What colleges are good that are near Arab Alabama?
1 week ago 5 Replies 0 Medals
Arriyanalol: i need help
1 week ago 6 Replies 2 Medals
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!