Ask your own question, for FREE!
Algebra 23 Online
OpenStudy (anonymous):

Okay now I have no clue how to solve this one. Rewrite the equation in Ax+By=C form. Use integers for A,B,C

OpenStudy (anonymous):

OpenStudy (anonymous):

arh...

OpenStudy (anonymous):

no biggie clear the fraction by multiplying EVERYTHING by 2

OpenStudy (anonymous):

Right, So how exactly? Do you make 1/2 to just 2?

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

no dear \(2\times (-\frac{1}{2})=-1\)

OpenStudy (anonymous):

but don't forget to multiply all by 2 all the way across left hand side and right hand side,each term

OpenStudy (anonymous):

Oh. okay let me try to solve it :)

OpenStudy (anonymous):

kk

OpenStudy (anonymous):

|dw:1442282234913:dw|

OpenStudy (anonymous):

Right?

OpenStudy (anonymous):

yesh...

OpenStudy (anonymous):

wait im completely confused

OpenStudy (anonymous):

lets go slow again you want to get rid of the fraction right? so that the coefficients are "integers"

OpenStudy (anonymous):

right

OpenStudy (anonymous):

you got \[y=-\frac{1}{2}x-3\]

OpenStudy (anonymous):

to get rid of the two in the denominator, multiply both sides by \(2\) \[2\times y=2\times (-\frac{1}{2}x-3)\]

OpenStudy (anonymous):

distribute on the right hand side

OpenStudy (anonymous):

2y=-1x-3

OpenStudy (anonymous):

hmm looks like you forgot the distributive property you also have to multiply \(-3\) by \(2\)

OpenStudy (anonymous):

ohh so 2y=-1x-6

OpenStudy (anonymous):

correct?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

now add \(x\) to both sides and you are done

OpenStudy (anonymous):

1x+2y=6

OpenStudy (anonymous):

yeah but drop the \(1\) or your teacher will think you are daft

OpenStudy (anonymous):

so just x+2y=6

OpenStudy (anonymous):

damn, dropped the minus sign again on the right didn't you?

OpenStudy (anonymous):

x-2y=6???

OpenStudy (anonymous):

a common mistake actually now on the right, it was \(-6\) so it is still \(-6\)

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!