simultaneous equation help

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July 3rd 2010, 05:58 AM
tylerdurden

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
July 3rd 2010, 06:04 AM
Prove It

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.
July 3rd 2010, 06:08 AM
eumyang

Quote:

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! ;)
July 3rd 2010, 06:20 AM
tylerdurden

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.
July 3rd 2010, 06:35 AM
Prove It

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

Expand it out.
July 3rd 2010, 06:57 AM
tylerdurden

4x2 + 6x + 6x + 9

thanks for the help!
July 3rd 2010, 07:55 AM
tylerdurden

not quite sure how you get from

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

to

x= -1 y = 7/5

and why

thanks
July 3rd 2010, 06:04 PM
Prove It

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 $5x+ 7 = 0$ .