SOLUTION: really need help
solve the system of nonlinear equations for x and y. sketch a graph of each equation showing the intersection points of the line and the parabola. x^2 + y=4 an
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solve the system of nonlinear equations for x and y. sketch a graph of each equation showing the intersection points of the line and the parabola. x^2 + y=4 an
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Question 62716: really need help
solve the system of nonlinear equations for x and y. sketch a graph of each equation showing the intersection points of the line and the parabola. x^2 + y=4 and 2x + y=1 thanks so much Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! really need help
solve the system of nonlinear equations for x and y. sketch a graph of each equation showing the intersection points of the line and the parabola.
x^2 + y = 4 and 2x + y = 1
Solve it like any system, using the elimination method
x^2 + y = 4
2x + y = 1
------------ subtract
x^2 - 2x + 0y = 3; y is eliminated so you have:
x^2 - 2x - 3 = 0
Factor this to:
(x-3)(x+1) = 0
x = +3 and x = -1, the solution to this system
:
To graph this (you know the intersection points will be x=+3 and x=-1 from the above solution)
:
First arrange the equations in the general form (y=)
x^2 + y = 4
y = 4 - x^2
:
2x + y = 1
y = 1 - 2x
:
Write up a table to plot, y = 4 - x^2
Start with x = -5: ie y = 4 - (-5^2) = - 21
:
x | y
-----
-5|-21
-4|-12
-3|-5
-2|0
-1|+3; point of intersection
0 |+4
+2|0
+3|-5; point of intersection
+4|-12
:
Do the same with y = 1 - 2x, it's a straight line so a couple plots will do it
x | y
-----
-3|+7
-1|+3; point of intersection
0 |+1
+1|-1
+3|-5; point of intersection
+4|-7
:
:
Your graph should look this
:
Do you see the relationship between our solutions and the graph intersections?
you can email me if you have a question on this?
