Algebra II???

Question

Find all solutions to the following system of equations algebraically.

 

Y+X=2

(X+4)²+(Y-2)²=16

Answer 1

Substitute the first equation into the second equation.

 

From the first equation,

 

X = 2 - Y

 

((2 - Y) + 4)² + (Y - 2)² = 16

 

 

Next, we can simplify and solve for Y.

 

(6 - Y)² + (Y - 2)² = 16

 

(36 - 12Y + Y²) + (Y² - 4Y + 4) = 16

 

2Y² - 16Y + 40 = 16

 

 

Subtract 16 on both sides of equation.

 

2Y² - 16Y + 24 = 0

 

 

Factor out a 2 on left side.

 

2(Y² - 8Y + 12) = 0

 

 

Factor completely.

 

2(Y - 2)(Y - 6) = 0

 

 

Y = 2       and       Y = 6

 

Plug in these Y values to solve for X. 

 

In the graphical sense, since you have a line intersecting a circle, you will have two points of intersection.  Thus, you shall have two solutions.