solve (x+2)^2+7y^2=5 and x^2+4x-y=6 by substitution or elimination.

Question

solve by substitution or elimination.

Answer 1

x^2+4x-y = 6

Subtract 6 from each side.

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

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

Add 10,y to each side.

x^2+4x+4 = y+10

(x+2)^2 = y+10 --------> (1)

(x+2)^2+7y^2 = 5 -----------> (2)

From equation (1) substitute the value of (x+2)^2 in equation (2)

y+10+7y^2 = 5

Subtract 5 from each side.

7y^2+y +10-5 = 0

7y^2+y+5 = 0

Compare the quadratic equatiion to ax^2+bx+c = 0

Roots are x = (-b±√(b^2-4ac))/2a.

y = [-1±√1-(4*7*5)]/2*7

y = (-1±√1-140)/14

y = (-1±√-139)/14

negitive square root is not possible.

So the system has no solution.