# Simultaneous equation

1. Jan 24, 2010

### Gringo123

I have just tried to solve the following simultaneous equation.

x + 2y = 5
x squared + y squared = 36

I guess that the 1st step is to rearrange the 1st equation to give:

x = 5 - 2y

From that point onwards I have no idea what to do! Can anybody guide me through this?
Thanks a lot!

2. Jan 24, 2010

### Fightfish

Substitute x with (5 - 2y) in the second equation to give:
$$(5 - 2y)^2 + y^2 = 36$$
That's how to solve almost all simultaneous equations; the idea is to eliminate one variable in one of the equations so as to obtain an equation solely in terms of one variable, which can then be solved.
Carry on.

3. Jan 24, 2010

### Gringo123

so...
(5 - 2y) x (5 - 2y) + y squared = 36
multipying out the brackets gives us:
(25 - 10y - 10y + 4y) + y squared = 36
so...
(25 - 16y) + y squared = 36
I've had a long hrad think at this stage and i'm stuck again! Am I going in the right direction?
Thanks for your help!

4. Jan 24, 2010

### Fightfish

Yes, you are going in the right direction...but you made a mistake in expanding the brackets.
$$(5 - 2y)^2 + y^2 = 36$$
$$(25 - 20y + 4y^2) + y^2 = 36$$
$$5y^2 - 20y + 25 = 36$$
$$5y^2 - 20y - 11 = 0$$
Now its just a matter of solving for y from this quadratic equation.

5. Jan 24, 2010

### Gringo123

just going back to the expanding of the brackets, i worked it outr as follows:
(5 - 2y) x (5 - 2y)
so..
5 x 5 = 25
5 x -2y = -10y
-2y x 5 = -10y
-2y x -2y = 4y (why does this equal 4y squared?)
By the way, how do you type the little 2 tyat indicated squared?
Thanks again

6. Jan 24, 2010

### Dollydaggerxo

because a minus x a minus is a plus. -2 x -2 = 4
y x y = y^2

so -2y x -2y = 4y^2

7. Jan 24, 2010

### tiny-tim

Hi Gringo123!

[NOPARSE]Type "y2",[/NOPARSE] and it comes out y2

(or use the X2 tag just above the Reply box )

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