Help with math problem!

[780blzit]

Hi, I'm having a little bit of trouble with a math problem and I have no idea how to solve it because I'm absolutely terrible at fraction things..

 

 

Any help is appreciated! 

 

[TheHolyCawCaw]

isolate y in both equations and then do y=Y and you will get x

 

PS: If you want me to write it say so

[780blzit]

isolate y in both equations and then do y=Y and you will get x

 

PS: If you want me to write it say so

 

I'd love for you to write it! As I said, I'm terrible with fractions..

[tonychau77]

Make x or y the subject of one of one equation and substitute it into the other equation, then substitute the value obtained into any one of the equations.

I don't have pen and paper so I can't write it for you. And copying solutions isn't the best way to learn.

[780blzit]

Make x or y the subject of one of one equation and substitute it into the other equation, then substitute the value obtained into any one of the equations. I don't have pen and paper so I can't write it for you. And copying solutions isn't the best way to learn.

 

The problem isn't the technique, I know how to solve these types of problems. But I find it impossible when it's fractions.

I've done other problems like this in the book and I can do them with no problem.

[TheHolyCawCaw]

The problem isn't the technique, I know how to solve these types of problems. But I find it impossible when it's fractions.

I've done other problems like this in the book and I can do them with no problem.

k gimme a sec

[as96]

The first one should be:

x/3 + y/4 = 1/12

 

4x/12 + 3y/12 = 1/12

 

4x + 3y = 1

 

4x = 3y + 1

 

x = 3y/4 + 1/4

 

Then use that to find the y in the second and use the y to find x in the first

 

edit --- ops it's 1 not -1

[TheHolyCawCaw]

  Some little steps to help ya

OH ffs why does it turn it around meh

[780blzit]

The first one should be:

x/3 + y/4 = 1/12

 

4x/12 + 3y/12 = 1/12

 

4x + 3y = 1

 

4x = 3y + 1

 

x = 3y/4 + 1/4

 

Then use that to find the y in the second and use the y to find x in the first

 

edit --- ops it's 1 not -1

 

 

Shouldn't it be 4x = 1-3y ?

 

Anyway, the x and y has the same value in both equations.

[as96]

Shouldn't it be 4x = 1-3y ?

 

Anyway, the x and y has the same value in both equations.

Yes -3y, sorry it's not easy to do math on the edit post of ltt

[780blzit]

  Some little steps to help ya

OH ffs why does it turn it around meh

 

You did it! Thank you soo much!

[tonychau77]

x/3 + y/4 = 1/12

12(x/3 + y/4) = 12(1/12)

4x + 3y = 1

x = (1 - 3y)/4 ...(1)

x/2 - y/6 = 2/3

6(x/2 - y/6) = 6(2/3)

3x - y = 4 ...(2)

Substitute (1) into (2)

3((1 - 3y)/4) - y = 4

4*3((1 - 3y)/4) - 4y = 16

3(1 - 3y) - 4y = 16

3 - 9y - 4y = 16

-13y = 16 - 3

y = -1

Substitute y = -1 into (1)

x = (1 - 3(-1))/4 = 1

Edit: ninjaed a few minutes ago

[Bloodyvalley]

invert everything and boom