simultaneous equation help

Jul 2010
4
0
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
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
I assume that the two equations are:

\(\displaystyle y = 2x + 3\) and \(\displaystyle x^2 + y^2 = 2\).

Substituting into equation 2 gives

\(\displaystyle x^2 + (2x + 3)^2 = 2\).

Now expand, collect like terms, and solve.
 
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Jan 2010
278
138
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?
\(\displaystyle y\,=\,2x\,+\,3\)
\(\displaystyle x^2\,+\,y^2\,=\,2\)

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

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

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

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


EDIT: Beaten to it! ;)
 
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Jul 2010
4
0
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.
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,897
5,001
Remember that \(\displaystyle (2x + 3)^2 = (2x + 3)(2x + 3)\).

Expand it out.
 
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Jul 2010
4
0
4x2 + 6x + 6x + 9

thanks for the help!
 
Last edited:
Jul 2010
4
0
not quite sure how you get from

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

to

x= -1 y = 7/5

and why

thanks
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,897
5,001
When you have two numbers being multiplied together to give \(\displaystyle 0\), that means at least one of them has to be \(\displaystyle 0\).

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