# Simultaneous Equations, 2 Problems.

• Feb 4th 2011, 12:27 PM
Simultaneous 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 PM
Quote:

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

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

Quote:

5x=8y
4x-3y+17=0
Get equal coefficients for y

$\displaystyle 5x=8y\Rightarrow\ 15x=24y$

$\displaystyle 4x-3y+17=0\Rightarrow\ 8(4x-3y+17)=4(0)=0$

Try the same technique as for the first part.
• Feb 4th 2011, 01:17 PM
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

I'm still unsure how to work out the previous ones using this method.
• Feb 4th 2011, 01:21 PM
pickslides
Quote:

Thanks for your help but in the back of my book it says equation 1 is x= 4/3 y= 2/3.

That is exactly the answer I get when using Archie's method!
• Feb 4th 2011, 01:36 PM
Wilmer
Quote:

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

I'm still unsure how to work out the previous ones using this method.

That's known as "elimination"
x+y=2
2x-y=2
3x = 4 ; OK?

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 PM
Quote:

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

I'm still unsure how to work out the previous ones using this method.

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 PM
Thanks guys, really appreciate your help!
• Feb 4th 2011, 04:54 PM
bjhopper
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 PM
Quote:

Originally Posted by bjhopper
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

The solution was correct, the signs are opposite.
• Feb 5th 2011, 05:18 AM
bjhopper
simultaneous 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