Question 1088793
.
sketch each system of equations. Then solve the system by the substitution method

x^2 + y^2=6
y=x^2
~~~~~~~~~~~~~~~~~~


<pre>
{{{x^2 + y^2}}} = 6,     (1)
y = {{{x^2}}}.        (2)


{{{graph( 330, 330, -5.5, 5.5, -5.5, 5.5,
          sqrt(6 - x^2), -sqrt(6 - x^2), x^2
)}}}


Plots {{{x^2 + y^2}}} = 6   and  y = {{{x^2}}}.



To solve the system, replace {{{x^2}}} in the equation (1) by y, according to (1). You will get

y + y^2 = 6,         (3)   or

{{{y^2 + y - 6}}} = 0,         (4)

(y+3)*(y-2) = 0  ====>  the roots are {{{y[1]}}} = -3  and  {{{y[2]}}} = 2.


Since y = {{{x^2}}},  y must be positive, so only the root y = 2 survives.


Then x = +/- {{{sqrt(2)}}}.


<U>Answer</U>. The solutions to the system  (1),(2)  are  (x,y) = ({{{sqrt(2)}}},{{{2}}})  and  (x,y) = ({{{-sqrt(2)}}},{{{2}}}).
</pre>

To see more solved samples of such systems, look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Systems-of-equations/Solving-the-system-of-algebraic-equations-of-degree-2.lesson>Solving the system of algebraic equations of degree 2</A> 

in this site.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic &nbsp;"<U>Systems of equations that are not linear</U>".