# Method of Substitution! Help!

• Apr 15th 2008, 05:34 PM
luigiandme
Method of Substitution! Help!
Use the method of substitution to solve the system

http://webwork.tuhsd.k12.az.us/webwo...5e21a21381.png

Answer for should be in (_,_),(_,_) form or just (_,_) form.

Thanks!
• Apr 15th 2008, 05:40 PM
o_O
Since $x + y = 1$, you can deduce that ${\color{blue}y = 1 - x}$.

Plug this to the original equation:
$x^{2} + {\color{blue} y^{2}} = 8$
$x^{2} + {\color{blue} (1-x)^{2}} = 8$

Continue on and solve for x then y.
• Apr 15th 2008, 05:47 PM
TheEmptySet
Quote:

Originally Posted by luigiandme
Use the method of substitution to solve the system

http://webwork.tuhsd.k12.az.us/webwo...5e21a21381.png

Answer for should be in (_,_),(_,_) form or just (_,_) form.

Thanks!

lets isloate one variable in the bottom equation lets say y.

$x+y=1 \iff y=1-x$

Now will will sub this into the first equation for y.

$x^2+(1-x)^2=8$ expand the left hand side

$x^2+1-2x+x^2=8 \iff 2x^2-2x=7$

$2(x^2+x+\frac{1}{4})=7+2\cdot \frac{1}{4}$

$2(x+\frac{1}{2})^2=\frac{15}{2} \iff (x+\frac{1}{2})^2=\frac{15}{4}$

$x=-\frac{1}{2} \pm \frac{\sqrt{15}}{2}$

plug this back in to find the values for y
• Apr 15th 2008, 09:03 PM
luigiandme
hmm.. im sorry but i just cant seem to be getting the correct answer after plugging it in! This is driving me nuts...
• Apr 15th 2008, 09:11 PM
o_O
Quote:

Originally Posted by TheEmptySet
lets isloate one variable in the bottom equation lets say y.

$x+y=1 \iff y=1-x$

Now will will sub this into the first equation for y.

$x^2+(1-x)^2=8$ expand the left hand side

$x^2+1-2x+x^2=8 \iff 2x^2-2x=7$

$2(x^2 {\color{red}-}x+\frac{1}{4})=7+2\cdot \frac{1}{4}$

$2(x{\color{red}-}\frac{1}{2})^2=\frac{15}{2} \iff (x{\color{red} -}\frac{1}{2})^2=\frac{15}{4}$

$x={\color{red}+}\frac{1}{2} \pm \frac{\sqrt{15}}{2}$

plug this back in to find the values for y

A little mistake ...