Asked by Kyle
Solve the system
x^2+y^2-4x+2y=20
4x +3y =5
x^2+y^2-4x+2y=20
4x +3y =5
Answers
Answered by
MathMate
Rewrite
x^2+y^2-4x+2y=20
as
(x-2)^2+(y+1)^2=25=5^2 ...(1)
represents a circle of radius 5 with centre (2,-1).
which intersects with the line
4x+3y=5 ...(2)
So write
y=(5-4x)/3 and substitute in the circle equation (1)
to get (x-2)^2+(((5-4x)/3)+1)^2=5^2
Solve for x in the above quadratic to get:
x=5 or x=-1
Find y accordingly.
x^2+y^2-4x+2y=20
as
(x-2)^2+(y+1)^2=25=5^2 ...(1)
represents a circle of radius 5 with centre (2,-1).
which intersects with the line
4x+3y=5 ...(2)
So write
y=(5-4x)/3 and substitute in the circle equation (1)
to get (x-2)^2+(((5-4x)/3)+1)^2=5^2
Solve for x in the above quadratic to get:
x=5 or x=-1
Find y accordingly.
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