Thanks for your help but in the back of my book it says equation 1 is x= 4/3 y= 2/3.
I'll give you an example of one I know how to work out: 2x-y=5, x-2y=4. (2,-1)
The method I have been taught is: 2(2x-y=5) x-2y=4
4x-2y=10 x-2y=4
3x=6
x=2
-y= 5-4
y=-1 (sorry about the mess)
I'm still unsure how to work out the previous ones using this method.
That's known as "elimination"
Your 1st problem:
x+y=2
2x-y=2
Add the 2 equations:
3x = 4 ; OK?
Your 2nd problem:
5x=8y
4x-3y+17=0
Rewrite:
5x - 8y = 0 [1]
4x - 3y = -17 [2]
[1] * 4, [2] * 5:
20x - 32y = 0
20x - 15y = -85
Finish it...
If you can't, you need classroom help...
It's very very similar.
You should understand why the steps are being taken.
It just makes the process so much easier.
Let's just examine the steps of the example you gave that you know how to do...
Solve for x, so to set that up, get the same amount of y in both,
hence multiply the first by 2...
Now we have
therefore subtract them to remove y
So that's why pickslides said it's the same thing.
However, you can choose to add or subtract.
You subtracted above because you had the same multiple of y.
We can do that also for the questions
Subtract