Could some one help with solving the addition method Please

[passionflower_40]

A system with fractions


x/3 - y/ 2 = - 5/6

 

[happy]

x/3 - y/ 2 = - 5/6

6x/3 - 6y/2 = (-5/6)6

2x- 3y = -5

solve for x or y?

 

[stapel]

A "system" requires at least two equations. You posted only one.

Please reply with clarification.

Thank you.

Eliz.

 

[happy]

 

[passionflower_40]

Reposting my help question again

stapels you is so right there was another fraction. I have been racking my brains in how to do it and been up all night to get this solve for today.

x/3 - y/2 = - 5/6

x/5 - y/3 = - 3/5

 

[passionflower_40]

x/3 - y/ 2 = - 5/6

6x/3 - 6y/2 = (-5/6)6

2x- 3y = -5

solve for x or y?

I want to thank you, I forgot the other fraction

it is x/5 - y/3 = - 3/5

 

[Denis]

x/3 - y/2 = - 5/6 [1]
x/5 - y/3 = - 3/5 [2]

Multiply [1] by 6:
2x - 3y = -5
2x = 3y - 5
x = (3y - 5) / 2 [3]

Multiply [2] by 15:
3x - 5y = -9 [4]

Substitute [3] in [4]:
3(3y - 5) / 2 - 5y = -9
Multiply by 2:
3(3y - 5) - 10y = -18
9y - 15 - 10y = -18
-y = -3
Multiply by -1:
y = 3

Substitute y=3 in [4]:
3x - 5(3) = -9
3x - 15 = -9
3x = 6
x = 6/3
x = 2

Let me know what you didn't understand...if anything.

 

[passionflower_40]

x/3 - y/2 = - 5/6 [1]
x/5 - y/3 = - 3/5 [2]

Multiply [1] by 6:
2x - 3y = -5
2x = 3y - 5
x = (3y - 5) / 2 [3]

Multiply [2] by 15:
3x - 5y = -9 [4]

Substitute [3] in [4]:
3(3y - 5) / 2 - 5y = -9
Multiply by 2:
3(3y - 5) - 10y = -18
9y - 15 - 10y = -18
-y = -3
Multiply by -1:
y = 3

Substitute y=3 in [4]:
3x - 5(3) = -9
3x - 15 = -9
3x = 6
x = 6/3
x = 2

Let me know what you didn't understand...if anything.

Dennis this really did help you broke it down for me to see how it is done I believe I am getting it. I rewrote it on paper to see if I came up witht he same as u did. You demonstrated to me a precise step-by-step that is better than what my math book had shown me. Thank you so much.