1. ## Simultaneous Equations

x^2+y^2=2
3x-2y=5

2. Originally Posted by jerrymalta
x^2+y^2=2
3x-2y=5

From the second equation:

$x=\dfrac{5+2y}{3}$

Now, substitute in the first one.

Fernando Revilla

3. I thought this would be relatively straightforward but I can't seem to solve myself.

x = (5 + 2y)/3

substitute into x^2 + y^2 = 2

((5+2y)/3)^2 + y^2 = 2

25/9 + 20/9y + 4/9y^2 + y^2 = 2

13/9y^2 + 20/9y + 7/9 = 0

x 9: 13y^2 + 20y + 7 = 0

using -b+-rt b^2 - 4ac
----------------
2a

a=13, b=20, c=7
y= -13/20 or y = -7/20

substituting into 3x-2y=5
when y=-13/20, x = 37/50
when y = -7/20, x = 43/30

But this is incorrect because putting these x and y values into circle equation don't produce 2. Can anyone see where I have gone wrong?

4. Originally Posted by angypangy
I thought this would be relatively straightforward but I can't seem to solve myself.

x = (5 + 2y)/3

substitute into x^2 + y^2 = 2

((5+2y)/3)^2 + y^2 = 2

25/9 + 20/9y + 4/9y^2 + y^2 = 2

13/9y^2 + 20/9y + 7/9 = 0

x 9: 13y^2 + 20y + 7 = 0

using -b+-rt b^2 - 4ac
----------------
2a

a=13, b=20, c=7

it's fine to here

y= -13/20 or y = -7/20

substituting into 3x-2y=5
when y=-13/20, x = 37/50
when y = -7/20, x = 43/30

But this is incorrect because putting these x and y values into circle equation don't produce 2. Can anyone see where I have gone wrong?
$\displaystyle\ 13y^2+20y+7=0\Rightarrow\ y=\frac{-20\pm\sqrt{400-364}}{26}$

$\displaystyle\Rightarrow\ y=\frac{-20\pm6}{26}$

5. Thanks. I realised I was being careless. I will have to learn to be more accurate.