Bill said that the line Y=6 cuts the curve x^2+y^2=25 at two points
1. by eliminating y show that bill is incorrect.
2.by eliminating y, find solutions to the simultaneous equations
x^2 +y^2=25
y=2x-2
2007-02-25 04:08:10 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1) x^2+6^2=25
x^2 = -11
No real x, thus Y=6 does not cut the curve.
2) Substitute for y
x^2+(2x-2)^2=25
5x^2 - 8x - 21=0
(5x + 7)(x - 3)=0
The two solutions are
x = -7/5, y = -24/5
x = 3, y = 4
2007-02-25 05:29:11 · answer #1 · answered by nor^ron 3 · 0⤊ 0⤋
sparkling up the equations first for y so that you'll graph/draw both lines y = -6/7x + 6 y = x - 2 After graphing locate the position both lines intersect. At this factor is the answer to the set of equations.
2016-12-04 22:29:17 · answer #2 · answered by ? 4 · 0⤊ 0⤋
x^2+6^2=25 for y=6
x^2+36=25
x^2= -11
which is not real.there the bill is incorrect.
x^2+(2x-2)^2=25
x^2+4x^2+4-8x=25
5x^2-8x--21=0
5x^2-15x+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
x= -7/5 ,3
2007-02-25 05:19:52 · answer #3 · answered by Diksha A 3 · 0⤊ 0⤋