Could someone please explain to me how to work out the following equations, I have tried various methods but can't come up with the correct answers.

x+y=2

2x-y=2

5x=8y

4x-3y+17=0

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- Feb 4th 2011, 12:27 PMhenriklarsson1Simultaneous Equations, 2 Problems.
Could someone please explain to me how to work out the following equations, I have tried various methods but can't come up with the correct answers.

x+y=2

2x-y=2

5x=8y

4x-3y+17=0 - Feb 4th 2011, 12:35 PMArchie Meade
- Feb 4th 2011, 01:17 PMhenriklarsson1
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. - Feb 4th 2011, 01:21 PMpickslides
- Feb 4th 2011, 01:36 PMWilmer
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... - Feb 4th 2011, 01:43 PMArchie Meade
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...

$\displaystyle 2x-y=5,\;\;\;x-2y=4$

Solve for x, so to set that up, get the same amount of y in both,

hence multiply the first by 2...

$\displaystyle 2x-y=5\Rightarrow\ 2(2x-y)=2(5)\Rightarrow\ 4x-2y=10$

Now we have

$\displaystyle 4x-2y=10,\;\;\;x-2y=4$

therefore subtract them to remove y

$\displaystyle (4x-2y)-(x-2y)=10-4\Rightarrow\ 4x-x-2y--2y=6$

$\displaystyle \Rightarrow\ 3x-2y+2y=6\Rightarrow\ 3x=6$

$\displaystyle 3x=3(2)\Rightarrow\ x=2$

$\displaystyle 2(2)-y=5\Rightarrow\ 4-y=5=4-(-1)\Rightarrow\ y=-1$

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

$\displaystyle x+y=2,\;\;\;2x-y=2\Rightarrow$

$\displaystyle 2(x+y)=2(2)\Rightarrow\ 2x+2y=4$

Subtract

$\displaystyle (2x+2y)-(2x-y)=4-2=2$

$\displaystyle 2x-2x+2y--y=2$

$\displaystyle 2y+y=2\Rightarrow\ 3y=2\Rightarrow\ y=\frac{2}{3}$ - Feb 4th 2011, 01:49 PMhenriklarsson1
Thanks guys, really appreciate your help!

- Feb 4th 2011, 04:54 PMbjhopper
Thanks to Wilmer for simpler solutions particularly problem 1. There is a sign error in your solution to 2.x and y must have the same signs.

bjh - Feb 4th 2011, 05:09 PMArchie Meade
- Feb 5th 2011, 05:18 AMbjhoppersimultaneous equations
Sorry for my mistake.I did this differently

5x=8y

4x-3y+17=0

x=8y/5

4*8y/5 -15y/5 =-17

32y/5 - 15y /5 = -17

17y/5 = -17

y =-5