Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Hi, it says to use the substitution method.... 3x+5y=3/5 y=5x-1

OpenStudy (anonymous):

I just want to know what to do. I have never done one before

OpenStudy (lgbasallote):

this kid's gonna help you real good dunworry :D taught him myself hahaha jk

OpenStudy (lgbasallote):

seems he's putting very much details into this..just be patient :D seems like it's gonna be good

OpenStudy (anonymous):

igba is right

OpenStudy (anonymous):

Method of substitution means that we have to put x=.... or y=.... onto one equation which is in terms of x and y to eliminate one of the variables in order to solve the equation So y=5x-1 can help us elimnate y in 3x+5y=3/5. Sub y=5x-1 into 3x+5y=3/5 3x+5(5x-1) =3/5 3x+25x-5=3/5 28x-5=3/5 28x=28/5 x=1/5 we can now get the value of x. Don't forget that we have to find y also. So, put x=1/5 into y=5x-1, y=5(1/5)-1=0 The answer is x=1/5 and y=0

OpenStudy (lgbasallote):

there we go :DD

OpenStudy (anonymous):

how did you get 28/5???

OpenStudy (anonymous):

28x-5=3/5 as we want to find x so we want the answer to be expressed in x=... first, to cancel -5 on LHS, 28x-5+5=3/5+5 28x = 28/5 there is 28 on LHS. We have to cancel it 28x/28=(28/5)/28 x=1/5

OpenStudy (lgbasallote):

3/5 + 25/5 = 28/5...additional info ^_^

OpenStudy (anonymous):

thank you so much

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!