# Thread: Method of Substitution! Help!

1. ## Method of Substitution! Help!

Use the method of substitution to solve the system

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

Thanks!

2. 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.

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

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

4. hmm.. im sorry but i just cant seem to be getting the correct answer after plugging it in! This is driving me nuts...

5. 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 ...