Help with systems of equations?

Question

How can i solve this system of equations algebraically?  Please show steps

x²-6x+y²+4y-10=0

y=x+2

 

Answer 1

Substitute x + 2 for the 'y' in the big equation.

 

x² - 6x + (x + 2)² + 4(x + 2) - 10 = 0

Square the binomial and use the distributive property for multiplying 4 times the binomial, x + 2

 

x² - 6x + x² + 4x + 4 + 4x + 8 - 10 = 0

Combine like terms.

2x² + 2x + 2 = 0

Divide each term by 2.

x² + x + 1 = 0

Factoring the trinomial is a problem unless you know how to use the Quadratic Formula. 

Are you sure you copied the problem correctly, Omar?

Mary Ann

 

Answer 2

I agree with Mary, the two equations never intersect, and therefore have no solution as a system. 

 

to continue:

 

X^2+x+1 = 0

X^2 + x + 1/4 + 3/4 = 0

(x+1/2)^2 = -3/4

is unsolvable, because any real number squared must be positive, but if sqrt(-1) is allowed, then it continues:

x+1/2 = (+ or -)sqrt(-3/4)

x = (+ or -)sqrt(-1)*sqrt(3/4) -1/2 

x = (+ or -)i*sqrt(3/4) -1/2                            ("i" is a common way to write sqrt(-1))

 

and since y = x+2,

y = (+ or -)i*sqrt(3/4) -1/2  + 2

y = (+ or -)i*sqrt(3/4) + 3/2

 

this can be shown by graphing, as one equation is a circle centered at (3,-2) with radius Sqrt(23), the other is a line of slope one and y-intersect 2. These paths never intersect. 

 

work:

x^2 - 6x + y^2 + 4y - 10 = 0

X^2 - 6x + 9 - 9 + y^2 + 4y + 4 - 4 - 10 = 0

(x-3)^2 - 9 + (y+2)^2 - 4 - 10 = 0

(x-3)^2 + (y+2)^2 = 23 // is a circle

 

Are you allowed to use imaginary numbers to solve?