Please Help

[jwt13]

Ok so I have my Algebra 1 exam on tuesday and I need help to understand how to do a problem like this.

 

2y=3/2X - 14

 

I need to know how to solve for Y I,ve learned it before it seems simple and was one of the first things we learned but i cant find nothing on it in my notes or my book so if you could explain please do so

[praguepride]

First you solve for x then you replace the x with the y and you can then solve it

[praguepride]

2y=3/2X - 14

 

2y + 14 = 3/2x

 

(4y + 28)/3 = x

 

 

Then you go back to the original equation and replace x with 3y + 21 so

 

2y = 3/2((4y + 28)/3) - 14

 

do the math...

 

 

http://www.jamesbrennan.org/algebra/systems/substitution_method.htm

[d_w_w]

Making it too complicated I think.

 

Just divide both sides of the equation by 2, no?

[d_w_w]

2y=3/2X - 14

 

2y + 14 = 3/2x

 

(4y + 28)/3 = x

 

 

Then you go back to the original equation and replace x with 3y + 21 so

 

2y = 3/2((4y + 28)/3) - 14

 

do the math...

 

If you do the math:

 

2y = (3/2)(4y/3) + (3/2)(28/3) - 14

2y = 2y + 14 - 14

2y = 2y

y = y

 

Can't use substitution with a single equation.

[jwt13]

Making it too complicated I think.

 

Just divide both sides of the equation by 2, no?

 

Thats what I'm thinking we did but im not 100% sure.

[Craig Edwards]

y=3x/4-7 is the answer. It is hard to write how to do it on these broads but basicly you divide both ends by 2 so you can get that 2 off the y and then times the 2 on the 3/2 that brings it to 4 an you always put the variable on the top. I'm not that good at explaning it but that is how you do it for that one problem