How to Solve a System with Two Equations and Two Unknowns?

[Raza]

I was reading the news and I found question 11 to be confusing. I never learned how to solve systems.

Question:
Solve the following system for x and y
1)$$\frac{x}{8}-\frac{y}{2}=1$$-------------------------2)$$\frac{x}{3}=y+\frac{2}{3}$$

WORK:

1)$$\frac{x}{8}-\frac{y}{2}=1$$-------------------------2)$$\frac{x}{3}-\frac{2}{3}=y$$

1)$$=\frac{x}{4}-y=2$$

$$x-y=8$$-------------------------$$x-2=3y$$

Edit: Changed the "news" link.

 

[arildno]

Not surprising, since what you've written is meaningless.

You've got TWO numbers, called x and y, that simultaneously satisfies TWO equations.

Please set up the ORIGINAL equations, your attempt of solving them is totally flawed.

 

[Raza]

I am sorry that I had the wrong link. Now, it's corrected.

 

[berkeman]

You've still got a lot of algebra errors in your GIF file and your post. It should look more like:

-1-
x/8 - y/2 = 1, which simplifies to x - 4y = 8

-2-
x/3 = y + 2/3, which simplifies to x - 3y = 2

Then you combine the simplified equations 1 & 2 to eliminate one variable, so you can solve for the other one. Do you have an idea of how to combine the simplified 1 & 2 to solve for y? Hint -- think about subtracting equations...

 

[HallsofIvy]

It looked like you were starting out correctly by solving the second equation for y: y= x/3- 2/3. But then your x/4- y= 2 is puzzling. I don't see where it comes from. Presumably the reason for solving for y was to substitute for y in the first equation: x/8- y/2= x/8 - (1/2)(x/3- 2/3)= x/8- x/6- 1/3= 1.
Then 3x/24- 4x/24- 1/3= -x/24- 1/3= 1. Adding 1/3 to both sides,
-x/24= 1+ 1/3= 4/3 so x= (4.24)/(3)= 8(4)= 32. Then y= 32/3- 2/3= 30/3= 10.

 

[Raza]

I finally found how to do this.
From equation 1, I found $$x=8+4y$$
I used that equation as a replacment of x in equation 2.
I got x= -16 and y= -6.

 

[mattmns]

Yep that works perfectly.