# Math Help - simultaneous equation help

1. ## simultaneous equation help

i'd be really grateful if someone could complete this simultaneous equation with working out explaining each stage of the process. thanks.

y = 2x + 3

x2 + y2 = 2

2. I assume that the two equations are:

$y = 2x + 3$ and $x^2 + y^2 = 2$.

Substituting into equation 2 gives

$x^2 + (2x + 3)^2 = 2$.

Now expand, collect like terms, and solve.

3. Originally Posted by tylerdurden
i'd be really grateful if someone could complete this simultaneous equation with working out explaining each stage of the process. thanks.

y = 2x + 3

x2 + y2 = 2
You mean this?
$y\,=\,2x\,+\,3$
$x^2\,+\,y^2\,=\,2$

Plug the first equation into the second where you see the y:
$x^2 + (2x + 3)^2 = 2$

Solve for x:
$x^2 + 4x^2 + 12x + 9 = 2$
$5x^2 + 12x + 7 = 0$
$(x + 1)(5x + 7) = 0$

Set each factor equal to 0 and you'll get
$x = -1$ or $x = {-\frac{7}{5}}$

Plug both values into the first equation to get the corresponding y values.

EDIT: Beaten to it!

4. cheers guys. appreciate the help.

the only bit i'm not too sure on is how to get from

x2 + (2x+3)2 = 2 to

x2 + 4x2 + 12x = 9 =2

would really appreciate if you could explain what i have to do.

thanks.

5. Remember that $(2x + 3)^2 = (2x + 3)(2x + 3)$.

Expand it out.

6. 4x2 + 6x + 6x + 9

thanks for the help!

7. not quite sure how you get from

(x+1) (5x +7) = 0

to

x= -1 y = 7/5

and why

thanks

8. When you have two numbers being multiplied together to give $0$, that means at least one of them has to be $0$.

So if $(x + 1)(5x + 7) = 0$ then $x + 1 = 0$ or $5x+ 7 = 0$.