Find the solution/point of intersection of the following quadratic-linear system. Show steps appropriately: y+x^2=-6x-3 and x+y-1=0. Solve using both an algebraic and graphical method.
To solve the quadratic-linear system consisting of equations y + x^2 = -6x - 3 and x + y - 1 = 0, we will first solve it algebraically and then verify the solution graphically.
Algebraic Solution:
Step 1: We can rewrite the first equation y + x^2 = -6x - 3 as y = -x^2 - 6x - 3 by moving all terms to one side of the equation.
Step 2: Substitute the expression for y from the first equation into the second equation:
x + (-x^2 - 6x - 3) - 1 = 0
Simplifying the equation:
x - x^2 - 6x - 4 = 0
Rearranging the terms:
-x^2 - 5x - 4 = 0
Step 3: Solve the quadratic equation -x^2 - 5x - 4 = 0 by factoring, completing the square, or using the quadratic formula.
By factoring, we can rewrite the equation as:
(-x - 4)(x + 1) = 0
Setting each factor equal to zero:
-x - 4 = 0 or x + 1 = 0
Solving for x:
x = -4 or x = -1
Step 4: Substitute the values of x back into either of the original equations to solve for y.
For x = -4:
y + (-4)^2 = -6(-4) - 3
y + 16 = 24 - 3
y + 16 = 21
y = 21 - 16
y = 5
The first solution is (x, y) = (-4, 5).
For x = -1:
y + (-1)^2 = -6(-1) - 3
y + 1 = 6 - 3
y + 1 = 3
y = 3 - 1
y = 2
The second solution is (x, y) = (-1, 2).
The quadratic-linear system has two solutions: (-4, 5) and (-1, 2).
Graphical Solution:
Step 1: Plot the graphs of the two equations on the same coordinate plane.
The first equation, y + x^2 = -6x - 3, represents a downward-facing parabola.
The second equation, x + y - 1 = 0, represents a straight line.
Step 2: Find the point(s) of intersection of the two graphs.
The points of intersection will represent the solution(s) to the quadratic-linear system.
Step 3: Verify that the coordinates of the points of intersection match the solutions obtained algebraically.
By both algebraic and graphical methods, we found that the quadratic-linear system has two solutions: (-4, 5) and (-1, 2).