.
sketch each system of equations. Then solve the system by the substitution method
x^2 + y^2=6
y=x^2
~~~~~~~~~~~~~~~~~~
= 6, (1)
y =
. (2)
Plots
= 6 and y =
.
To solve the system, replace
in the equation (1) by y, according to (1). You will get
y + y^2 = 6, (3) or
= 0, (4)
(y+3)*(y-2) = 0 ====> the roots are
= -3 and
= 2.
Since y =
, y must be positive, so only the root y = 2 survives.
Then x = +/-
.
Answer. The solutions to the system (1),(2) are (x,y) = (
,
) and (x,y) = (
,
).
To see more solved samples of such systems, look into the lesson
- Solving the system of algebraic equations of degree 2
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of equations that are not linear".